Centered Hardy-Littlewood maximal functions on Heisenberg type … · 2018-11-16 · TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 366, Number 3, March 2014, Pages 1497–1524

TRANSACTIONS OF THEAMERICAN MATHEMATICAL SOCIETYVolume 366, Number 3, March 2014, Pages 1497–1524S 0002-9947(2013)05965-XArticle electronically published on September 26, 2013

CENTERED HARDY-LITTLEWOOD MAXIMAL FUNCTIONS

ON HEISENBERG TYPE GROUPS

HONG-QUAN LI AND BIN QIAN

Abstract. In this paper, by establishing uniform lower bounds for the Poisson

kernel and (−Δ)−12 on the Heisenberg type group H(2n,m) with m ≥ 2, which

follow from the various properties of Bessel functions and Legendre functions,we prove that there exists a constant A > 0 such that, for all f ∈ L1(H(2n,m))and all n,m ∈ N∗ satisfying 4 ≤ m2 � logn, we have ‖MKf‖L1,∞ ≤ An‖f‖1,where MK denotes the centered Hardy-Littlewood maximal function definedby the Koranyi norm. For the centered Hardy-Littlewood maximal functionMCC defined by the Carnot-Carathedory distance, we prove ‖MCCf‖L1,∞ ≤A(m)n‖f‖1 holds for some constant A(m) independent of n.

1. Introduction

Consider the standard centered Hardy-Littlewood maximal function, MRn , inR

n (n ∈ N∗), i.e.

MRnf(x) = supr>0

1

|BRn(x, r)|

∫BRn (x,r)

|f(y)| dy, x ∈ Rn, f ∈ L1

loc(Rn),

where dy is the Lebesgue measure and |BRn(x, r)| is the volume of the euclideanball with the center x ∈ Rn and the radius r > 0.

By the tripling property of the volume, i.e.

|BRn(x, 3r)| ≤ 3n|BRn(x, r)|, ∀x ∈ Rn, r > 0,

one deduces from the Vitali covering lemma that MRn satisfies the weak type (1, 1)bounds with

‖MRn‖L1−→L1,∞ ≤ 3n.

However, applying the Hopf-Dunford-Schwartz maximal ergodic theorem, Steinand Stromberg obtained in [20] via the heat kernel that there exists a constantA > 0 such that

‖MRn‖L1−→L1,∞ ≤ Aφ(n), ∀n, with φ(n) = n.(1.1)

In the setting of the Heisenberg groups, H(2n, 1), the estimate of type (1.1) hasbeen obtained for the centered Hardy-Littlewood maximal function defined by theCarnot-Caratheodory distance or by the Koranyi norm. The proof is based on auniform lower estimate of the Poisson kernel (the integral kernel of the Poissonsemigroup; there is no relation with the one in [4]); see [10] for details.

Received by the editors February 10, 2012.2010 Mathematics Subject Classification. Primary 42B25, 43A80 .

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1498 HONG-QUAN LI AND BIN QIAN

For the maximal functionMG associated to the Carnot-Caratheodory distance orthe pseudo-distance induced by the fundamental solution to the Grushin operator

ΔG =

n∑i=1

∂2

∂x2i

+ (

n∑i=1

x2i )

∂2

∂u2,

the first named author has obtained estimate (1.1) for MG in [12].As mentioned in [12], the above three results can be explained roughly by an

estimate of the following type:

infn≥3,h>0,g �=g′∈B(g,h)

φ(n)n

h2|B(g, h)|(−Δ)−1(g, g′) > 0, with φ(n) = n,(1.2)

in the euclidean spaces, Heisenberg groups and for the Grushin operators. Further-more, we believe that there is a close relation between the estimate of type (1.1) (ofcourse, the volume of the ball and the dimension play their roles) and the Greenfunction. In fact, the work [12] is motivated by the estimate (1.2). Also, the resultsin [20], [10] and [12] can be explained by an estimate of the following type:

infn≥3,h>0,g �=g′∈B(g,h)

φ(n)

√n

h|B(g, h)|(−Δ)−

12 (g, g′) > 0.(1.3)

Following the above idea and applying various properties of the real hyperbolicspaces of dimension n (n ≥ 2) which are measured metric spaces of exponentialvolume growth, Li and Lohoue showed in [15] that (1.1) holds with φ(n) = n lnn inthis case. A similar method works for the harmonic AN groups. For other O(n lnn)results, see [20] and [17].

We remark that up to a universal constant, the two terms nh2 and

√nh , which can

be found in (1.2) and (1.3) respectively, are optimal. Note that it suffices to takeh = 1 in the above three models, i.e. Rn, H(2n, 1) and ΔG (thanks to the dilationstructure); see [10] and [12] for details.

The purpose of this paper is to use this idea to show (1.1) holds on the Heisenbergtype groups H(2n,m). We suppose m ≥ 2 in what follows, rather than the case m =1 which has been treated in [10]. We will see it is necessary to distinguish the caseof m = 1 from the one of m ≥ 2. The proof for m = 1 is very natural, however theone for m ≥ 2 is much more technical (it is based on the various properties of Besselfunctions and Legendre functions), and the technique used for m ≥ 2 does not workfor m = 1 because of the properties of Legendre functions. The method here followsthe ones in [10,12], but it is not enough to get the desired result only by using the

Poisson kernel or (−Δ)−12 (g, g′) and (−Δ)−

12 exp

(−ω dK(g,g′)√

n

√−Δ

)(g, g′) alone.

Here we must take both of them into consideration. To this end, we should makegood use of various properties of Bessel functions and Legendre functions. Notethat using the same method in [10], Zhao and Song in [25] obtained (1.1) in thespecial case H(2n, 3) for the centered Hardy-Littlewood maximal functions definedby the Koranyi norm.

However, (1.1) has been obtained in the setting of Sn−1 (n ≥ 2), the unit sphereof dimension n − 1 (i.e. the n − 1 dimensional, simply connected Riemannianmanifold of constant sectional curvature 1); see [9] and [14]. In this case, we can’t

use (1.2) or (1.3) to explain this bound since (−ΔSn−1)−1 and (−ΔSn−1)−12 do not

exist.Let us first introduce the concept of the Heisenberg type groups H(2n,m) and

centered Hardy-Littlewood maximal functions.


CENTERED HARDY-LITTLEWOOD MAXIMAL FUNCTIONS 1499

1.1. Heisenberg-type groups and centered Hardy-Littlewood maximalfunctions. Recall that a Heisenberg type group (for short, H-type) can be consid-ered as H(2n,m) = R

2n × Rm (m,n ∈ N

∗) with the group law (see e.g. TheoremA.2, p. 199 in [3], and one can refer to [8] for the original definition)

(x, t) · (x′, t′) = (x+ x′, t+ t′ + 2−1〈x, Ux′〉),with x = (x1, . . . , x2n) ∈ R2n, t = (t1, . . . , tm) ∈ Rm and

〈x, Ux′〉 = (〈x, U (1)x′〉, . . . , 〈x, U (m)x′〉) ∈ Rm,

where the matrices U (1), · · · , U (m) satisfy the following two conditions:1. U (j) is a 2n× 2n anti-symmetric and orthogonal matrix, for all j = 1, 2, · · · ,m;2. U (i)U (j) + U (j)U (i) = 0, for all i, j ∈ {1, · · · ,m} with i = j.

Let U (j) = (U(j)k,l )k,l≤2n (1 ≤ j ≤ m). The canonical sub-Laplacian on H(2n,m)

can be written as Δ =∑2n

l=1 X2l , where Xl (1 ≤ l ≤ 2n) are the left invariant vector

fields on H(2n,m), defined by

Xl =∂

∂xl+

1

2

m∑j=1

( 2n∑k=1

xkU(j)k,l

) ∂

∂tj.

Notice that H(2n, 1) is the exact Heisenberg group of dimension 2n + 1. Recallthat, as in [8], (2n,m) satisfies the following condition: 2n = (2a + 1)24p+q witha, p ∈ N∗ and 0 ≤ q < 3, and it implies

m < ρ(2n) = 8p+ 2q.(1.4)

In particular, the following estimate will be used in this paper:

m ≤ 2 log2 (2n).(1.5)

Let o = (0, 0) denote the identity of H(2n,m) and g = (x, t) ∈ R2n ×Rm denotea point of H(2n,m). We use the following notation through the whole paper:

|x|2 =

2n∑k=1

x2k, |t|2 =

m∑j=1

t2j and t · λ =

m∑j=1

tjλj for λ ∈ Rm.

Recall that the Haar measure dg on H(2n,m) is the Lebesgue measure.There are two standard distances on H(2n,m): one is the Carnot-Caratheodory

distance dCC (associated to {X1, · · · , X2n}; see [23]), and the other one is definedby the Koranyi norm dK , which is associated to the fundamental solution of Δ (i.e.the Green function; see for example [8]). Moreover, one has

dCC(gg1, gg2) = dCC(g1, g2), dK(gg1, gg2) = dK(g1, g2), ∀g, g1, g2 ∈ H(2n,m).

By convention, denote dK(g) = dK(g, o) and dCC(g) = dCC(g, o). Recall that(see for example [8])

dK(x, t) =(|x|4 + 16|t|2

) 14

, ∀(x, t) ∈ H(2n,m) = R2n × R

m.(1.6)

Denote μ(ϕ) = (2ϕ−sin 2ϕ)/(2 sin2 ϕ) : ]−π, π[−→ R and let μ−1 be the inversefunction. One has d2CC(0, t) = 4π|t| and for x = 0 (see [19] or [22], pp. 90-91),

d2CC(x, t) = (θ/ sin θ)2|x|2 with θ = μ−1(4|t|/|x|2).(1.7)

In what follows, one denotes BK(g, r) (g ∈ H(2n,m), r > 0) the open ball withthe center g and the radius r induced by the Koranyi norm dK , and BCC(g, r)



induced by the Carnot-Caratheodory distance dCC . For a measurable set E, let|E| denote the volume and χE the characteristic function. For f ∈ L1

loc(H(2n,m)),one can define two centered Hardy-Littlewood maximal functions respectively by

MKf(g) = supr>0

|BK(g, r)|−1

∫BK(g,r)

|f(g′)| dg′, ∀g ∈ H(2n,m),

MCCf(g) = supr>0

|BCC(g, r)|−1

∫BCC(g,r)

|f(g′)| dg′, ∀g ∈ H(2n,m).

The main result is the following

Theorem 1.1. There exists a constant A > 0 such that for all n,m ∈ N∗ satisfyingm2 � log n, we have

‖MKf‖L1,∞ ≤ An‖f‖1, ∀f ∈ L1(H(2n,m)).(1.8)

Furthermore, for any fixed m ∈ N∗, we have

‖MCCf‖L1,∞ ≤ L(m)n‖f‖1, ∀n, ∀f ∈ L1(H(2n,m)),(1.9)

for some constant L(m) > 0 independent of n.

Remark 1.2. (1) Notice that (1.2) holds for all H(2n,m) (see section 3 below). Webelieve it is most possible to show (1.8) holds for all H(2n,m). By applying moreproperties of Bessel functions and Legrendre functions, we may prove it.

(2) Moreover, by a certain property of Beta function, we could get an upperbound for L(m) in (1.9).

(3) From the result obtained by Naor and Tao [17] or by [20], we have

‖Mf‖L1→L1,∞ ≤ L(2n+ 2m) ln(2n+ 2m), ∀(2n,m)(1.10)

holds with M = MK or M = MCC for some positive constant L. For m fixed, asn → ∞, (1.8) and (1.9) are better than (1.10). When m → ∞, for M = MCC ,(1.9) is not always better.

(4) Observe that there exists a bound of type limn−→+∞ ‖MCube‖L1−→L1,∞ =+∞ for the centered maximal functions associated to cubes in Rn; see [1] for details.

Outline of the proof

The main idea of the proof has been pointed out in [12]. More precisely, bycertain results in [10], we need only to prove (1.8); by the recursion formula obtainedin [11], it is enough to consider the case of m ≥ 3 odd. To this end, we divide theproof into two cases: in the first case, we get the uniform lower bound for thePoisson kernel; in the other case, we obtain good estimations of (−Δ)−

12 (g) and

(−Δ)−12 exp

(−ω dK(g)√

n

√−Δ

)(g).

This paper is organized as follows: some properties of Bessel functions and Le-gendre functions are presented in section 2. In section 3 we review the heat kernel,the Poisson kernel and Green function on H(2n,m). In section 4, we prove Theorem1.1 for m ≥ 3 odd and dK metric, and Theorem 1.1 for m ≥ 2 even and dK metricis proved in section 5. In section 6, we prove Theorem 1.1 for dCC metric.



Notation

We shall use C,C ′, A,A′, etc. to denote absolute positive constants whose valuemay differ at each occurrence.

For two functions f and g, we denote f = O(g) if there exists a constant c > 0

such that |f | ≤ c|g|, f = o(g) if lim fg = 0, and f ∼ g if there exists an A > 1 such

that A−1f ≤ g ≤ Af .

2. Review of Bessel functions and Legendre functions

In what follows, we denote Jν the Bessel function, for �ν > − 12 and −π <

arg z < π, defined by (see for example [16], p. 65 and p. 79)

Jν(z) =

+∞∑k=0

(−1)k(z/2)ν+2k

k!Γ(ν + k + 1)=

2√πΓ( 12 + ν)

(z2

)ν∫ 1

0

(1− h2)ν−12 cos (zh) dh.

(2.1)

In particular, we have the following estimate which will be used repeatedly:

(z/2)−ν |Jν(z)| ≤1

Γ(ν + 1), z > 0.(2.2)

Recall that (see for example §3.1.1 in [16], p. 67, and §8.472 in [6], p. 926, or §3.2in [24], p. 46)

1

z

d

dz

(z−νJν(z)

)= −z−ν−1Jν+1(z).(2.3)

In the case of m ≥ 3 odd, Jm−22

(λ) is a simple function; see for example [16],

p. 72, or §8.46 in [6], pp. 924-925, or [24], p. 53. More precisely,

Jm−22

(z) = T1(z) + T2(z),(2.4)

where

T1(z) =1√2πz

e−iπ2

m−12 eiz

m−32∑

k=0

ikΓ(m−12 + k)

k!Γ(m−12 − k)(2z)k

,

T2(z) =1√2πz

eiπ2

m−12 e−iz

m−32∑

k=0

(−i)kΓ(m−12 + k)

k!Γ(m−12 − k)(2z)k

.

Recall that for �(μ+ ν) > −1 and �s > |�a|, one has (see [16], p. 446)∫ +∞

0

λμJν(aλ)e−sλ dλ = Γ(μ+ ν + 1)(s2 + a2)−

12 (μ+1)P−ν

μ (s√

s2 + a2),(2.5)

where P−νμ (r) (−1 < r < 1) is the Legendre function (“on the cut”) with the

parameters μ and −ν.When n ∈ N∗ and m ≥ 2 satisfies (1.4), one has for all 0 < θ ≤ π

2 (see [16],p. 188 and p. 203)∣∣∣P−m−2

2

n+m−12

(cos θ)∣∣∣ < 4

Γ(n+ 32 )

Γ(n+ m−12 + 1)

π− 12 (n+

m− 1

2)−

12 (sin θ)−

m−12 .(2.6)



Throughout this paper, we will adopt the following estimate (see for example[16], p. 12):

Γ(s) =√2πs−

12

(se

)s[1 +O(s−1)

], s � 1.(2.7)

3. Review of the heat kernel, Poisson kernel

and Green function on H(2n,m)

Let ph = p(2n,m)h (h > 0) be the heat kernel (i.e. the kernel of ehΔ) on H(2n,m),

Ph = P(2n,m)h the Poisson kernel (i.e. the kernel of e−h

√−Δ) and G = G(2n,m) the

Green function (i.e. the kernel of (−Δ)−1). By convention, one denotes ph(g) =ph(g, o), Ph(g) = Ph(g, o), P (g) = P1(g) and G(g) = G(g, o), where o is the identityof H(2n,m).

The heat kernel. We will need the following expression of p(2n,m)h (x, t), due to

Randall ([18], p. 292):

ph(x, t) = 2(4π)−n−m2 h−n−m

×∫ +∞

0

rm−1e−|x|2r4h coth r

( r

sinh r

)n(r|t|2h

)−m−22

Jm−22

(r|t|h

)dr.(3.1)

In particular,

ph(x, t) = h−n−mp(x/√h, t/h).(3.2)

For fixed m,n ∈ N∗, w1, w2 ≥ 0, denote

p(n,m,w1, w2) = 2(4π)−n−m2

×∫ +∞

0

rm−1 exp(−w1r

4coth r

)( r

sinh r

)n (w2r

2

)−m−22

Jm−22

(w2r)dr.(3.3)

It follows that

p(x, t) = p(n,m, |x|2, |t|), ∀(x, t) ∈ H(2n,m).

By (2.3), we have (see also (1.12) in [11])

∂

∂ω2p(n,m, ω1, ω2) = −2πω2p(n,m+ 2, ω1, ω2).(3.4)

Notice that, for m,n ∈ N∗, for all (x, t) ∈ H(2n,m) with |t| = 0, we have (see(1.13) in [11])

p(n,m, |x|2, |t|) = 2|t|∫ ∞

1

w√w2 − 1

p(n,m+ 1, |x|2, w|t|)dw.(3.5)

The Poisson kernel. By the formula of subordination, i.e.

e−√−Δ =

1

2√π

∫ +∞

0

h− 32 e−

14h ehΔ dh,

it follows from (3.1) and the Fubini theorem that

P (x, t) = 2(4π)−n−m+12

∫ +∞

0

rm−1( r

sinh r

)n

T dr,(3.6)



where

T =

∫ +∞

0

h−n−m− 32 e−

1+|x|2r coth r4h

(r|t|2h

)−m−22

Jm−22

(r|t|h

) dh.(3.7)

For ω1 ≥ 0, ω2 ∈ R+, denote

P (n,m, ω1, ω2) =1

2√π

∫ ∞

0

h−n−m− 32 e−

14h p(n,m,w1/h, w2/h)dh;

it follows that P (2n,m)(x, t) = P (n,m, |x|2, |t|). Applying (3.5), one gets

P (n,m, |x|2, |t|) = 2|t|∫ ∞

1

w√w2 − 1

P (n,m+ 1, |x|2, w|t|)dw.(3.8)

Let

U =1 + |x|2r coth r

4, V =

√(1 + |x|2r coth r4

)2

+ (|t|r)2.

By (3.6) and (2.5), one can rewrite

P (n,m, |x|2, |t|) = 2(4π)−n−m+12 Γ(n+m+

1

2)

×∫ +∞

0

rm−1( r

sinh r

)n(r|t|2

)−m−22

V−(n+m+12 +1)P

−m−22

n+m+12

(U

V

)dr.(3.9)

Let Q = 2n+2m be the homogenous dimension of H(2n,m). We have the followingtime scaling property for the Poisson kernel:

Ph(x, t) = h−QP (x/h, t/h2), ∀h > 0, (x, t) ∈ R2n × R

m.(3.10)

Note that for m ≥ 3 odd, by (2.4) and (3.7), one has

T =

∫ +∞

0

h−n−m− 32 e−

1+|x|2r coth r4h (

r|t|2h

)−m−2

2

(T1(

r|t|h

) + T2(r|t|h

))dh

=2

m−32

√π(r|t|)m−1

2

m−32∑

k=0

Γ(m−12 + k)

k!Γ(m−12 − k)2k

(|t|r)−k

×∫ +∞

0

hk−n−m2 −2e−

1+|x|2r coth r4h

[e

i(|t|r)h −iπ

2m−1

2 ik + e−i(|t|r)

h +iπ2m−1

2 (−i)k]dh

=2

m−32

√π(r|t|)m−1

2

m−32∑

k=0

Γ(m−12 + k)

k!Γ(m−12 − k)2k

(|t|r)−kΓ(n+m

2+ 1− k)

×{e−iπ

2m−1

2 ik[1 + |x|2r coth r

4− i(|t|r)

]k−n−m2 −1

+ eiπ2

m−12 (−i)k

[1 + |x|2r coth r4

+ i(|t|r)]k−n−m

2 −1}.

Substituting into (3.6), it yields

P (n,m, |x|2, |t|) = (4π)−n−m+12 2

m−12 e−iπ

2m−1

2

√π|t|m−1

2

m−32∑

k=0

Γ(m−12 + k)ikΓ(n+ m

2 + 1− k)

k!Γ(m−12 − k)2k|t|k

×∫R

rm−1

2 −k( r

sinh r

)n[1 + |x|2r coth r4

− i(|t|r)]k−n−m

2 −1

dr.(3.11)



Green function. For (x, t) ∈ H(2n,m) with |t| = 0, the Green function G(x, t) isthe kernel of the operator (−Δ)−1 which can be expressed as

G(x, t) =

∫ ∞

0

ph(x, t)dh

= 2(4π)−n−m2

∫ ∞

0

h−n−m dh

×∫ ∞

0

rm−1 exp

(−|x|2r coth r

4h

)( r

sinh r

)n(r|t|2h

)−m−22

Jm−22

(r|t|h

) dr.

Denote

G(n,m,w1, w2) =

∫ ∞

0

h−n−mp(n,m,w1/h, w2/h) dh, ω1 ≥ 0, ω2 ∈ R+,(3.12)

thus G(x, t) = G(n,m, |x|2, |t|). For the Heisenberg group H(2n, 1), i.e. m = 1, wehave (see [5, 7])

G(2n,1)(x, t) = G(n, 1, |x|2, |t|)

= (2π)−1(4π)−nΓ(n)B

(n

2,1

2

)((|x|2/4)2 + |t|2

)−n2

.(3.13)

By (3.4), it follows that

G(2n,2k+1)(x, t) = G(n, 2k + 1, |x|2, |t|) =(− 1

π

∂

∂|t|2

)k

G(n, 1, |x|2, |t|).

Combining this with (3.13), through direct computation, yields

G(2n,2k+1)(x, t) = G(n, 2k + 1, |x|2, |t|)

=1

Q− 24m− 1

2Γ(n)Γ((Q+ 2)/4)

πn+m2 Γ(n+1

2 )d2−QK (x, t),m = 2k + 1,(3.14)

which has been obtained by [7]; see also [8]. By (3.5), we have, for (x, t) ∈ H(2n, 2k),

G(2n,2k)(x, t) = 2|t|∫ ∞

1

w√w2 − 1

G(n, 2k + 1, |x|2, w|t|)dw.(3.15)

Note that

|BK(o, 1)| = 2πm2

mΓ(m2 )

∫|x|<1

(1− |x|4

16

)m2

dx

=π

m+2n2

m4mΓ(m2 )Γ(n)B(m2

+ 1,n

2

).(3.16)

Together with (3.14) and (3.15), (1.2) follows easily by direct calculation.

4. Proof of Theorem 1.1 for m ≥ 3 odd and dK metric

The proof follows the idea for the case of Heisenberg group (see [10]). More

precisely, we consider the Poisson semigroup on H(2n,m), e−h√−Δf = f ∗ Ph

(h > 0).As

Ph(x, t) ≥ 0, ∀(x, t) ∈ R2n × R

m, ‖Ph‖1 = 1, ∀h > 0,



by the Hopf-Dunford-Schwartz maximal ergodic theorem, one has∣∣∣{g; sups>0

1

s

∫ s

0

e−h√−Δf(g) dh > λ

}∣∣∣ ≤ 2

λ‖f‖1, ∀λ > 0, f ∈ L1(H(2n,m)).

To prove Theorem 1.1 for M = MK , one needs only to prove that there exists aconstant A > 0 such that for all n,m ∈ N∗ verifying m2 � log n, one has

MKf(g) ≤ An sups>0

1

s

∫ s

0

e−h√−Δf(g) dh, ∀g ∈ H(2n,m), 0 ≤ f ∈ L1(H(2n,m)).

For go ∈ H(2n,m), denote Lgo the left translation operator defined by Lgof(g) =

f(gog); one has LgoM = MLgo(M = Mk or M = MCC) and Lgoe−h

√−Δ =

e−h√−ΔLgo . By the dilation structure on H(2n,m), it suffices to find some s(n) > 0

such that for all g = (x, t) = o, one has

|BK(o, 1)|−1χBK(o,1)(g) ≤ An1

s(n)

∫ s(n)

0

Ph(g) dh

= An1

s(n)

∫ s(n)

0

h−QP

(x

h,t

h2

)dh

= An1

s(n)

∫ s(n)

0

h−QP

(n,m,

|x|2h2

,|t|h2

)dh.(4.1)

To this end, we shall divide the proof into two cases: a)√nφ ≥ C∗m with φ

defined by (4.2) below and C∗ � 1 to be determined later; b)√nφ < C∗m. For

case a) we adapt the method in [10] to prove (4.1). The key point here is to get theuniform lower bound for the Poisson kernel. For case b) we adapt the method in

[12] to prove (4.1). The key point here is to get good estimations for (−Δ)−12 (g, g′)

and (−Δ)−12 exp

(−ω dK(g,g′)√

n

√−Δ

)(g, g′).

For M = MCC , we will prove the desired result by comparing with the case ofMK .

4.1. Lower bounds for the Poisson kernel P (x, t). In what follows, one denotesφ ∈ [0, π

2 ] satisfying

e−iφ = d−2K (x, t)(|x|2 − i4|t|).(4.2)

For ω1, ω2 ≥ 0, r ∈ C, let

f(ω1, ω2, r) :=ω21

4r coth r − i(ω2r).

For any (x, t) ∈ H(2n,m), we have that �f(|x|, |t|, r) ≥ 0 holds for all 0 ≤ �r ≤π2 . In fact, for all 0 ≤ � ≤ π

2 and for ι ∈ R, one has (see [2], p. 645)

�f(|x|, |t|, i�) =�

sin�cos�

|x|24

+ |t|� ≥ 0

and

�{f(|x|, |t|, ι+ i�)− f(|x|, |t|, i�)

}=

|x|24

sinh2 ι

sinh2 ι+ sin2 �(ι coth ι−� cot�) ≥ 0.



Let

F (r) = rm−1

2 −k( r

sinh r

)n(1

4+

|x|24

r coth r − i|t|r)k−n−m

2 −1

.

It is clear that F is analytic in

Ω = {r ∈ C; 0 < �r <π

2},

and continuous on Ω. Moreover, one has

limr∈Ω,|r|−→+∞

|F (r)| = 0.

By Cauchy’s fundamental theorem, one gets∫R

F (r) dr =

∫R

F (r + iφ) dr = W.(4.3)

By (4.2), it follows that

f(|x|, |t|, r) = r

4 sinh rd2K(x, t) cosh (r − iφ).

Denote

K = (r + iφ)m−1

2 −k( sinh (r + iφ)

r + iφ

)m2 +1−k

×(14

sinh (r + iφ)

r + iφ+

1

4d2K(x, t) cosh r

)k−n−m2 −1

;

it follows that

W =

∫R

K dr.

Give

σ = σ(m) > 0

which will be determined later (we remark here that we can choose σ = m− 12 ). Let

W1 =

∫ σ

−σ

K dr, W2 =

∫|r|≥σ

K dr.

We have W = W1 +W2. First let us estimate W2. Note that

(4.4) max{| sinh (r + iφ)|,

∣∣∣ sinh (r + iφ)

r + iφ

∣∣∣} ≤ cosh r, ∀0 ≤ φ ≤ π

2, r ∈ R.

Hence, as n −→ +∞ and dK(x, t) � n12 , one has

|W2| ≤ 4n+m2 +1−k

∫|r|≥σ

[(d2K(x, t)− 1) cosh r

]k−n−m2 −1

(cosh r)m2 +1−k dr

≤ 2 · 4n+m2 +1−k(d2K(x, t)− 1)k−n−m

2 −1

∫ +∞

σ

(cosh r)−n sinh r(sinhσ)−1 dr

=2 · 4n+m

2 +1−k

dK(x, t)2(n+m2 +1−k)

e−(n+m2 +1−k) ln (1−d−2

K (x,t)) (coshσ)1−n

n− 1(sinhσ)−1

=2 · 4n+m

2 +1−k

dK(x, t)2(n+m2 +1−k)

(coshσ)1−n

n− 1(sinhσ)−1

(1 +O(

n

d2K(x, t))).

Now let us estimate W1 :



By Taylor’s formula, we have∣∣∣∣∣( sinh (r + iφ)

r + iφ

)m2 +1−k

−( sinφ

φ

)m2 +1−k

− (m

2+ 1− k)r

(sinφφ

)m2 −k d

ds

∣∣∣s=0

( sinh (s+ iφ)

s+ iφ

)∣∣∣∣∣≤ r2

2sup

η∈[0,r]

∣∣∣∣ d2dη2

( sinh (η + iφ)

η + iφ

)m2 +1−k

∣∣∣∣ .Note that there exists a constant C > 0 such that for all −m− 1

2 ≤ η ≤ m− 12 ,

one has

(cosh η)m2 = e

m2 ln [1+(cosh η−1)] ≤ e

m2 (cosh η−1) ≤ C.

Thus, by (4.4), there exists a constant C1 > 0 such that for all m ≥ 3 odd,

0 ≤ k ≤ m−32 , 0 ≤ φ ≤ π

2 and −m− 12 ≤ r ≤ m− 1

2 , one has

( sinh (r + iφ)

r + iφ

)m2 +1−k

=( sinφ

φ

)m2 +1−k

− i[(m

2+ 1− k)r

( sinφφ

)m2 −k( sinφ

φ

)′]+ E(m, k, φ; r)(4.5)

and

|E(m, k, φ; r)| ≤ C1m2r2.

Moreover, in the case of nd2K(x,t)

� 1, by the fact that m < 3 log2 (2n), one has

(14

sinh (r + iφ)

r + iφ+

1

4d2K(x, t) cosh r

)k−n−m2 −1

= 4n+m2 +1−k

(d2K(x, t) cosh r

)k−n−m2 −1

e−(n+m

2 +1−k) ln (1+ 1

d2K

(z,t) cosh r

sinh (r+iφ)r+iφ )

= 4n+m2 +1−kdK(x, t)−2(n+m

2 +1−k)(cosh r

)k−n−m2 −1

S,

where

S = e−(n+m

2 +1−k) ln (1+ 1

d2K

(z,t) cosh r

sinh (r+iφ)r+iφ )

= 1 +O(n

d2K(x, t)).

For 0 ≤ j ≤ m−12 − k, set

W1,1(j) =( sinφ

φ

)m2 +1−k

∫ σ

−σ

rm−1

2 −k−j(cosh r

)k−n−m2 −1

S dr,

W1,2(j) = −i[(m

2+ 1− k)

( sinφφ

)m2 −k( sinφ

φ

)′]

×∫ σ

−σ

rm−1

2 −k−j+1(cosh r

)k−n−m2 −1

S dr,

W1,3(j) =

∫ σ

−σ

rm−1

2 −k−jE(m, k, φ; r)(cosh r

)k−n−m2 −1

S dr.



One can write

W1 = 4n+m2 +1−kdK(x, t)−2(n+m

2 +1−k)

×m−1

2 −k∑j=0

Cjm−1

2 −k(iφ)j

(W1,1(j) +W1,2(j) +W1,3(j)

).(4.6)

Let us first estimate W1,1(m−12 − k) :

Note that

W1,1(m− 1

2− k) = 2

( sinφφ

)m2 +1−k[

1 +O(n

d2K(x, t))]

×{∫ +∞

0

−∫ +∞

σ

}(cosh r

)k−n−m2 −1

dr.

Combining this with the facts that

∫ +∞

0

(cosh r

)k−n−m2 −1

dr =

∫ +∞

0

(cosh r

)k−n−m2 −2

d(sinh r)

=1

2B(

n+ m2 + 2− k

2− 1

2,1

2), (see [16], pp. 6-7),∫ +∞

σ

(cosh r

)k−n−m2 −1

dr ≤ 1

sinh σ

∫ +∞

σ

(cosh r

)k−n−m2 −1

sinh r dr

=1

n+ m2 − k

1

sinh σ

(cosh σ

)k−n−m2

,

and

B(n+ m

2 + 1− k

2,1

2) = Γ(

1

2)Γ(

n+m2 +1−k

2 )

Γ(n+m

2 +2−k

2 )=

√π(n+ m

2 + 2− k

2

)− 12

(1 + o(1)),

n −→ +∞ (see [16], p. 6 and p. 12),

one has

W1,1(m− 1

2− k) = B(

n+ m2 + 1− k

2,1

2)( sinφ

φ

)m2 +1−k

×[1 +O(

1√nσ

)][1 +O(

n

d2K(x, t))],

d2K(x, t)

n� 1, n � 1.(4.7)

On the other hand, observe that for 1 ≤ α ≤ m−12 + 2,

∫ σ

−σ

|rα|(cosh r

)k−n−m2 −1

|S| dr

≤ 2[1 +O(

n

d2K(x, t))] ∫ +∞

0

sinhα r(cosh r

)k−n−m2 −1

dr

= B(α+ 1

2,n+ m

2 + 1− k − α

2)[1 +O(

n

d2K(x, t))].



Together with (2.7), (4.7) and the fact 0 ≤ φ ≤ π2 , one has for 0 ≤ j ≤ m−1

2 − k,

with the notation α(j) := m−12 − k − j,

|W1,2(j)| = W1,1(m− 1

2− k)m2

α(j)+22 Γ(

α(j) + 2

2)n−α(j)+1

2 O(1),

|W1,3(j)| = W1,1(m− 1

2− k)m2

( φ

sinφ

)m2

2α(j)+3

2 Γ(α(j) + 3

2)n−α(j)+2

2 O(1),

and for 0 ≤ j ≤ m−12 − k − 1,

|W1,1(j)| = W1,1(m− 1

2− k)2

α(j)+12 Γ(

α(j) + 1

2)n−α(j)

2 O(1),

and O(1) ≤ C2 independent of (2n,m, k, j).Note that

m ≤ 2 log2 (2n) + 4,

as n → +∞. It follows thatm√n= o(1),

and for 0 ≤ φ ≤ π2 , one has

m2(

φsinφ

)m2

2n≤

m2(

π2

)m2

2n= 22 log2 m+m

2 log2 (π2 )−log2 (2n) = o(1).

Thus, one has for 0 ≤ j ≤ m−12 − k − 1,

|W1,1(j)|+ |W1,2(j)|+ |W1,3(j)|

= W1,1(m− 1

2− k)2

α(j)+12

α(j) + 1

2Γ(

α(j) + 1

2)n−α(j)

2 O(1)

and

W1,1(m− 1

2− k) +W1,2(

m− 1

2− k) +W1,3(

m− 1

2− k)

= W1,1(m− 1

2− k)(1 + o(1)).

Recall that (see for example [16], p. 3)

Γ(2r) = π− 12 22r−1Γ(r)Γ(

1

2+ r).

One has for 0 ≤ j ≤ m−12 − k − 1,

Cjm−1

2 −kφj2

α(j)+12

α(j) + 1

2Γ(

α(j) + 1

2)n−α(j)

2

=(m−1

2 − k)!

j!

2α(j)+1

2α(j)+1

2 Γ(α(j)+12 )

Γ(2α(j)+12 )

n−m−1

2−k

2 (√nφ)j

=(m−1

2 − k)!

j!n−

m−12

−k

2 (√nφ)j

√π

1 + α(j)

Γ(α(j)2 + 1)2−

1+α(j)2 .

Therefore, when√nφ

m� 1



holds and σ = 1√m, combined with (4.6) and (4.7), one has

W1 = 4n+m2 +1−kdK(x, t)−2(n+m

2 +1−k)W1,1(m− 1

2− k)

× (iφ)m−1

2 −k[1 +O(

m√nφ

)]

=√2π4n+

m2 +1−kdK(x, t)−2(n+m

2 +1−k)( sinφ

φ

) 32

n− 12 (i sinφ)

m−12 −k

× (1 + o(1))[1 +O(

m√nφ

)][1 +O(

n

d2K(x, t))].(4.8)

Together with (3.11), (4.3) and the fact that 4|t| = d2K(x, t) sinφ, we have thefollowing

Proposition 4.1. We have

P (x, t) = 23m2 π−n−m+1

2 d−Q−1K (x, t)

(sinφ

φ

) 32

n− 12 (1 + o(1))

[1 +O(

m√nφ

)]

×[1 +O(

n

d2K(x, t))] m−1

2∑k=0

Γ(m−12 + k)

k!Γ(m−12 − k)2k

(sinφ)−2kΓ(n+m

2+ 1− k),

for all (x, t) ∈ H(2n,m) satisfying√nφ � m and d2K(x, t) � n, where o(1) → 0 as

n → ∞. In particular, there exist two positive constants C∗, C∗ � 1 such that for

all (x, t) ∈ H(2n,m) satisfying√nφ ≥ C∗m and dK(x, t) > C∗n

12 , we have

P (x, t) ≥ 2−123m2 π−n−m+1

2 d−Q−1K (x, t)

(sinφ

φ

) 32

n− 12Γ(n+

m

2+ 1).

4.2. Estimate of (−Δ)−12 exp

(−ω dK(g)√

n

√−Δ

)(g) (ω ≥ 0) for

√nφ ≤ C∗m.

Throughout this subsection, we assume φ defined in (4.2) satisfies√nφ ≤ C∗m,

which will be used repeatedly.For ω ≥ 0, g = (x, t) ∈ H(2n,m), one has

(−Δ)−12 exp

(−ωdK(g)n− 1

2 (−Δ)12

)(g)

=1√π

∫ ∞

0

h−1/2ehΔ(g) exp

(−ω2d2K(g)

4nh

)dh.(4.9)

In particular,

(−Δ)−12 (g) =

1√π

∫ ∞

0

h−1/2ehΔ(g) dh.

Substituting (3.1) into (4.9) and applying the Fubini theorem, it follows that

(−Δ)−12 exp

(−ωn− 1

2 dK(g)√−Δ

)(g)

= 2π− 12 (4π)−n−m

2

∫ ∞

0

rm−1( r

sinh r

)n

Ξdr,(4.10)



where

Ξ =

∫ ∞

0

h−n−m− 12 exp

(−|x|2r coth r + ω2 d2

K(g)n

4h

)(r|t|2h

)−m−22

Jm−22

(r|t|h

)dh

=

(r|t|2

)−m−22

∫ ∞

0

exp

(−|x|2r coth r + ω2 d2

K(g)n

4u

)un+m

2 − 12 Jm−2

2(r|t|u)du.

Thanks to (2.5), one has

Ξ =

(r|t|2

)−m−22

Γ(n+m− 1

2)V −(n+m

2 + 12 )P

−m−22

n+m2 − 1

2

(U

V

),

where

U =|x|2r coth r + ω2 d2

K(g)n

4, V =

√U2 + (r|t|)2.

Substituting into (4.10), we have, for g = (x, t) = o,

(−Δ)−12 exp

(−ω

dK(g)√n

√−Δ

)(g)

= 2π− 12 (4π)−n−m

2 Γ(n+m− 1

2)

×∫ ∞

0

rm−1( r

sinh r

)n(r|t|2

)−m−22

V −(n+m2 + 1

2 )P−m−2

2

n+m2 − 1

2

(U

V

)dr.(4.11)

Recall that (see (4.2))

cosφ = d−2K (x, t)|x|2, sinφ = 4d−2

K (x, t)|t|.This yields

U =d2K(x, t)r coth r cosφ

4+

ω2d2K(x, t)

4n(4.12)

and

V =√U2 + (r|t|)2 =

rd2K(x, t)√cos2 φ+ sinh2 r + U

4 sinh r,(4.13)

where we denote

U =ω4 sinh2 r

n2r2+

2ω2 sinh r cosh r cosφ

nr(which is increasing for r ≥ 0).(4.14)

Also, we denote

U0 =ω4

n2+

2ω2 cosφ

n.(4.15)

For ϑ ∈ [0, π2 ] satisfying

cosϑ =U

V=

cosh r cosφ+ ω2 sinh rnr√

cos2 φ+ sinh2 r + U

,(4.16)

this implies

sinϑ =sinh r sinφ√


.(4.17)



For simplicity, denote

R := rm−1( r

sinh r

)n(r|t|2

)−m−22

V −(n+m2 + 1

2 ),(4.18)

M1 :=

∫ ζ

0

RP−m−2

2

n+m2 − 1

2

(cosϑ)dr,(4.19)

and

M2 :=

∫ ∞

ζ

RP−m−2

2

n+m2 − 1

2

(cosϑ)dr,(4.20)

where ζ = nν0− 12 with ν0 ∈ (0, 12 ); we shall find ν0 = 1

4 works. To estimate M1,M2,we need the following estimations:

4.2.1. Estimate of P−m−2

2

n+m2 − 1

2

(cos θ) (m ≥ 3, 0 ≤ θ ≤ π2 ) for m sin2 θ ≤ 1. The main

result in this subsection is the following lemma.

Lemma 4.2. There exists some function E(n,m; θ) = O( mΓ(m−1

2 )(sin θ)

m2 +1), such

that for all m ≥ 3 and 0 ≤ θ ≤ 1 satisfying m sin2 θ ≤ 1, we have

P−m−2

2

n+m2 − 1

2

(cos θ) = (n+m

2)−

m−22 Jm−2

2

((n+

m

2) sin θ

)+ E(n,m; θ).(4.21)

Proof. Recall that for m ≥ 2, we have (see [16]. p. 188 or [6], p. 961):

Γ(m− 1

2)P

−m−22

n+m2 − 1

2

(cos θ)

=(π2

)−1/2

(sin θ)−m−2

2

∫ θ

0

(cos t− cos θ)m−3

2 cos[(n+m

2)t]dt

=(π2

)−1/2

(sin θ)−m−2

2 2m−3

2

∫ θ

0

(sin2

θ

2− sin2

t

2

)m−32

cos[(n+m

2)t]dt.

By change of the variable with t = y sin θ, one has

Γ(m− 1

2)P

−m−22

n+m2 − 1

2

(cos θ) =(π2

)−1/2

(sin θ)−m−4

2 2m−3

2

(sin

θ

2

)m−3

×∫ θ

sin θ

0

(1−

sin2 y sin θ2

sin2 θ2

)m−32

cos((n+

m

2)y sin θ

)dy

:=(π2

)−1/2

(sin θ)−m−4

2 2m−3

2

(sin

θ

2

)m−3

L.(4.22)

To estimate P−m−2

2

n+m2 − 1

2

(cos θ), one needs only to estimate L. To this end, let

L1 =

∫ 1

0

(1−

sin2 y sin θ2

sin2 θ2

)m−32

cos((n+

m

2)y sin θ

)dy

and

L2 =

∫ θsin θ

1

(1−

sin2 y sin θ2

sin2 θ2

)m−32

cos((n+

m

2)y sin θ

)dy;



thus

L = L1 + L2.(4.23)

Note that one has

|L2| ≤θ

sin θ− 1 = O(sin2 θ), ∀0 < θ ≤ π

2, ∀(2n,m).(4.24)

For 0 ≤ y ≤ 1, by the differential mean value theorem, we have(1−

sin2 y sin θ2

sin2 θ2

)(m−3)/2

= (1− y2)(m−3)/2 + E1(m, y, θ),

where E1(m, y, θ) = O(m sin2 θ). Hence

L1 =

∫ 1

0

(1− y2)(m−3)/2 cos((n+

m

2)y sin θ

)dy

+

∫ 1

0

E1(m, y, θ) cos((n+

m

2)y sin θ

)dy

=1

2π1/2

((n+ m

2 ) sin θ

2

)−m−22

Γ(m− 1

2)Jm−2

2

((n+

m

2) sin θ

)+ E2(m, θ, n+

m

2),(4.25)

with E2(m, θ, n+ m2 ) = O(m sin2 θ), where the last equality follows from (2.1).

For m ≥ 3, combining this with (4.22), (4.23), (4.24) and (4.25), one has

P−m−2

2

n+m2 − 1

2

(cos θ) = (n+m

2)−

m−22 Jm−2

2

((n+

m

2) sin θ

)

+O(2−m−3

2 m

Γ(m−12 )

(sin θ)m2 +1(cos

θ

2)3−m

)

+ (n+m

2)−

m−22

[(cos

θ

2

)3−m

− 1]Jm−2

2

((n+

m

2) sin θ

).

Notice that for 0 < mθ2 ≤ 1,

1 ≤ (cosθ

2)3−m ≤ e−m ln (1−2 sin2 θ

4 ).

By (2.2), we get the desired result. �

4.2.2. Estimates of M1 and M2. Our aim in this subsection is now to estimate M1

and M2.Notice that for 0 ≤ φ ≤ C∗ m√

nand 0 < r ≤ ζ = nν0− 1

2 , by (4.17), we have for n

large enough,

m sin2 ϑ ≤ 1.

Substituting (4.21) into the expression (4.19), we have

M1 =

∫ ζ

0

R[(n+

m

2)−

m−22 Jm−2

2

((n+

m

2) sinϑ

)+ E(n,m;ϑ)

]dr

=

(sinφ

2

)−m−22

4n+m− 12 d

−(2n+2m)+1K (x, t)

(M11 + (n+

m

2)−

m−22 ·M12

),(4.26)



where

M11 =

∫ ζ

0

rm2

( r

sinh r

)−(m2 + 1

2 )

(cos2 φ+ sinh2 r + U)− 1

2 (n+m2 + 1

2 )E(n,m;ϑ)dr

and

M12 =

∫ ζ

0

rm2

( r

sinh r

)−(m2 + 1

2 ) Jm−22

((n+ m

2 ) sinϑ)

(cos2 φ+ sinh2 r + U)12 (n+

m2 + 1

2 )dr.

We have the following estimates for M11 and M12. The proof is postponed untilthe Appendix.

Lemma 4.3. Given ω ≥ 0 and C∗ > 0, there exists a constant C(ω,C∗) > 0 suchthat

M12 = e−ω2

((n+ m

2 ) sinφ

2

) 12 (n+m− 5

2 ) (cosφ)−(n− 12 )

Γ( 12 (n+m− 12 ))

×K 12 (n−

12 )

((n+

m

2) sinφ

) [1 +O

(n−1eC(ω,C∗)m2

)](4.27)

and

|M11| ≤ (n+m

2)−

m−22 M12O

(m

32

neC(ω,C∗)m2

).(4.28)

Here Kν(z) is the modified Bessel function defined by

Kν(z) = K−ν(z) =

∫ ∞

0

e−z cosh t cosh(νt) dt, �z > 0.

Hence we obtain from (4.26),

M1 = 4n+m− 12 e−ω2

d−Q+1K (x, t)

((n+ m

2 ) sinφ

2

) 12 (n−

12 )

× (cosφ)−(n− 12 )

Γ( 12 (n+m− 12 ))

K 12 (n−

12 )

((n+

m

2) sinφ

) [1 +O(n−1eC(ω,C∗)m2

)].(4.29)

Furthermore one has the following estimate by (7.7) in the Appendix, for m2 �log n:

M1 ≥ 1

9e−ω2

4n+m− 12 d−Q+1

K (x, t)Γ( 12 (n− 1

2 ))

Γ( 12 (n+m− 12 ))

.(4.30)

Recall that

M2 =

∫ ∞

ζ

RP−m−2

2

n+m2 − 1

2

(cosϑ)dr



(see (4.20)), where R is defined in (4.18). Combining with (2.6) and (4.14), we have

|M2| ≤ 2m+2

2 4n+m+ 12 d−Q+1

K (x, t)Γ(n+ 3

2 )

Γ(n+ m+12 )

π− 12 (n+

m− 1

2)−

12 (sinφ)−m+ 3

2

×∫ ∞

ζ

r−12 sinh r(cos2 φ+ sinh2 r + U)

− 12 (n+1)dr

≤ c′2m+2

2 4n+m+ 12 d−Q+1

K (x, t)Γ(n+ 3

2 )

Γ(n+ m+12 )

(n+m

2)−

12 (sinφ)−m+ 3

2

× (cos2 φ+ U0 + sinh2 ζ)−12 (n−1)

≤ C(ω, ν0)2m+2

2 4n+m+ 12 d−Q+1

K (x, t)e−ω2

(sinφ)−m+ 32 e−

14n

2ν0,

where the last inequality follows from (2.7) and

(cos2 φ+ U0 + sinh2 ζ)−12 (n−

12 ) ≤ e−

n−12 ln [(cosφ+ω2

n )2+ζ2]

≤ C(ω, ν0)e−ω2

e−14n

2ν0.(4.31)

Notice that for n large enough,

Γ(12 (n− 1

2 ))

Γ( 12 (n+m− 12 ))

∼(n2

)−m2

.

By (4.29) and (4.30), it follows, for some constant C(ω,C∗) ∈ R+, that

|M2| ≤ M1O(e−C(ω,C∗)n2ν0).(4.32)

Together with (4.29), (4.30) and choosing ν0 = 14 , we have the following.

Proposition 4.4. Given ω ≥ 0, C∗ > 0. There exists a constant c(ω,C∗) > 0such that for all g = (x, t) ∈ H(2n,m) satisfying

√nφ ≤ C∗m,

(−Δ)−12 exp

(−ω

dK(g)√n

√−Δ

)(g)

= 2mπ−n−m2 − 1

2 e−ω2

Γ(n+m− 1

2)

× (dK(g))−(2n+2m)+1

((n+ m

2 ) sinφ

2

) 12 (n−

12 ) (cosφ)−(n− 1

2 )

Γ( 12 (n+m− 12 ))

×K 12 (n−

12 )

((n+

m

2) sinφ

) [1 +O(n−1ec(ω,C∗)m2

)].(4.33)

Moreover, if we assume m2 � log n holds, by (4.30), we have the followingestimate:

(−Δ)−12 (g) ≥ 1

92mπ− 2n+m+1

2 (dK(g))1−(2n+2m)Γ(12 (n− 1

2 ))Γ(n+m− 1

2 )

Γ( 12 (n+m− 12 ))

.(4.34)

4.3. Proof of Theorem 1.1 for m odd. First let us prove the following.

Lemma 4.5. There exists a constant C > 0 such that for all (2n,m) satisfyingm2 � log n with m odd, we have

n32 |BK(o, 1)| (−Δ)−

12 (g) > C, ∀g = (x, t) ∈ BK(o, 1) \ {o} ⊂ H(2n,m)

satisfying√nφ ≤ C∗m.



Proof. Recall that (see (3.16)),

|BK(o, 1)| = πm+2n

2

m4mΓ(m2 )Γ(n)B(m2

+ 1,n

2

).

Combining this with the estimate (4.34) and the property Γ(r+1) = rΓ(r)(r > 0),one has

n32 |BK(o, 1)| (−Δ)−

12 (g) ≥ 1

18π− 1

2 2−mn32

Γ(m+ n− 12 )Γ(

12 (n− 1

2 ))Γ(n2 )

Γ( 12 (m+ n− 12 ))Γ(n)Γ(

n2 + m

2 + 1).

Thanks to the property Γ(2z) = π− 12 22z−1Γ(z)Γ(z + 1

2 ) (z > 0) (see [16], p. 3),

n32 |BK(o, 1)| (−Δ)−

12 (g) ≥

√2

36π− 1

2n32Γ(n2 − 1

4 )Γ(n2 + m

2 + 14 )

Γ(n2 + 12 )Γ(

n2 + m

2 + 1)

∼ 1

9π− 1

2

(1 +

m+ 1

n

)− 34

;

the last inequality follows from (2.7). The desired conclusion follows. �

Now we are in a position for the

Proof of (4.1). To prove the desired result, we divide it into two cases.

Case (a).√nφ > C∗m. For 0 < dK(x, t) < 1, we have∫ 1

C∗√

n

0

Ph(x, t)dh

≥∫ dK (x,t)

C∗√

n

0

Ph(x, t)dh =

∫ dK (x,t)

C∗√

n

0

h−QP (x

h,t

h2)dh

≥ 2−123m2 π−n−m+1

2

(sinφ

φ

) 32

n− 12Γ(n+

m

2+ 1)

∫ dK (x,t)

C∗√

n

0

h−Qd−Q−1K (

x

h,t

h2)dh,

where the last inequality follows from Proposition 4.1. By the scaling property ofKoranyi norm dK , it yields∫ dK (x,t)

C∗√

n

0

Ph(x, t)dh

≥ 2−223m2 π−n−m+1

2 n− 32

(sinφ

φ

) 32

Γ(n+m

2+ 1)d−Q+1

K (x, t)C−2∗ .

Together with (3.16), it follows for all g = (x, t) ∈ BK(o, 1) that

n32 |BK(o, 1)|

∫ 1C∗

√n

0

Ph(x, t)dh

≥ 2−3π− 12

(sinφ

φ

) 32

C−2∗ 2−

m2

Γ(n2 )

Γ(n2 + m2 + 1)

Γ(n+ m2 + 1)

Γ(n).

Thanks to (2.7),

n32 |BK(o, 1)|

∫ 1C∗

√n

0

Ph(x, t)dh ≥ 2−4π− 12

(sinφ

φ

) 32

C−2∗ .



Choose s(n) = 1C∗

√n; (4.1) follows directly.

Case (b).√nφ ≤ C∗m. Put s(n) = 100√

n. By Proposition 4.4, there exists c > 0

such that for 0 < dK(g) < 1, we have∫ 100√n

0

e−h√−Δ(g)dh >

∫ 100√ndK(g)

0

e−h√−Δ(g)dh

= (−Δ)−12 (g)− (−Δ)−

12 e

−100dK (g)√

n(−Δ)−

12(g)

≥ c(−Δ)−12 (g).

Combining this with Lemma 4.5, it is clear that (4.1) holds.In all, if we choose s(n) = 100√

n, then (4.1) holds. �

Remark 4.6. From the above proof, we remark that there exists a constant c > 0

such that for log n � l2 with l odd, we have ∀0 < (|x|4 + 16|t|2) 14 < 4,

n32

∫ 400√n

0

h−(2n+2l)P (n, l,|x|2h2

,|t|h2

)dh

≥ cπ− 2n+l2 4l

Γ(n)Γ(n2 + l2 + 1)

Γ(n2 )

[|x|4 + 16|t|2

] 1−(2n+2l)4

.(4.35)

5. Proof of Theorem 1.1 for m even and dK metric

Proof of (4.1) for m even. For g = (x, t) ∈ H(2n,m) with dK(g) < 1 and |t| = 0,thanks to (3.8), one has

n32

∫ 400√n

0

h−(2n+2m)P (n,m,|x|2h2

,|t|h2

)dh

= 2|t|n 32

∫ 400√n

0

h−(2n+2m+2)dh

∫ ∞

1

w√w2 − 1

P

(n,m+ 1,

|x|2h2

,w|t|h2

)dw

> 2|t|∫ 2

|t|

1

w√w2 − 1

{n

32

∫ 400√n

0

h−(2n+2m+2)P (n,m+ 1,|x|2h2

,w|t|h2

)dh}dw

≥ cπ− 2n+m+12 4m+1Γ(n)Γ(

n2 + m+1

2 + 1)

Γ(n2 )|t|

×∫ 2

|t|

1

w√w2 − 1

[|x|4 + 16(w|t|)2

] 1−(2n+2m+2)4

dw.

However, by the change of variable√w2 − 1 = s/|t|, we have

|t|∫ 2

|t|

1

w√w2 − 1

[|x|4 + 16(w|t|)2

] 1−(2n+2m+2)4

dw

>

∫ 1

0

[(|x|4 + 16|t|2) + s2

] 1−(2n+2m+2)4

ds

= (|x|4 + 16|t|2)1−(2n+2m)

4

∫ (|x|4+16|t|2)−12

0

(1 + τ2)1−(2n+2m+2)

4 dτ.



Observe that∫ (|x|4+16|t|2)−12

0

(1 + τ2)1−(2n+2m+2)

4 dτ

>

∫ +∞

0

(1 + τ2)1−(2n+2m+2)

4 dτ −∫ +∞

1

τ (1 + τ2)1−(2n+2m+2)

4 dτ

= B(1

2,n+m

2− 1

4)− o

((n+m)−2

)>

1

2B(

1

2,n+m

2− 1

4), for all n+m � 1.

By Stirling’s formula, we obtain (4.1) with s(n) = 400√nfor m even immediately.

�

Remark 5.1. From the proof of Theorem 1.1, it is easy to see that (1.3) holds.

6. Proof of Theorem 1.1 for dCC metric

We will need two lemmas as follows.

Lemma 6.1. For all g = (x, t) ∈ H(2n,m), we have dK(g) ≤ dCC(g).

Proof. The proof follows from [10]. �

Lemma 6.2. There exists a constant c > 0 (independent of m and n) such that

|BCC(o, 1)| ≥ cm|BK(o, 1)|, for all n ∈ N∗.

Proof (essential due to [10]). Clearly we have

|BCC(o, 1)| =∫

θsin θ |x|<1,−π<θ<π,

4|t|=μ(θ)|x|2dxdt = Ωm

∫|x|<1

(μ(θ0)|x|2

4

)m

dx,

where θ0 ∈ [0, π) satisfies sin θ0θ0

= |x| and Ωm = 2πm2

mΓ(m2 ) . It follows that

|BCC(o, 1)| = Ωm

∫|x|<1

(2θ0 − sin 2θ0

8θ20

)m

dx.(6.1)

Notice that

|BK(o, 1)| = Ωm

∫|x|<1

(1− |x|4

16

)m2

dx(6.2)

= Ωm

∫|x|<1

1

4m

[1−

(sin θ0θ0

)4]m

2

dx.

Denote

c := infθ0∈[0,π)

2θ0 − sin(2θ0)

2θ20

(1−

(sin θ0θ0

)4)− 1

2

,

which is positive; see [10]. Combining with (6.1) and (6.2), we complete the proof.�

Remark 6.3. By certain properties of the Bessel functions, we can improve theconstant c.



Proof of Theorem 1.1 for M = MCC . For m fixed, the desired conclusion followsdirectly from the one with M = MK and the above two lemmas. �

7. Appendix: Proof of Lemma 4.3

Recall that 0 ≤ φ � 1, ω ≥ 0, ζ = nν0− 12 with ν0 ∈ (0, 12 ), which will be

determined later, and

U =ω4 sinh2 r

n2r2+

2ω2 sinh r cosh r cosφ

nr, sinϑ =

sinh r sinφ√cos2 φ+ sinh2 r + U

,

M12 =

∫ ζ

0

sinhm2 r

√sinh r

r Jm−22

((n+ m

2 ) sinϑ)

(cos2 φ+ sinh2 r + U)12 (n+

m2 + 1

2 )dr.

Let

Θ := (cos2 φ+ sinh2 r + U)− 1

2 (n+m2 + 1

2 )Jm−22

((n+

m

2) sinϑ

)(4.17)=

((n+

m

2) sinφ sinh r

)m−22


2 (n+m− 12 )

×(

(n+ m2 ) sinφ sinh r√


)−m−22

Jm−22

((n+ m

2 ) sinφ sinh r√cos2 φ+ sinh2 r + U

)

=:((n+

m

2) sinφ sinh r

)m−22


2 (n+m− 12 ) · I2.(7.1)

Applying the differential mean value theorem to the function z → z−νJν(z), onegets

I2 =

((n+ m

2 ) sinφ sinh r√cos2 φ+ U0

)−m−22

Jm−22

((n+ m

2 ) sinφ sinh r√cos2 φ+ U0

)+Υ,

where

Υ =d

dz

(z−

m−22 Jm−2

2(z)

)(ξ)(n+

m

2) sinφ sinh r

×(

1√cos2 φ+ U0

− 1√cos2 φ+ sinh2 r + U

)

(2.3)= −ξ · ξ−m

2 Jm2(ξ)(n+

m

2) sinφ sinh r

×(

1√cos2 φ+ U0

− 1√cos2 φ+ sinh2 r + U

)

for some ξ ∈(

(n+m2 ) sinφ sinh r√

cos2 φ+sinh2 r+U�

,(n+m

2 ) sinφ sinh r√cos2 φ+U0

).

Set

M121 = (cos2 φ+ U0)m−2

4

∫ +∞

0

um2 Jm−2

2

((n+m

2 ) sinφ√cos2 φ+U0

u)

(cos2 φ+ U0 + u2)12 (n+m− 1

2 )du

M122 = (cos2 φ+ U0)m−2

4

∫ ∞

sinh ζ

um2 Jm−2

2

((n+m


u)

(cos2 φ+ U0 + u2)12 (n+m− 1

2 )du,



E10 =

∫ ζ

0

(sinh r)m2

(√sinh r

r− cosh r

)·Θdr,

E11 =((n+

m

2) sinφ

)m−22

∫ ζ

0

(sinh r)m−1 cosh r

(cos2 φ+ sinh2 r + U)12 (n+m− 1

2 )Υ dr,

and

E12 = (cos2 φ+ U0)m−2

4

∫ ζ

0

(sinh r)m2 cosh rJm−2

2

((n+m


sinh r)

(cos2 φ+ sinh2 r + U0)12 (n+m− 1

2 )

×(e− 1

2 (n+m− 12 ) ln [1+

U�−U0cos2 φ+sinh2 r+U0

] − 1

)dr.

Observe that

M12 = (M121 −M122) + E10 + E11 + E12.(7.2)

We have the following (see for example [16], p. 105 or [6], p. 678):

M121 = (cos2 φ+ U0)m−2

4

((n+ m

2 ) sinφ

2√cos2 φ+ U0

) 12 (n+m− 5

2 )

× (cos2 φ+ U0)− 1

4 (n−12 )

Γ( 12 (n+m− 12 ))

K− 12 (n−

12 )

((n+

m

2) sinφ

)

= e−ω2

((n+ m

2 ) sinφ

2

) 12 (n+m− 5

2 ) (cosφ)−(n− 12 )

Γ( 12 (n+m− 12 ))

×K 12 (n−

12 )

((n+

m

2) sinφ

)[1 +O(

m2

n)

],(7.3)

where Kν(z) is modified Bessel function, and the last equality follows from thefollowing fact:

(cos2 φ+ U0

)− 12 (n−

12 ) = (cosφ)−(n− 1

2 )

(1 +

ω2

n cosφ

)−(n− 12 )

= e−ω2

(cosφ)−(n− 12 )

[1 +O(

m2

n)

].

Lemma 7.1. For m3 � n,

M121 ≥ 1

9e−ω2

(1

2(n+

m

2) sinφ

)m−22 Γ

(12 (n− 1

2 ))

Γ( 12 (n+m− 12 ))

.(7.4)



Proof. Denote ν = 12 (n− 1

2 ), z = (n+ m2 ) sinφ. Recall that

(cosφ)−(n− 12 )K 1

2 (n−12 )

((n+

m

2) sinφ

)

= exp

{−(n− 1

2 )

2ln(1− sin2 φ)

}∫ ∞

0

e−z2 (u+

1u )uν−1du

≥ 1

3e

n2 sin2 φ

∫ ∞

0

e−z2 (u+

1u )uν−1du

≥ 1

3e

n2 sin2 φ− z2

4ν

∫ ∞

(2ν)/z

e−zu2 uν−1du

=1

3e−

sin2 φ2 O(m)

(2

z

)ν ∫ ∞

ν

e−uuν−1du

=1

3

(1 +O(

m3

n)

)(2

z

)ν

Γ(ν, ν),(7.5)

where Γ(ν, ν) is an incomplete gamma function. Recall that (see [16], p. 340, or [6],p. 901)

Γ(ν + 1, ν) = νΓ(ν, ν) + ννe−ν ,

and by the following asymptotic estimate for Γ(ν + 1, ν) (see (1.4) in [21]):

Γ(ν + 1, ν) = Γ(ν + 1)

(1

2+

1

3

(2

νπ

) 12

)+O(

1

ν), ν → ∞,

it follows that

Γ(ν, ν) ≥ ν−1

(Γ(ν + 1)

2− ννe−ν

)=

Γ(ν)

2− νν−1e−ν

≥ 2Γ(ν)

5,(7.6)

where the last inequality follows from (2.7).Combining with (7.5), (7.6), one has, for m3 � n,

(7.7) (cosφ)−θK 12 θ

((n+

m

2) sinφ

)≥ 2

15

(2

(n+ m2 ) sinφ)

) 12 θ

Γ(1

2θ),

where θ = n− 12 . The desired result follows easily. �

Lemma 7.2. There exists some positive constant C(ω, ν0),

|E10|+ |E11|+ |E12| = M121O(mn

exp{m2C(ω, ν0)

}).

Proof. Since the proof is similar for E10, E11, E12, we only prove

E10 = M121O(mn

exp{m2C(ω, ν0)

}).

Notice that ∣∣∣∣∣√

sinh r

r− cosh r

∣∣∣∣∣ ≤ c sinh2 r, ∀0 < r < 1.



(2.2) implies

|E10| ≤1

Γ(m2 )

((n+ m

2 ) sinφ

2

)m−22

∫ ζ

0

(sinh r)m+1

(cos2 φ+ U0 + sinh2 r)12 (n+m− 1

2 )dr

≤ 1

Γ(m2 )

((n+ m

2 ) sinφ

2

)m−22

∫ ∞

0

(sinh r)m+1

(cos2 φ+ U0 + sinh2 r)12 (n+m− 1

2 )d sinh r

=1

2Γ(m2 )

((n+ m

2 ) sinφ

2

)m−22

(cos2 φ+ U0)− 1

2 (n−52 )B

(m+ 2

2,1

2(n− 5

2)

)

≤ m

4

((n+ m

2 ) sinφ

2

)m−22

(cos2 φ+ U0)− 1

2 (n−52 )

Γ( 12 (n− 52 ))

Γ( 12 (n+m− 12 ))

.

(7.8)

Combining this with (7.4), (2.7) and the fact that√nφ ≤ C∗m, we get the

desired result. �

Now let us turn to

Proof of Lemma 4.3. By (2.2), one has

M122 ≤ 1

Γ(m2 )

((n+ m

2 ) sinφ

2

)m−22

∫ ∞

sinh ζ

um−1(cos2 φ+ U0 + u2)−12 (n+m− 1

2 )du

≤ 1

2Γ(m2 )

((n+ m

2 ) sinφ

2

)m−22

∫ ∞

sinh ζ

(cos2 φ+ U0 + u2)−12 (n+

32 )du2

=1

Γ(m2 )

((n+ m

2 ) sinφ

2

)m−22 1

n− 12

(cos2 φ+ U0 + sinh2 ζ)−12 (n−

12 ).

By (7.4) and (4.31), it is easy to see that

M122 = M121O(c(ω,C∗, νo)n

−2).(7.9)

Combining this with (7.2), (7.9), and the above two lemmas, (4.27) holds.Now let us proof (4.28). Obverse that

|M11| ≤ cm

Γ(m−12 )

(sinφ)m2 +1

∫ ζ

0

r−12 (sinh r)m+ 3

2

(cos2 φ+ sinh2 r + U)12 (n+m+ 3

2 )dr

≤ cm

Γ(m−12 )

(sinφ)m2 +1

∫ +∞

0

(sinh r)m+1

(cos2 φ+ sinh2 r + U0)12 (n+m+ 3

2 )d(sinh r)

= cm

2Γ(m−12 )

(sinφ)m2 +1(cos2 φ+ U0)

− 12 (n−

12 )B

(m

2+ 1,

n

2− 1

4

).

Now

B

(m

2+ 1,

n

2− 1

4

)=

Γ(m2 + 1)Γ(n2 − 14 )

Γ(n2 + m2 + 3

4 )=

2Γ(m2 + 1)

n+m− 12

Γ( 12 (n− 12 ))

Γ( 12 (n+m− 12 ))

.

(7.4) implies (4.28). �



Acknowledgement

This work was partially supported by the NSF of China 11171070 and 11201040,NCET-09-0316 and “The Program for Professor of Special Appointment (EasternScholar) at Shanghai Institutions of Higher Learning”. Part of this work was ob-tained in February 2009 during the first author’s visit to Toulouse University. Hewould like to express his sincere thanks to Professor D. Bakry for the kind invita-tion.

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School of Mathematical Sciences, Fudan University, 220 Handan Road, Shanghai

200433, People’s Republic of China

E-mail address: hongquan [email protected]

E-mail address: hong [email protected]

School of Mathematics and Statistics, Changshu Institute of Technology 215500,

Changshu, People’s Republic of China

E-mail address: [email protected] address: [email protected]


















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Centered Hardy-Littlewood maximal functions on Heisenberg type … · 2018-11-16 · TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 366, Number 3, March 2014, Pages 1497–1524