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Journal of Inequalities and Applications
 
J Inequal Appl. 2017; 2017(1): 312.
Published online 2017 December 19. doi:  10.1186/s13660-017-1584-8
PMCID: PMC5736795

Commutators associated with Schrödinger operators on the nilpotent Lie group

Abstract

Assume that G is a nilpotent Lie group. Denote by L = −Δ + W the Schrödinger operator on G, where Δ is the sub-Laplacian, the nonnegative potential W belongs to the reverse Hölder class Bq1 for some q1D2 and D is the dimension at infinity of G. Let ℛ = ∇(−Δ+W)−½ be the Riesz transform associated with L. In this paper we obtain some estimates for the commutator [h, ℛ] for hLipνθ, where Lipνθ is a function space which is larger than the classical Lipschitz space.

Keywords: commutator, Lipschitz space, nilpotent Lie groups, reverse Hölder inequality, Riesz transform, Schrödinger operator

Introduction

Assume G to be a connected and simply connected nilpotent Lie group and 𝔤 to be its Lie algebra identified with the space of left invariant vector fields. Given X = {X1, …, Xl} ⊆ 𝔤, a Hörmander system of left invariant vector fields on G. Let Δ=i=1lXi2 be the sub-Laplacian on G associated with X and the gradient operator [nabla] be denoted by ∇ = (X1, …, Xl). Following [1], one can define a left invariant metric d associated with X which is called the Carnot-Carathéodory metric: let xy ∈ G, and

d(xy) = inf{δ∣∃γ:[0, δ] → Gγ(0) = xγ(δ) = y}, 

where γ is a piecewise smooth curve satisfying

γ(s)=i=1lai(s)Xi(γ(s)),s[0,δ],i=1l|ai(s)|21.

If x ∈ G and r > 0, we will denote by B(xr) = {y ∈ Gd(xy) < r} the metric balls. Assume dx to be the Haar measure on G. Then, for every measurable set E ⊆ G, |E| denotes the measure of E. Suppose e to be the unit element of G. Note that V(r) = |B(er)| = |B(xr)| for any x ∈ G and r > 0. It follows from in [2] or [3] that there exists a constant C1 > 0 such that

C11rdV(r)C1rd,0r1;
1

C11rDV(r)C1rD,1r<,
2

where d and D denote the local dimension and the dimension at infinity of G, and there is D ≥ d > 0. At this time, the Lie group G is also called a Lie group of polynomial growth. If G is a stratified Lie group, then Dd (cf. [1]). Also, there exist positive constants C2C3 > 1 such that

C21(Rr)dV(R)V(r)C2(Rr)D,0<r<R<;
3

V(2r) ≤ C3V(r).
4

Throughout this paper, we always assume that d ≥ 2 .

Let L = −Δ + W be the Schrödinger operator, where Δ is the sub-Laplacian on G and the nonnegative potential W belongs to the reverse Hölder class Bq1 for some q1D2 and D > 3. The Riesz transform associated with the Schrödinger operator L is defined by

ℛ = ∇(−Δ+W)−½.
5

Let b be a locally integrable function on G and T be a linear operator. For a suitable function f, the commutator is defined by [bT]fbT(f)−T(bf). Many researchers have paid attention to the commutator on n. It is well known that Coifman, Rochberg and Weiss [4] proved that [bT] is a bounded operator on Lp for 1 < p < ∞ if and only if b ∈ BMO(ℝn), when T is a Calderón-Zygmund operator. Janson [5] proved that the commutator is bounded from Lp(ℝn) into Lq(ℝn) if and only if b ∈ Lipν(ℝn) with ν=(1p1q)n, where Lipν(ℝn) is the Lipschitz space. Sheng and Liu [6] proved the boundedness of the commutator [b, ℛ] from the Hardy space HLp(Rn) into Lq(ℝn) when b belongs to a larger Lipschitz space. Comparatively, there has been much less research on the commutator on nilpotent Lie groups. The goal of this paper will be to obtain some estimates for the commutator related to the Schrödinger operator on nilpotent Lie groups. The complicated structure of nilpotent Lie group will bring some essential difficulties to our estimates in the following sections.

Note that a non-negative locally Lq integrable function W on G is said to belong to Bq (1 < q < ∞) if there exists C > 0 such that the reverse Hölder inequality

(1|B|BW(y)qdy)1qC(1|B|BW(y)dy)
6

holds for every ball B in G.

We first introduce an auxiliary function as follows.

Definition 1

Let W ∈ Bq for some qD2. For x ∈ G, the function m(xW) is defined by

1m(x,W)=ρ(x):=supr>0{r:r2V(r)B(x,r)W(y)dy1}.

Now we define the space Lipνθ(G) on the nilpotent Lie group.

Definition 2

Let θ > 0 and 0 < ν < 1, the space Lipνθ(G) consists of the functions f satisfying

|f(x)f(y)|Cdν(x,y)(1+d(x,y)ρ(x)+d(x,y)ρ(y))θ

holds true for all xy ∈ G, x ≠ y. The norm on Lipνθ(G) is defined as follows:

fLipνθ(G)supx,yG,xy(|f(x)f(y)|dν(x,y)(1+d(x,y)ρ(x)+d(x,y)ρ(y))θ)<.

It is easy to see that this space is exactly the Lipschitz space when θ = 0 if G is a stratified Lie group (cf. [7] and [8]).

We also introduce the following maximal functions.

Definition 3

Let fLloc1(G). For 0 < γ < D, the fractional maximal operator is defined by

Mγf(x)=supxBrγ|B|B|f(y)|dy,xG,

where the supremum on the right-hand side is taken over all balls B ⊆ G and r is the radius of the ball B.

Definition 4

Given α > 0, the maximal functions for fLloc1(G) and x ∈ G are defined by

Mρ,αf(x)=supxBB1|B|B|f(y)|dy

and

Mρ,αf(x)=supxBB1|B|B|f(y)fB|dy,

where ρ,α = {B(yr):y ∈ G and r ≤ αρ(y)}.

We are in a position to give the main results in this paper.

Theorem 1

Assume W ∈ Bq1 for some q1D2, where D denotes the dimension at infinity of the nilpotent Lie group G. Let

1q2={1,if q1D,11q1+1D,if D2q1<D.
7

Denote the adjoint operator of by R˜=(Δ+W)12. Then, for any hLipνθ(G), 0 < ν < 1, the commutator [h,R˜] is bounded from Lq(G) into Lp(G), 1p=1qνD, if q2 < p < ∞.

We immediately deduce Corollary 1 by duality.

Corollary 1

Assume W ∈ Bq1 for some q1D2, where D denotes the dimension at infinity of the nilpotent Lie group G. Let

1q2={1,if q1D,11q1+1D,if D2q1<D.

Then, for any hLipνθ(G), 0 < ν < 1, the commutator [h, ℛ] is bounded from Lp(G) into Lq(G), 1p=1qνD, if 1<p<q2.

Throughout this paper, unless otherwise indicated, C will be used to denote a positive constant that is not necessarily the same case at each occurrence and it depends at most on the constants in (3) and (6). We always denote δ=2Dq1. By A ∼ B, we mean that there exist constants C > 0 and c > 0 such that cABC.

Estimates for the kernels of and R˜

In this section we recall some estimates for the kernels of Riesz transform and the dual Riesz transform R˜, which have been proved in [3].

Lemma 1

W ∈ Bq is a doubling measure, that is, there exists a constant C > 0 such that

B(x,2r)W(y) dy ≤ CB(x,r)W(y) dy.

Lemma 2

There exists C > 0 such that, for 0 < r < R < ∞,

r2V(r)B(x,r)W(y)dyC(Rr)Dq2r2V(R)B(x,R)W(y)dy.

Lemma 3

If rρ(x), then

r2V(r)B(x,r)W(y)dy=1.

Moreover,

r2V(r)B(x,r)W(y)dy1if and only ifr1m(x,W).

Lemma 4

There exist constants Cl0 > 0 such that

1C(1+d(x,y)ρ(x))l0l0+1ρ(x)ρ(y)C(1+d(x,y)ρ(x))l0.

In particular, ρ(x) ∼ ρ(y) if d(xy) ≤ Cρ(x).

Lemma 5

There exist constants C > 0 and l1 > 0 such that

B(x,R)d2(x,y)W(y)V(d(x,y))dyCR2V(R)B(x,R)W(y)dyC(1+Rρ(x))l1.

Using Lemma 4, we immediately have the following lemma.

Lemma 6

There exist l0 > 0, C > 0 such that, for any x and y in G,

1+d(x,y)ρ(x)+d(x,y)ρ(y)C(1+d(x,y)ρ(x))l0+1.

Let Γ(xyλ) denote the fundamental solution for the operator −Δ + Wλ, namely, [−Δ + Wλ]Γ(xyλ) = δ(y−1x), where δ is the Dirac function and λ ∈ [0, ∞). Markedly, Γ(xyλ) = Γ(yxλ).

Lemma 7

Let N be a positive integer.

  • (i)
    Suppose W ∈ Bq1 for some q1D2. Then there exists a constant CN > 0 such that, for x ≠ y,
    |Γ(x,y,λ)|CN(1+d(x,y)λ12)N(1+d(x,y)/ρ(x))Nd(x,y)V(d(x,y)).
  • (ii)
    Suppose W ∈ Bq1 for some q1D2. Then there exists a constant CN > 0 such that
    |xΓ(x,y,λ)|CN(1+d(x,y)λ12)N(1+d(x,y)/ρ(x))N×(d2(x,y)V(d(x,y))B(y,d(x,y))W(z)d(z,y)dzV(d(z,y))+d(x,y)V(d(x,y))).
    Particularly, if W ∈ Bq1 for some q1 ≥ D, then there exists CN > 0 such that, for x ≠ y,
    |xΓ(x,y,λ)|CN(1+d(x,y)λ12)N(1+d(x,y)/ρ(x))Nd2(x,y)V(d(x,y)).

By the functional calculus, we may write

(Δ+W)12=1π0λ12(Δ+W+λ)1dλ.
8

Let fC0(G). From (−Δ+W+λ)−1f(x) = ∫GΓ(xyλ)f(y) dy, it follows that

f(x) = ∫G𝒦(xy)f(y) dy
9

where

K(x,y)=1π0λ12xΓ(x,y,λ)dλ.
10

Similarly, the adjoint operator of is defined to be

R˜f(x)=GK˜(x,y)f(y)dy,
11

where

K˜(x,y)=1π0λ12yΓ(y,x,λ)dλ.
12

We recall estimates of the kernels for and R˜ (cf. [3]).

Lemma 8

Suppose W ∈ Bq1 for some D2q1<D. For any integer N > 0, there exists CN > 0 such that

|K(x,y)|CN(1+d(x,y)/ρ(x))N(d(x,y)V(d(x,y))B(y,d(x,y))W(z)d(z,y)dzV(d(z,y))+1V(d(x,y)))
13

and

|K(x,y)K(x,y)|CNdδ(y,y)(1+d(x,y)/ρ(x))Nd1δ(x,y)V(d(x,y))(B(y,d(x,y))W(z)d(z,y)dzV(d(z,y))+1d(x,y))
14

for some δ > 0 and 0<d(y,y)<d(x,y)16. If W ∈ Bq1 for some q1 ≥ D, then

|K(x,y)|CN(1+d(x,y)/ρ(x))N1V(d(x,y))
15

and

|K(x,y)K(x,y)|CN(1+d(x,y)/ρ(x))Ndδ(y,y)V(d(x,y))dδ(x,y).
16

Lemma 9

Suppose W ∈ Bq1 for some D2q1<D. For any integer N > 0, there exists CN > 0 such that

|K˜(x,y)|CN(1+d(x,y)/ρ(x))N(d(x,y)V(d(x,y))B(y,d(x,y))W(z)d(z,y)dzV(d(z,y))+1V(d(x,y)))
17

and

|K˜(x,y)K˜(x,y)|CNdδ(x,x)(1+d(x,y)/ρ(x))Nd1δ(x,y)V(d(x,y))(B(y,d(x,y))W(z)d(z,y)dzV(d(z,y))+1d(x,y))
18

for some δ > 0 and 0<d(x,x)<d(x,y)16. If W ∈ Bq1 for some q1 ≥ D, then

|K˜(x,y)|CN(1+d(x,y)/ρ(x))N1V(d(x,y))
19

and

|K˜(x,y)K˜(x,y)|CN(1+d(x,y)/ρ(x))Ndδ(x,x)V(d(x,y))dδ(x,y).
20

Some technical lemmas and propositions

Proposition 1

Let θ > 0 and 1 ≤ s < ∞. If hLipνθ(G), then there exists a positive constant C such that, for all BB(xr) with x ∈ G and r > 0,

(1|B|B|h(y)hB|sdy)1/sChLipνθ(G)rν(1+rρ(x))(l0+1)θ.

Proof

Since hLipνθ(G), then

(1|B(x,r)|B(x,r)|h(y)hB|sdy)1s(1|B(x,r)|B(x,r)|h(y)h(x)|sdy)1s+(1|B(x,r)|B(x,r)|h(x)hB|sdy)1s2C(1|B(x,r)|B(x,r)|hLipνθ(G)dν(x,y)(1+d(x,y)ρ(x)+d(x,y)ρ(y))θ|sdy)1sChLipνθ(G)rν(1+rρ(x)+rρ(y))θChLipνθ(G)rν(1+rρ(x))(l0+1)θ,

where we have used Lemma 6 in the penultimate inequality.

Similar to the proof of Proposition 1, we immediately get the following.

Lemma 10

Let hLipνθ(G), BB(xr) and s ≥ 1. Then there exists a positive constant C such that, for all k ∈ N,

(1|2kB|2kB|h(y)hB|sdy)1sChLipνθ(G)2kνrν(1+2krρ(x))(l0+1)θ.

Proposition 2

Let W ∈ Bq1 for q1D2. Let

1q2={1,if q1D,11q1+1D,if D2q1<D.

Then there exists C > 0 such that, for any fC0(G),

∥ℛf(x)∥Lp(G) ≤ CfLp(G)

where 1<p<q2.

Proposition 3

Let W ∈ Bq1 for q1D2. Let

1q2={1,if q1D,11q1+1D,if D2q1<D.

Then there exists C > 0 such that, for any fC0(G),

R˜f(x)Lp(G)CfLp(G),

where q2 < p < ∞.

For the proofs of Proposition 2 and Proposition 3, one can refer to [3].

Proposition 4

There exists a sequence of points {xk}k=1 in G, so that the set of critical balls QkB(xkρ(xk)), k ≥ 1, satisfies

  • (i)
    kQkG;
  • (ii)
    There exists N such that, for every k ∈ N, ♯{j:4Qj ∩ 4Qk ≠ ∅} ≤ N.

Proposition 5

For 1 < p < ∞, there exist positive constants C, α and β such that if {Qk}k=1 is a sequence of balls as in Proposition  4, then

G|Mρ,αf(x)|pdxCG|Mρ,βf(x)|pdx+Ck|Qk|(1|Qk|2Qk|f(x)|dx)p

for all fLloc1(G).

The above propositions have been proved in [9] and [10] in the case of a homogeneous space, respectively.

Lemma 11

Let W ∈ Bq1 for q1D2, and let

1q2={1,if q1D,11q1+1D,if D2q1<D

and hLipνθ(G). Then, for q2 < m < ∞, there exists a positive constant C such that

1|Q|Q|[h,R˜]f(x)|dxChLipνθ(G)infyQ{Mmν(|f|m)(y)}1m

holds true for all fLlocm(G) and every ball QB(x0ρ(x0)), where Mmν is a fractional maximal operator.

Proof

Throughout the proof of the lemma, we always assume D2q1<D. Let f ∈ Lp(G) and QB(x0ρ(x0)). For

[h,R˜]f=(hhQ)R˜fR˜((hhQ)f),
21

we need to consider the average on Q for each term. By the Hölder inequality with m > q2 and Proposition 1,

1|Q|Q|(hhQ)R˜f(x)|dx(1|Q|Q|(h(x)hQ)|mdx)1/m(1|Q|Q|R˜f(x)|mdx)1/mChLipνθ(G)(ρ(x0))ν(1|Q|Q|R˜f(x)|mdx)1/m.

If we write ff1f2 with f1fχ2Q, due to Proposition 3, we get

(ρ(x0))ν(1|Q|Q|R˜f1(x)|mdx)1/mC(ρ(x0))ν(1|2Q|2Q|f(x)|mdx)1/mCinfyQ{Mmν(|f|m)(y)}1m
22

for x ∈ Q, and using (17) in Lemma 9, we split R˜f2(x) into two parts

|R˜f2(x)|=|d(x0,z)>2ρ(x0)K˜(x,z)f(z)dz|C{I1(x)+I2(x)},

where

I1(x)=d(x0,z)>2ρ(x0)|f(z)|V(d(x,z))(1+d(x,z)/ρ(x))Ndz

and

I2(x)=d(x0,z)>2ρ(x0)d(x,z)|f(z)|V(d(x,z))(1+d(x,z)/ρ(x))NB(z,d(x,z)4)d(u,z)W(u)V(d(u,z))dudz.

To deal with I1(x), noting that ρ(x) ∼ ρ(x0) and d(zx) ∼ d(zx0), we split d(zx0) > 2ρ(x0) into annuli to obtain

(ρ(x0))νI1(x)=(ρ(x0))νd(x0,z)>2ρ(x0)|f(z)|V(d(x,z))(1+d(x,z)/ρ(x))NdzCk12Nk(ρ(x0))νV(2kρ(x0))d(x0,z)<2kρ(x0)|f(z)|dzCinfyQ{Mmν(|f|m)(y)}1m.
23

Secondly, we consider the term I2(x). We have, for x ∈ Q,

(ρ(x0))νI2(x)=(ρ(x0))νd(x0,z)>2ρ(x0)d(x,z)|f(z)|V(d(x,z))(1+d(x,z)/ρ(x))N×B(z,d(x,z)4)d(u,z)W(u)V(d(u,z))dudzC(ρ(x0))νd(x0,z)>2ρ(x0)d(x,z)|f(z)|V(d(x,z))(1+d(x,z)/ρ(x))N×B(z,4d(x0,z))d(u,z)W(u)V(d(u,z))dudzC(ρ(x0))νk12Nk2kρ(x0)V(2kρ(x0))d(x0,z)<2k+1ρ(x0)|f(z)|×B(z,2k+3d(x0,z))d(u,z)W(u)V(d(u,z))dudzC(ρ(x0))νk12Nk2kρ(x0)V(2kρ(x0))×d(x0,z)<2k+1ρ(x0)|f(z)|I1(WχB(x0,2kρ(x0)))(z)dz.

Let q2 < m < D. Using the Hölder inequality and the boundedness of the fractional integral 1:Lm ↦ Lq1 with

1q1=1m1β,

where

β={d,if 2kρ(x0)<1,D,if 2kρ(x0)1

(cf. Theorem 1.6 in [11]), we obtain

d(x0,z)<2k+1ρ(x0)|f(z)|I1(WχB(x0,2kρ(x0)))(z)dzfχB(x0,2kρ(x0))mI1(WχB(x0,2kρ(x0)))mfχB(x0,2kρ(x0))mWχB(x0,2kρ(x0))q1.

Since W ∈ Bq1, we obtain

WχB(x0,2kρ(x0))q1C(V(2kρ(x0)))1q1B(x0,2kρ(x0))WC(2kρ(x0))2(V(2kρ(x0)))11q1(2kρ(x0))2V(2kρ(x0))B(x0,2kρ(x0))WC(2k)l12(V(2kρ(x0)))11q1(ρ(x0))2,

where in the last two inequalities we have used doubling measure and the definition of ρ, respectively. Therefore,

I2(x)V(ρ(x0))1/q11k12k(N+l11d/q1)fχB(x0,2kρ(x0))m.

Finally, observing that

fχB(x0,2kρ(x0))m(V(2kρ(x0)))1minfyQMmf(y)

and using that 1q11m=1D or 1d, we have

(ρ(x0))νI2(x)CinfyQ{Mmν(|f|m)(y)}1m

by choosing N large enough.

So far, we have solved the term (h − hQ)ℛf, now we want to control ∫ℛ[(h − hQ)f] dx by the term CinfyQ{Mmν(|f|m)(y)}1m. We still split ff1f2. Choose q2<m˜<m and set t=m˜mmm˜. Using the boundedness of R˜ on Lm˜(G) and the Hölder inequality, we get

1|Q|Q|R˜(hhQ)f1(x)|dx(1|Q|Q|R˜(hhQ)f1(x)|m˜dx)1/m˜C(1|Q|Q|(hhQ)f(x)|m˜dx)1/m˜C(1|Q|2Q|f(x)|mdx)1/m(1|Q|2Q|h(x)hQ|tdx)1/tC((ρ(x0))mν|Q|2Q|f(x)|mdx)1/mhLipνθ(G)(1+ρ(x0)ρ(x0))(l0+1)θChLipνθ(G)infyQ{Mmν(|f|m)(y)}1m,

where we have applied Proposition 1 to the last but one inequality. Similarly, for x ∈ Q and using (17) in Lemma 9, we have

|R˜(hhQ)f2(x)|=|d(x0,z)>2ρ(x0)K˜(x,z)(h(z)hQ)f(z)dz|C{I˜1(x)+I˜2(x)},

where

I˜1(x)=d(x0,z)>2ρ(x0)|(hhQ)f(z)|V(d(x,z))(1+d(x,z)/ρ(x))Ndz

and

I˜2(x)=d(x0,z)>2ρ(x0)|(hhQ)f(z)|V(d(x,z))(1+d(x,z)/ρ(x))NB(z,d(x,z)4)d(u,z)W(u)V(d(u,z))dudz.

We start by observing that for 1m˜<m, t=m˜mmm˜, and by Lemma 6,

[(hhQ)f]χB(x0,2kρ(x0))m˜fχB(x0,2kρ(x0))m(hhQ)χB(x0,2kρ(x0))tC(B(x0,2kρ(x0))|f(y)|mdy)1/mhLipνθ(G)(2kρ(x0))ν(1+2k)(l0+1)θV1/t(2kρ(x0))C2k(l0+1)θV1/m˜(2kρ(x0))hLipνθ(G)infyQ{Mmν(|f|m)(y)}1m.
24

For I˜1(x), using (24) with m˜=1, we have

I˜1(x)Ck12NkV(2kρ(x0))d(x0,z)>2kρ(x0)|h(z)hQ||f(z)|dzChLipνθ(G)infyQ{Mmν(|f|m)(y)}1mk12k(N+(l0+1)θ)ChLipνθ(G)infyQ{Mmν(|f|m)(y)}1m

if we choose N sufficiently large.

To deal with I˜2(x), we discuss as in the estimate for I2(x) with (h − hQ)f instead of f and m and q1˜ instead of m and q1, but we cannot avoid to discuss the different cases where 2kρ(x0) ≥ 1 and 2kρ(x0) < 1. Let

1m˜={1q1˜1d,if 2kρ(x0)<1,1q1˜1D,if 2kρ(x0)1.

Using (24), we similarly have

I˜2(x)ChLipνθ(G)infyQ{Mmν(|f|m)(y)}1m,

where we choose N large enough to ensure the above series converges.

Lemma 12

Let R˜=(Δ+W)12 be the adjoint operator of the Riesz transform . Then there exists C > 0 such that, for any fLlocm(G) and hLipνθ(G),

Mρ,α([h,R˜]f)(x)ChLipνθ(G)({Mmν(|f|m)(y)}1m+{Mmν(|R˜f|m)(x)}1m).
25

Proof

Let fLlocm(G), x ∈ G and a ball BB(x0r) with x ∈ B and r < ϵρ(x0), ϵ > 0, we need to control J=1|B|B|[h,R˜]f(y)c|dy by the right-hand side of (25) for some constant c, which will be designated later. Let ff1f2, where f1fχ2B and f2f − f1. Then

[h,R˜]f=[hh2B,R˜]f=(hh2B)R˜fR˜(hh2B)f1R˜(hh2B)f2A1f+A2f+A3f.

Take c=d(x0,z)2rK˜(x0,z)(h(z)h2B)f(z)dz. Then we have

J1|B|B|A1f(y)|dy+1|B|B|A2f(y)|dy+1|B|B|A3f(y)c|dyJ1+J2+J3.

At first, we consider J1. Note that d(xx0) ≤ r < ϵρ(x0) implies ρ(x) ∼ ρ(x0). By the Hölder inequality, Proposition 1 and Proposition 3, we have

J1=1|B|B|A1f(y)|dy=1|B|B|(h(y)h2B)R˜f(y)|dyC|B|[B|h(y)h2B|mm1dy]m1m[B|R˜f(y)|mdy]1mC|B|[2B|h(y)h2B|mm1dy]m1m[B|R˜f(y)|mdy]1mC|B|hLipνθ(G)(1+2rρ(x0))(l0+1)θrν|B|m1m[B|R˜f(y)|mdy]1mC|B|hLipνθ(G)(1+2ϵρ(x0)ρ(x0))(l0+1)θrν|B|m1m[B|R˜f(y)|mdy]1mChLipνθ(G)(rmν|B|B|R˜f(y)|mdy)1mChLipνθ(G){Mmν(|R˜f|m)(x)}1m

for m > q2.

For J2, by the Hölder inequality and Proposition 3,

J21|B|B|A2f(y)|dy(1|B|B|A2f(y)|m˜dy)1m˜C(1|B|)1m˜(2B|(h(y)h2B)f(y)|m˜dy)1m˜C|B|1m˜(2B|h(y)h2B|mm˜mm˜dy)1m˜1m(2B|f(y)|mdy)1mChLipνθ(G)rν(1+2rρ(x0))(l0+1)θ(1|B|)1m(2B|f(y)|mdy)1mChLipνθ(G)rν(1+2ϵρ(x0)ρ(x0))(l0+1)θ(1|B|)1m(2B|f(y)|mdy)1mChLipνθ(G)(rmν|B|2B|f(y)|mdy)1mChLipνθ(G)((2r)mν|2B|2B|f(y)|mdy)1mChLipνθ(G){Mmν(|f|m)(y)}1m,

where q2<m˜<m.

Finally, we consider J3.

Case of q1 ≥ D: By Lemmas 9 and 6, we have

(2krd(x0,z)<2k+1r|K˜(y,z)K˜(x0,z)|mdz)1m(2krd(x0,z)<2k+1r|h(z)h2B|sdz)1sCNrδ(1+2kr/ρ(x0))N(2krd(x0,z)<2k+1r1dmδ(x0,z)Vm(d(x0,z))dz)1m×(d(x0,z)<2k+1r|h(z)h2B|sdz)1sCN(1+2kr/ρ(x0))Nrδ(2kr)δV(m1)/m(2kr)hLipνθ(G)2kνrν(1+2krρ(x0))(l0+1)θV1s(2kr)hLipνθ(G)2kνrνCN(1+2kr/ρ(x0))N(l0+1)θrδ(2kr)δV1/m(2kr),
26

where 1m+1m+1s=1 and 1s+1s=1. Therefore, via the Hölder inequality,

J3=1|B|B|A3f(y)c|dy1|B|B|d(x0,z)>2r(K˜(y,z)K˜(x0,z))(h(z)h2Q)f(z)dz|dy1|B|Bk=12krd(x0,z)<2k+1r|(K˜(y,z)K˜(x0,z))(h(z)h2Q)f(z)|dzdyhLipνθ(G)|B|Bk=12kνrνC(1+2kr/ρ(x0))N(l0+1)θrδ(2kr)δV1m1s(2kr)×(d(x0,z)<2k+1r|f(z)|mdz)1mChLipνθ(G)k=1CN(1+2kr/ρ(x0))N(l0+1)θrδ(2kr)δV1/m(2kr)(2kr)ν(2k+1r)νV1m(2k+1r)×((2k+1r)mνV(2k+1r)d(x0,z)<2k+1r|f(z)|mdz)1mChLipνθ(G)k=12kδCN(1+2kr/ρ(x0))N(l0+1)θ{Mmν(|f|m)(y)}1mChLipνθ(G){Mmν(|f|m)(y)}1m

if we choose N sufficiently large.

Case of D2q1<D: By Lemmas 9 and 6, we have

(2krd(x0,z)<2k+1r|K˜(y,z)K˜(x0,z)|pdz)1p(2krd(x0,z)<2k+1r|h(z)h2B|tdz)1tCNrδ(2kr)1δV(2kr)(1+2kr/ρ(x0))N(2krd(x0,z)<2k+1r|B(z,2k+3r)d(z,y)W(y)V(z,y)dy|pdz)1p×(2krd(x0,z)<2k+1r|h(z)h2B|tdz)1t+CNhLipνθ(G)2lνrν(1+2kr/ρ(x0))N(l0+1)θrδ(2kr)δV1/m(2kr)CNhLipνθ(G)2lνrν(1+2kr/ρ(x0))N(l0+1)θ×[rδ(2kr)1δVt1t(2kr)(B(x0,2k+3r)Wq1(z)dz)1q1+rδ(2kr)δV1/m(2kr)]CNhLipνθ(G)2lνrν(1+2kr/ρ(x0))N(l0+1)θ×[rδV1q1(2kr)(2kr)1δVt1t(2kr)((2k+3r)2V(2k+3r)B(x0,2k+3r)W(z)dz)+rδ(2kr)δV1/m(2kr)]CNhLipνθ(G)2lνrν(1+2kr/ρ(x0))Nl1(l0+1)θ(rδV1q1(2kr)(2kr)1δVt1t(2kr)+rδ(2kr)δV1/m(2kr))ChLipνθ(G)2lνrν(1+2kr/ρ(x0))Nl1(l0+1)θrδ(2kr)δV1/m(2kr),

where 1m+1p+1t=1 and 1q1=1p+1D. Therefore, for m > q2,

J3=1|B|B|A3f(y)c|dy1|B|B|d(x0,z)>2r(K˜(y,z)K˜(x0,z))(h(z)h2Q)f(z)dz|dy1|B|Bk=12krd(x0,z)<2k+1r|(K˜(y,z)K˜(x0,z))(h(z)h2Q)f(z)|dzdyhLipνθ(G)|B|Bk=12kνrνC(1+2kr/ρ(x0))Nl1(l0+1)θrδ(2kr)δV1m1s(2kr)×(d(x0,z)<2k+1r|f(z)|mdz)1mChLipνθ(G)k=1CN(1+2kr/ρ(x0))Nl1(l0+1)θrδ(2kr)δV1/m(2kr)(2kr)ν(2k+1r)νV1m(2k+1r)×((2k+1r)mνV(2k+1r)d(x0,z)<2k+1r|f(z)|mdz)1mChLipνθ(G)k=12kδCN(1+2kr/ρ(x0))Nl1(l0+1)θ{Mmν(|f|m)(y)}1mChLipνθ(G){Mmν(|f|m)(y)}1m,

if we choose N sufficiently large.

Proof of the main result

Proof of Theorem 1

Suppose hLipνθ(G). We choose m such that it satisfies q2 < m < p. We conclude from Proposition 4, Lemmas 5, 11 and 12 that

[h,R˜]fLp(G)pG|Mρ,α([h,R˜]f)(x)|pdxCG|Mρ,β([h,R˜]f)(x)|pdx+Ck|Qk|(1|Qk|2Qk|[h,R˜]f(x)|dx)pCG|Mρ,β([h,R˜]f)(x)|pdx+ChLipνθ(G)pk2Qk|{Mmν(|f|m)}1m(x)|pdxChLipνθ(G)(Mmν(|f|m))1mLp(G)pChLipνθ(G)fLq(G)p,

where 1p=1qνD.

Conclusions

We prove the Lp → Lq boundedness for the commutator which is generated by the Riesz transform and the function hLipνθ. We generalize the corresponding results on the Euclidean space in [6] to the nilpotent Lie group, and they may have some applications in harmonic analysis and PDE on the Lie group.

Acknowledgements

Research supported by the National Natural Science Foundation of China (No. 11671031, 11471018), the Fundamental Research Funds for the Central Universities (No. FRF-BR-17-004B).

Authors’ contributions

Authors’ contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Notes

Competing interests

The authors declare that they have no competing interests.

Footnotes

Publisher’s Note

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Contributor Information

Tianzhen Ni, moc.361@32184010871.

Yu Liu, nc.gro.ukp@57uyuil.

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