PANEITZ OPERATORS ON HYPERBOLIC SPACES …PANEITZ OPERATORS ON HYPERBOLIC SPACES AND HIGH ORDER HARDY-SOBOLEV-MAZ’YA INEQUALITIES ON HALF SPACES GUOZHEN LU AND QIAOHUA YANG Abstract

PANEITZ OPERATORS ON HYPERBOLIC SPACES AND HIGH ORDERHARDY-SOBOLEV-MAZ’YA INEQUALITIES ON HALF SPACES

GUOZHEN LU AND QIAOHUA YANG

Abstract. Though there has been extensive study on Hardy-Sobolev-Maz’ya inequalitieson upper half spaces for first order derivatives, whether an analogous inequality for higherorder derivatives holds has still remained open. By using, among other things, the Fourieranalysis techniques on the hyperbolic space which is a non-compact complete Riemannianmanifold, we establish the Hardy-Sobolev-Maz’ya inequalities for higher order derivativeson half spaces. Moreover, we derive sharp Poincare-Sobolev inequalities (namely, Sobolevinequalities with a substraction of a Hardy term) for the Paneitz operators on hyperbolicspaces which are of their independent interests and useful in establishing the sharp Hardy-Sobolev-Maz’ya inequalities. Our sharp Poincare-Sobolev inequalities for the Paneitz oper-ators on hyperbolic spaces improve substantially those Sobolev inequalities in the literature.The proof of such Poincare-Sobolev inequalities relies on hard analysis of Green’s functionsestimates, Fourier analysis on hyperbolic spaces together with Beckner’s Hardy-Littlewood-Sobolev inequality on hyperbolic spaces. Finally, we show the sharp constant in the Hardy-Sobolev-Maz’ya inequality for the bi-Laplacian in the upper half space of dimension fivecoincides with the best Sobolev constant. This is an analogous result to that of Benguria,Frank and Loss’ work concerning the sharp constant in the first order Hardy-Sobolev-Maz’yainequality in the three dimensional upper half space.

1. Introduction

Let n ≥ 3, 2 < p ≤ 2nn−2

and γ = (n−2)p2− n. The Hardy-Sobolev-Maz’ya inequality on the

half space Rn+ = {(x1, · · · , xn) : x1 > 0} reads as follows (see [50], Section 2.1.6):

(1.1)

∫Rn+|∇u|2dx− 1

4

∫Rn+

u2

x21

dx ≥ C

(∫Rn+xγ1 |u|pdx

) 2p

, u ∈ C∞0 (Rn+),

2000 Mathematics Subject Classification. Primary 35J20; 46E35.Key words and phrases. Hardy inequalities; Adams’ inequalities; Hyperbolic spaces; Hardy-Sobolev-

Maz’ya inequalities, Paneitz and GJMS operators, Hardy-Littlewood-Sobolev inequality on hyperbolicspaces, Fourier transforms on hyperbolic spaces.

The first author was partly supported by a US NSF grant and a grant from the Simons Foundation andthe second author’s research was partly supported by the National Natural Science Foundation of China(No.11201346).

1

2 GUOZHEN LU AND QIAOHUA YANG

where C is a positive constant which is independent of u. In particular, for γ = 0 andp = 2n

n−2we have

(1.2)

∫Rn+|∇u|2dx− 1

4

∫Rn+

u2

x21

dx ≥ Cn

(∫Rn+|u|

2nn−2dx

)n−22

, u ∈ C∞0 (Rn+).

It has been shown by R. D. Benguria, R. L. Frank and M. Loss ([10]) that the sharp constantC3 in (1.2) for n = 3 coincides with the corresponding best Sobolev constant. In the paper[47], G. Mancini and K. Sandeep showed inequality (1.1) is equivalent to the followingPoincare-Sobolev inequalities on hyperbolic space Hn (n ≥ 3):

(1.3)

∫Hn|∇Hu|2dV −

(n− 1)2

4

∫Hnu2dV ≥ C

(∫Hn|u|pdV

) 2p

, u ∈ C∞0 (Hn),

where 2 < p ≤ 2nn−2

, ∇H is the hyperbolic gradient and dV is the hyperbolic volume element.We refer to [11, 12] for Poincare-Hardy inequalities on Hn (n ≥ 3). In the case n = 2,Beckner [7] obtained a sharp Poincare-Sobolev inequalities on the two dimensional hyperbolicspace H2. Furthermore, there holds some first order Trudinger-Moser type inequalities onhyperbolic spaces [42, 43] and Hardy-Trudinger-Moser inequality on H2 (see [48, 44, 57]) andHardy-Adams inequality on H4 and Hn for all even dimensions n ≥ 4 ([45, 37]). We note thatour recent work [45, 37] on Hardy-Adams inequality on four and higher even dimensionalhyperbolic spaces can be regarded as a limiting case of our higher order Hardy-Sobolev-Maz’ya inequality in this paper in even dimensions. More recently, J. Li and the authorshave succeeded in establishing sharp Adams and Hardy-Adams inequalities of any fractionalorder in all dimensions [38]. For more information about first order Hardy-Sobolev-Maz’yain equalities, we refer to [17, 18, 19, 51, 56].

Though there has been an extensive study on first order Hardy-Sobolev-Maz’ya inequal-ities (1.1) and (1.3) in the past decades, whether an analogous inequality for higher orderderivatives holds has still remained open. In this paper we shall show that high order Hardy-Sobolev-Maz’ya inequalities indeed hold.

To state our results, let us introduce some conventions. It is known that hyperbolic spacehas its the Poincare ball model and the Poincare half space model and both models areequivalent. We denote by Bn the Poincare ball model. It is the unit ball

Bn = {x = (x1, · · · , xn) ∈ Rn||x| < 1}

equipped with the usual Poincare metric

ds2 =4(dx2

1 + · · ·+ dx2n)

(1− |x|2)2.

The hyperbolic volume element is dV =(

21−|x|2

)ndx and the distance from the origin to

x ∈ Bn is ρ(x) = log 1+|x|1−|x| . Then Bn is a complete non-compact Riemannian manifold of

PANEITZ OPERATORS ON HYPERBOLIC SPACES AND HIGH ORDER HSM INEQUALITIES 3

constant negative curvature. The associated Laplace-Beltrami operator is given by

∆H =1− |x|2

4

{(1− |x|2)

n∑i=1

∂2

∂x2i

+ 2(n− 2)n∑i=1

xi∂

∂xi

}.

The spectral gap of −∆H on L2(Bn) is (n−1)2

4(see e.g. [47]), i.e.

(1.4)

∫Bn|∇Hu|2dV ≥

(n− 1)2

4

∫Bnu2dV, u ∈ C∞0 (Bn).

The the GJMS operators on Bn is defined as follows (see [24], [32])

(1.5) Pk = P1(P1 + 2) · · · · · (P1 + k(k − 1)), k ∈ N,

where P1 = −∆H− n(n−2)4

is the conformal Laplacian on Bn. The sharp Sobolev inequalitieson Bn read as follows (see [29] for k = 1 and [5] and [41] for 2 ≤ k < n

2):

(1.6)

∫Bn

(Pku)udV ≥ Sn,k

(∫Bn|u|

2nn−2kdV

)n−2kn

, u ∈ C∞0 (Bn), 1 ≤ k <n

2,

where Sn,k is the best k-th order Sobolev constant. We remark that the above sharp Sobolevinequality on hyperbolic spaces (1.6) actually follows from the Sharp Sobolev inequality onspheres due to Beckner [5] (see Section 7 for another proof). We also refer the reader tothe recent works on Sobolev type inequalities for Paneitz operators on compact Riemannianmanifolds by Hang and Yang ([26, 27, 28]). On the other hand, by (1.4) and (1.5), we havethe following Poincare inequality

(1.7)

∫Bn

(Pku)udV ≥k∏i=1

(2i− 1)2

4


As we will show in the proof of Theorem 1.3 (see Section 5), inequality (1.7) is equivalent tothe Hardy inequality on the upper half space∫

Rn+|∇ku|2dx ≥

k∏i=1

(2i− 1)2

4

∫Rn+

u2

x2k1

dx, u ∈ C∞0 (Rn+),

and the constantk∏i=1

(2i−1)2

4is sharp (see [52]).

Next we define another 2k-th order operator Qk with k ≥ 2:

Qk =

(−∆H −

(n− 1)2

4

)(P1 + 2) · · · · · (P1 + k(k − 1))

=Pk −1

4(P1 + 2) · · · · · (P1 + k(k − 1)).

(1.8)

To this end, we have the following Sobolev type inequalities for Qk.


Theorem 1.1. Let 2 ≤ k < n2

and 2 < p ≤ 2nn−2k

. There exists a positive constant C suchthat for each u ∈ C∞0 (Bn),

(1.9)

∫Bn

(Qku)udV ≥ C

(∫Bn|u|pdV

) 2p

.

Notice that, by (1.4),

(1.10)

∫Bn

((P1 + 2) · · · · · (P1 + k(k − 1))u)udV ≥k∏i=2

(2i− 1)2

4


Combining (1.10), (1.7) and (1.21) yields∫Bn

(Pku)udV −k∏i=1

(2i− 1)2

4

∫Bnu2dV ≥

∫Bn

(Qku)udV.

We note that the proof of Theorem 1.1 uses the Hardy-Littlewood-Sobolev inequality onhyperbolic spaces (see Theorem 4.1). However, Theorem 4.1 alone is not sufficient to establishthe above sharp Sobolev inequality on hyperbolic spaces (i.e., Theorem 1.1). More delicateGreen’s function estimates for the kernels of powers of fractional Laplacians and Fourieranalysis on hyperbolic spaces are required. Sections 3 and 4 devote to such estimates andanalysis.

As an application of Theorem 1.1, we have the following Poincare-Sobolev inequalities forhigher order derivatives.


and 2 < p ≤ 2nn−2k

. There exists a positive constant C =C(n, p) such that for each u ∈ C∞0 (Bn),

(1.11)

∫Bn

(Pku)udV −k∏i=1

(2i− 1)2

4

∫Bnu2dV ≥ C

(∫Bn|u|pdV

) 2p

.

This improves substantially the Poincare-Sobolev inequalities (1.6) for higher order deriva-tives on the hyperbolic spaces Bn established by Beckner [5] and Liu [41].

If p = 2nn−2k

, then by (1.6), the sharp constant in (1.11) is less than or equal to the bestk-th order Sobolev constant.

As an application of Theorem 1.2, we have the following Hardy-Sobolev-Maz’ya inequali-ties for higher order derivatives:


and 2 < p ≤ 2nn−2k

. There exists a positive constant C suchthat for each u ∈ C∞0 (Rn

+),

(1.12)

∫Rn+|∇ku|2dx−

k∏i=1

(2i− 1)2

4

∫Rn+

u2

x2k1

dx ≥ C

(∫Rn+xγ1 |u|pdx

) 2p

,

where γ = (n−2k)p2− n.


In terms of the Poincare ball model Bn, inequality (1.11) can be written as follows:

(1.13)

∫Bn|∇ku|2dx−

k∏i=1

(2i− 1)2

∫Bn

u2

(1− |x|2)2kdx ≥ C

(∫Bn

(1− |x|2)γ|u|pdx) 2

p

.

In some special cases, we can compute the best constant of Poincare-Sobolev inequalities offractional order on hyperbolic spaces. The first one is on the hyperbolic spaces of dimension3 where we have the following:

Theorem 1.4. Let 12≤ s < 3

2. Then∫

B3

|(−1−∆H)s2u|2dV ≥ S3,s

(∫B3

|u|6

3−2sdV

) 3−2s3

, u ∈ C∞0 (Bn),

where S3,s = 22sπsΓ( 3+2s

2)

Γ( 3−2s2

)

(Γ( 3

2)

Γ(3)

) 2s3

is the best Sobolev constant of order s in dimension 3.

Choosing s = 1 in Theorem 1.4, we recover the following sharp Poincare-Sobolev inequalitywhich was first proved by Benguria, Frank and Loss [10]:∫

B3

|∇Hu|2dV −∫B3

|u|2dV ≥ S3,1

(∫B3

|u|6dV) 1

3

.(1.14)

Though it is not known in general yet whether the constant C on the right hand sides of theinequalities (1.11), (1.12) and (1.13) in Theorems 1.2 and 1.4 respectively is the same as thebest Sobolev constant Sn,k, we will show in this paper in the case of n = 5 and k = 2 thatthe best constant C of Poincare-Sobolev inequality, as well as the Hardy-Sobolev-Maz’yainequality in (1.11), (1.12) and (1.13) coincides with the best Sobolev constant Sn,k. In fact,we have the following:

Theorem 1.5. There holds, for each u ∈ C∞0 (B5),

(1.15)

∫B5

(P2u)udV − 9

16

∫B5

u2dV ≥ S5,2

(∫Bn|u|10dV

) 15

.

In terms of the Poincare ball model B5 and the Poincare half space model H5, respectively,inequality (1.15) is equivalent to the follows:∫

B5

|∆f |2dx− 9

∫B5

f 2

(1− |x|2)4dx ≥ S5,2

(∫B5

|f |10dx

) 15

, f ∈ C∞0 (B5);

∫R5+

|∆g|2dx− 9

16

∫R5+

g2

x41

dx ≥ S5,2

(∫R5+

|g|10dx

) 15

, g ∈ C∞0 (R5+).

However, it seems that the best constant for the sharp Poincare-Sobolev inequality or theHardy-Sobolev-Maz’ya inequality does not coincide with the best Sobolev constant for n = 6and k = 2 (see Remark 8.3). Hong has informed us that in a recent work [33] she proved thatthe sharp constant of Poincare-Sobolev inequality and the Hardy-Sobolev-Maz’ya inequality


for n = 7 and k = 3 also coincides with the Sobolev constant S7,3, using the same argu-ment as we do in Section 8 of our current paper. We remark that we can continue to showthat such constants coincide in higher dimension n by proving it one by one for each givendimension n. However, when n and k are larger, the argument and computation becomesincreasingly more difficult. We do not have a uniform proof which covers all the dimensionsn and the derivatives of all order k. Nevertheless, through some preliminary work we havebeen able to make the following conjecture.

Conjecture 1.6. Let n ≥ 9 be odd. There holds, for each u ∈ C∞0 (Bn),∫Bn

(P(n−1)/2u)udV −(n−1)/2∏i=1

(2i− 1)2

4

∫Bnu2dV ≥ Sn,(n−1)/2

(∫Bn|u|2ndV

) 1n

.

In terms of the Poincare ball model and the Poincare half space model respectively, theinequality above is equivalent to the following:∫Bn|∇

n−12 f |2dx−

(n−1)/2∏i=1

(2i−1)2

∫Bn

f 2

(1− |x|2)n−1dx ≥ Sn,(n−1)/2

(∫Bn|f |2ndx

) 1n

, f ∈ C∞0 (Bn);

∫Rn+|∇

n−12 g|2dx−

(n−1)/2∏i=1

(2i− 1)2

4

∫Rn+

g2

xn−11

dx ≥ Sn,(n−1)/2

(∫Rn+|g|2ndx

) 1n

, g ∈ C∞0 (Rn+).

We end this introduction with the following remark.The Hardy-Sobolev-Mazya inequality

(1.16)

∫Bn

(Pku)udV −k∏i=1

(2i− 1)2

4

∫Bnu2dV ≥ C

(∫Bn|u|pdV

) 2p

, 2 < p ≤ 2n

n− 2k,

corresponds to the Euler-Lagrange equation:

(1.17) Pku−k∏i=1

(2i− 1)2

4u = µup−1,

for a constant µ > 0. In fact,

µ =

∫Bn(Pku)udV −

k∏i=1

(2i−1)2

4

∫Bn u

2dV∫Bn |u|pdV

> 0.

Note the above equation is equivalent to

(1.18) Pkf −k∏i=1

(2i− 1)2

4f = fp−1.


By considering the Poincare half space model of the hyperbolic space and noting the identity(see Lemma 5.1)

(1.19) xn2

+k

1 (−∆)k ◦ xk−n2

1 = Pk,

we can see that the above equation can be converted to

(1.20) (−∆)kv −k∏i=1

(2i− 1)2

4

v

x2k1

= µvp

xγ1, γ =

(n− 2k)p

2− n.

Using an appropriate transform we can convert µ in the above equation (1.20) to 1, and thenconvert (1.17) to (1.18).

Therefore, we can consider the following higher order Brezis-Nirenberg problem [13] onthe hyperbolic spaces by studying the existence of positive solutions to the following higherorder equation

Pku− λu = up−1, λ ≤k∏i=1

(2i− 1)2

4, 2 < p ≤ 2n

n− 2k.

When k = 1, the above equation reduces to the one considered in [47]:

−∆Hu−(n(n− 2)

4+ λ

)u = up−1, λ ≤ 1

4, 2 < p ≤ 2n

n− 2.

We note the higher order Brezis-Nirenberg problem in Euclidean spaces has been consideredby numerous authors (see e.g. [21, 53] and references therein).

We will return to the study of the above problem in the future.After the paper was accepted for publication, the authors have confirmed the above con-

jecture 1.6. Furthermore, the authors have proved in [46] the following Poincare-Sobolevinequalities on Bn, as well as the Hardy-Sobolev-Maz’ya inequalities on half spaces.

Theorem 1.7. Let 2 ≤ k < n2. Suppose that there exists λ ∈ R such that for any u ∈

C∞0 (Bn), ∫Bn

(Pku)udV + λ

∫Bnu2dV ≥ Sn,k

(∫Bn|u|

2nn−2kdV

)n−2kn

.

If n ≥ 4k, then λ ≥ 0. If 2k + 2 ≤ n < 4k, then

λ ≥ −Γ(n/2)Γ(k)

k−1∑j=0

Γ(j+n−2k2

)

Γ(j+1)Γ(n−2k2

)

2n+2k

2 Γ(n−2k2

)∫ 1

0[r2k−n −

k−1∑j=0

Γ(j+n−2k2

)

Γ(j+1)Γ(n−2k2

)(1− r2)j]2 rn−1dr

(1−r2)2k

.(1.21)

In the case of P1 and n ≥ 4, the above Theorem 1.7 is in the spirit of the following resultdue to E. Hebey [29].


Theorem 1.8. ([29]) Let n ≥ 4. Suppose that there exists λ ∈ R such that for any u ∈C∞0 (Bn), ∫

Bn(P1u)udV + λ

∫Bnu2dV ≥ Sn,1

(∫Bn|u|

2nn−2dV

)n−2n

,

then λ ≥ 0.

The organization of this paper is as follows. In Section 2, we recall some necessary prelim-inary facts of hyperbolic spaces. Sharp estimates of Green’s functions of kernels of certainfractional Laplacians on hyperbolic spaces are given in Section 3. We shall prove Theorem1.1 and Theorem 1.2 in Section 4. The higher order Hardy-Sobolev-Maz’ya inequalities inall dimension n and all derivatives of order k, namely Theorem 1.3, are proved in Section 5.In Section 6, we prove the sharp constant in Hardy-Sobolev-Maz’ya inequality of fractionalorder s in case n = 3, namely Theorem 1.4. In Section 7, we show that, as in the caseof Euclidean space, the Hardy-Littlewood-Sobolev inequlity on hyperbolic space implies thesharp Sobolev inequalities on hyperbolic spaces. This latter result follows from the sharpSobolev inequality on the sphere obtained by Beckner [5] (see also [41]). In the last section,we prove that the sharp constant for the second order Hardy-Sobolev-Maz’ya inequality indimension 5 is the same as the best Sobolev constant S5,2, namely Theorem 1.5.

Acknowledgement: The authors wish to thank the referees for their very helpful com-ments and suggestions which improve the exposition of the paper. Indeed, thanks to theinsightful comments by the referees, we have discussed in more details about the validity ofTheorem 1.5 in general dimension n and include Conjecture 1.6. We have also included adiscussion about the relevance of our sharp higher order Hardy-Sobolev-Maz’ya inequalitieson hyperbolic spaces to the higher order analogue of the Brezis-Nirenberg problem for thehigher order PDEs. The authors also wish to thank W. Beckner for his interest in our workand his encouragement.

2. Notations and preliminaries

We begin by quoting some preliminary facts which will be needed in the sequel and referto [1, 22, 30, 31, 34, 40] for more information about this subject.

It is well known that hyperbolic space Bn is a noncompact Riemannian symmetric spaceof rank one that has a constant negative curvature −1. In fact, Bn is isomorphic toSO0(n, 1)/SO(n) and in particular, B2 is also isomorphic to SL(2,R)/SO(2), in which sev-eral sharp geometric inequalities have been established by Beckner [6, 7].

There are several models of hyperbolic space, for example, the Poincare half space modeland the Poincare ball model.

2.1. The Poincare half space model Hn. It is given by R+ × Rn−1 = {(x1, · · · , xn) :

x1 > 0} equipped with the Riemannian metric ds2 =dx21+···+dx2n

x21. The induced Riemannian

measure can be written as dV = dxxn1

, where dx is the Lebesgue measure on Rn. The hyperbolic


gradient is ∇H = x1∇ and the Laplace-Beltrami operator on Hn is given by

(2.1) ∆H = x21∆− (n− 2)x1

∂

∂x1

,

where ∆ =∑n

i=1∂2

∂x2iis the Laplace operator on Rn.

2.2. The Poincare ball model Bn. It is given by the unit ball

Bn = {x = (x1, · · · , xn) ∈ Rn||x| < 1}

equipped with the usual Poincare metric

ds2 =4(dx2

1 + · · ·+ dx2n)

(1− |x|2)2.

The hyperbolic gradient is ∇H = 1−|x|22∇ and the Laplace-Beltrami operator is given by

∆H =1− |x|2

4

{(1− |x|2)

n∑i=1

∂2

∂x2i

+ 2(n− 2)n∑i=1

xi∂

∂xi

}.

2.3. Mobius transformations. For each a ∈ Bn, we define the Mobius transformations Taby (see e.g. [1, 34])

Ta(x) =|x− a|2a− (1− |a|2)(x− a)

1− 2x · a+ |x|2|a|2,

where x ·a = x1a1 +x2a2 + · · ·+xnan denotes the scalar product in Rn. It is known that themeasure on Bn is invariant with respect to the Mobius transformations. A simple calculationshows

Ta(Ta(x)) =x;

1− |Ta(x)|2 =(1− |a|2)(1− |x|2)

1− 2x · a+ |x|2|a|2;

|Ta(x)| = |x− a|√1− 2x · a+ |x|2|a|2

;

sinhρ(Ta(x))

2=

|Ta(x)|√1− |Ta(x)|2

=|x− a|√

(1− |a|2)(1− |x|2);

coshρ(Ta(x))

2=

1√1− |Ta(x)|2

=

√1− 2x · a+ |x|2|a|2√(1− |a|2)(1− |x|2)

.

(2.2)

Using the Mobius transformations, we can define the distance from x to y in Bn as follows

ρ(x, y) = ρ(Tx(y)) = ρ(Ty(x)) = log1 + |Ty(x)|1− |Ty(x)|

.


Also using the Mobius transformations, we can define the convolution of measurable functionsf and g on Bn by (see e.g. [40])

(2.3) (f ∗ g)(x) =

∫Bnf(y)g(Tx(y))dV (y)

provided this integral exists. It is easy to check that

f ∗ g = g ∗ f.Furthermore, if g is radial, i.e. g = g(ρ), then (see e.g. [40], Proposition 3.15)

(2.4) (f ∗ g) ∗ h = f ∗ (g ∗ h)

provided f, g, h ∈ L1(Bn)

2.4. Fourier transform on hyperbolic spaces. In this subsection we recall some basicsof Fourier analysis on hyperbolic spaces and refer the reader to [20, 30, 31, 35, 55] for moreinformation about Fourier analysis on Riemannian symmetric spaces of noncompact type.

Set

eλ,ζ(x) =

(√1− |x|2|x− ζ|

)n−1+iλ

, x ∈ Bn, λ ∈ R, ζ ∈ Sn−1.

The Fourier transform of a function f on Bn can be defined as

f(λ, ζ) =

∫Bnf(x)e−λ,ζ(x)dV

provided this integral exists. If g ∈ C∞0 (Bn) is radial, then

(f ∗ g) = f · g.Moreover, the following inversion formula holds for f ∈ C∞0 (Bn) (see e.g. [40]):

f(x) = Dn

∫ +∞

−∞

∫Sn−1

f(λ, ζ)eλ,ζ(x)|c(λ)|−2dλdσ(ς),

where Dn = 123−nπ|Sn−1| and c(λ) is the Harish-Chandra c-function given by (see e.g. [40])

c(λ) =2n−1−iλΓ(n/2)Γ(iλ)

Γ(n−1+iλ2

)Γ(1+iλ2

).

Similarly, there holds the Plancherel formula:

(2.5)

∫Bn|f(x)|2dV = Dn

∫ +∞

−∞

∫Sn−1

|f(λ, ζ)|2|c(λ)|−2dλdσ(ς).

Since eλ,ζ(x) is an eigenfunction of ∆H with eigenvalue − (n−1)2+λ2

4, it is easy to check that,

for f ∈ C∞0 (Bn),

∆Hf(λ, ζ) = −(n− 1)2 + λ2

4f(λ, ζ).


Therefore, in analogy with the Euclidean setting, we define the fractional Laplacian onhyperbolic space as follows:

(2.6) (−∆H)γf(λ, ζ) =

((n− 1)2 + λ2

4

)γf(λ, ζ), γ ∈ R.

For more information about fractional Laplacian on hyperbolic space, we refer to [2, 4].

3. Sharp Estimates of Green’s functions

In what follows, a . b will stand for a ≤ Cb and a ∼ b will stand for C−1b ≤ a ≤ Cb witha positive constant C.

Let n ≥ 2. Denote by et∆H the heat kernel on Bn. It is well known that et∆H depends onlyon t and ρ(x, y). In fact, et∆H is given explicitly by the following formulas (see e.g. [15, 25]):

• If n = 2m, then

et∆H =(2π)−n+12 t−

12 e−

(n−1)2

4t

∫ +∞

ρ

sinh r√cosh r − cosh ρ

(− 1

sinh r

∂

∂r

)me−

r2

4t dr;(3.1)

• If n = 2m+ 1, then

et∆H =2−m−1π−m−1/2t−12 e−

(n−1)2

4t

(− 1

sinh ρ

∂

∂ρ

)me−

ρ2

4t .(3.2)

An explicit expression of Green’s function (−∆H + λ)−1 with λ > − (n−1)2

4is given by (see

[36, 49])

(λ−∆H)−1 =(2π)−n2 (sinh ρ)−

n−22 e−

(n−2)π2

iQn−22

θn(λ)(cosh ρ), n ≥ 3,(3.3)

where

θn(λ) =

√λ+

(n− 1)2

4− 1

2

and Qn−22

θn(λ)(cosh ρ) is the Legendre function of second type defined by (see [16], page 155)

Qµν (z) =ei(πµ)2−ν−1 Γ(ν + µ+ 1)

Γ(ν + 1)(z2 − 1)−µ/2

∫ π

0

(z + cos t)µ−ν−1(sin t)2ν+1dt,

Reν > −1, Re(ν + µ+ 1) > 0.

(3.4)

Therefore, for n ≥ 3,

(λ−∆H)−1 =An

(sinh ρ)n−2

∫ π

0

(cosh ρ+ cos t)n−42−θn(λ)(sin t)2θn(λ)+1dt,(3.5)

where

An = (2π)−n2

Γ(n2

+ θn(λ))

2θn(λ)+1Γ(θn(λ) + 1).


Lemma 3.1. Let n ≥ 3. There holds, for λ > − (n−1)2

4and ρ > 0,

(λ−∆H)−1 .

(1

sinh ρ2

)n−2(1

cosh ρ2

)1+2

√λ+

(n−1)2

4

.(3.6)

Proof. Set, for 0 ≤ α < 1,

f(α) =

∫ π

0

(1 + α cos t)n−42−θn(λ)(sin t)2θn(λ)+1dt.

Then f is a continuous function on [0, 1) and

limα→1−

f(α) =

∫ π

0

(1 + cos t)n−42−θn(λ)(sin t)2θn(λ)+1dt

=

∫ π

0

(2 cos2 t

2

)n−42−θn(λ)(

2 sint

2cos

t

2

)2θn(λ)+1

dt

=2n−22

+θn(λ)

∫ π

0

(cos

t

2

)n−3(sin

t

2

)2θn(λ)+1

dt

=2n−22

+θn(λ)

∫ π

0

(cos

t

2

)n−3(sin

t

2

)2

√λ+

(n−1)2

4

dt

≤2n−22

+θn(λ)π.

Therefore, there exists a constant C > 0 such that for each α ∈ [0, 1), |f(α)| = f(α) ≤ C.Thus, by (3.5),

(λ−∆H)−1 =An(cosh ρ)

n−42−θn(λ)

(sinh ρ)n−2f

(1

cosh ρ

).

(cosh ρ)n−42−θn(λ)

(sinh ρ)n−2=

(cosh ρ)n−42−θn(λ)

(2 sinh ρ2

cosh ρ2)n−2

∼(

1

sinh ρ2

)n−2(1

cosh ρ2

)2+2θn(λ)

=

(1

sinh ρ2

)n−2(1

cosh ρ2

)1+2

√λ+

(n−1)2

4

.

To get the last inequality, we use the fact cosh ρ ∼ cosh2 ρ2, ρ ≥ 0. �

Next we shall give the estimates of limiting case of Green’s function, namely λ = (n−1)2

4.

We compute, by (3.1) and (3.2),


• If n = 2m, then

(−∆H −

(n− 1)2

4

)−1

=

∫ ∞0

et(∆H+(n−1)2

4)dt

=1

2(2π)n+12

∫ +∞

ρ


(− 1

sinh r

∂

∂r

)m−1(r

sinh r

∫ ∞0

t−32 e−

r2

4t dt

)dr

=

√π

(2π)n+12

∫ +∞

ρ


(− 1

sinh r

∂

∂r

)m−11

sinh rdr

(3.7)

To get the last equation, we use the fact

(3.8)

∫ ∞0

t−32 e−

r2

4t dt =2

r

∫ ∞0

t−32 e−

1t dt =

2

r

∫ ∞0

t−12 e−tdt =

2

rΓ(1/2) =

2√π

r.

• If n = 2m+ 1, then

(−∆H −

(n− 1)2

4

)−1

=

∫ ∞0

et(∆H+(n−1)2

4)dt

=2−m−2π−m−1/2

(− 1

sinh ρ

∂

∂ρ

)m−1(ρ

sinh ρ

∫ ∞0

t−32 e−

ρ2

4t dt

)=2−m−1π−m

(− 1

sinh ρ

∂

∂ρ

)m−11

sinh ρ.

(3.9)

To get the last equation, we also use (3.8).

Lemma 3.2. Let k be a nonnegative integer. Then there exist constants {ai}ki=0 such that

(3.10)

(− 1

sinh ρ

∂

∂ρ

)2k1

sinh ρ=

k∑i=0

ai

(1

sinh ρ

)2i+2k+1

.

Moreover, a0 = (2k)!.

Proof. We shall prove by induction. It is easy to see that (3.10) is valid for k = 0. Nowsuppose that equation (3.10) is valid for k = l, i.e.

(− 1

sinh ρ

∂

∂ρ

)2l1

sinh ρ=

l∑i=0

ai

(1

sinh ρ

)2i+2l+1


and a0 = (2l)!. Then(− 1

sinh ρ

∂

∂ρ

)2l+21

sinh ρ=

(− 1

sinh ρ

∂

∂ρ

)2 l∑i=0

ai

(1

sinh ρ

)2i+2l+1

=l∑

i=0

(2i+ 2l + 1)ai

[−(

1

sinh ρ

)2i+2l+3

+ (2i+ 2l + 3)

(1

sinh ρ

)2i+2l+5

cosh2 ρ

]

=l∑

i=0

(2i+ 2l + 1)ai

[−(

1

sinh ρ

)2i+2l+3

+ (2i+ 2l + 3)

(1

sinh ρ

)2i+2l+5

(1 + sinh2 ρ)

]

=l∑

i=0

(2i+ 2l + 1)ai

[(2i+ 2l + 2)

(1

sinh ρ

)2i+2l+3

+ (2i+ 2l + 3)

(1

sinh ρ

)2i+2l+5]

=l+1∑i=0

a′i

(1

sinh ρ

)2i+2l+3

,

where

a′0 =a0(2l + 1)(2l + 2) = (2l + 2)!;

a′i =(2i+ 2l + 1)(2i+ 2l + 2)ai + (2i+ 2l − 1)(2i+ 2l + 1)ai−1, i = 1, 2, · · · , l;a′l+1 =(4l + 1)(4l + 3)al.

The desired result follows. �

Lemma 3.3. Let m be a nonnegative integer. There holds, for ρ > 0,

(3.11)

∣∣∣∣(− 1

sinh ρ

∂

∂ρ

)m1

sinh ρ

∣∣∣∣ . ( 1

sinh ρ2

)2m+11

cosh ρ2

.

Proof. If m is even, namely m = 2k for some nonnegative integer k, then by Lemma 3.2,∣∣∣∣(− 1

sinh ρ

∂

∂ρ

)m1

sinh ρ

∣∣∣∣ =

∣∣∣∣∣(− 1

sinh ρ

∂

∂ρ

)2k1

sinh ρ

∣∣∣∣∣.

k∑i=0

(1

sinh ρ

)2i+2k+1

.

(1

sinh ρ

)4k+1

+

(1

sinh ρ

)2k+1

∼(

1

sinh ρ2

)4k+11

cosh ρ2

[(1

cosh ρ2

)4k

+sinh2k ρ

2

cosh2k ρ2

]

.

(1

sinh ρ2

)4k+11

cosh ρ2

.


If m is odd, namely m = 2k + 1 for some nonnegative integer k. Also by Lemma 3.2,

∣∣∣∣(− 1

sinh ρ

∂

∂ρ

)m1

sinh ρ

∣∣∣∣ =

∣∣∣∣∣(− 1

sinh ρ

∂

∂ρ

)2k+11

sinh ρ

∣∣∣∣∣=

∣∣∣∣∣k∑i=0

ai(2i+ 2k + 1)

(1

sinh ρ

)2i+2k+3

cosh ρ

∣∣∣∣∣.

k∑i=0

(1

sinh ρ

)2i+2k+3

cosh ρ

.

(1

sinh ρ

)4k+3

cosh ρ+

(1

sinh ρ

)2k+3

cosh ρ

∼(

1

sinh ρ2

)4k+31

cosh ρ2

[cosh ρ

cosh4k+2 ρ2

+sinh2k ρ

2

cosh2k+2 ρ2

cosh ρ

].

(3.12)

Notice that cosh ρ ∼ cosh2 ρ2, ρ ≥ 0. We have, by (3.12),

∣∣∣∣(− 1

sinh ρ

∂

∂ρ

)m1

sinh ρ

∣∣∣∣ .( 1

sinh ρ2

)4k+31

cosh ρ2

.

These complete the proof. �

Lemma 3.4. Let n ≥ 3. There holds, for ρ > 0,

(3.13)

(−∆H −

(n− 1)2

4

)−1

.

(1

sinh ρ2

)n−21

cosh ρ2

.

Proof. If n is even, namely n = 2m for some positive integer m ≥ 2. Then by (3.7) andLemma 3.3,

(−∆H −

(n− 1)2

4

)−1

.∫ +∞

ρ


(1

sinh r2

)2m−11

cosh r2

dr

≤ 1

cosh ρ2

∫ +∞

ρ


(1

sinh r2

)2m−1

dr.


Using the substitution t =√

cosh r − cosh ρ/√

2 =√

sinh2 r2− sinh2 ρ

2, we have

(−∆H −

(n− 1)2

4

)−1

.1

cosh ρ2

∫ +∞

0

(1

t2 + sinh2 ρ2

) 2m−12

dt

=1

cosh ρ2

(1

sinh ρ2

)2m−2 ∫ +∞

0

(1

t2 + 1

) 2m−12

dt

∼ 1

cosh ρ2

(1

sinh ρ2

)n−2

.

If n is odd, namely n = 2m+1 for some positive integer m ≥ 1. Then by (3.9) and Lemma3.3, (

−∆H −(n− 1)2

4

)−1

∼(− 1

sinh ρ

∂

∂ρ

)m−11

sinh ρ.

(1

sinh ρ2

)n−21

cosh ρ2

.

Thus, we have completed the proof of the lemma.�

4. Proofs of Theorem 1.1: Sharp Poincare-Sobolev inequalities for Paneitzoperators

The main purpose of this section is to establish the Sharp Poincare-Sobolev inequalitiesfor Paneitz operators on hyperbolic spaces, namely Theorem 1.1.

For the sake of completeness, we first provide a slightly different proof of the Hardy-Littlewood-Sobolev inequality on Bn, which was first proved by Beckner on hyperbolic upperhalf spaces [8].

Theorem 4.1. Let 0 < λ < n and p = 2n2n−λ . Then for f, g ∈ Lp(Bn),

(4.1)

∣∣∣∣∣∣∣∫Bn

∫Bn

f(x)g(y)(2 sinh ρ(Ty(x))

2

)λdVxdVy∣∣∣∣∣∣∣ ≤ Cn,λ‖f‖p‖g‖p,

where

(4.2) Cn,λ = πλ/2Γ(n/2− λ/2)

Γ(n− λ/2)

(Γ(n/2)

Γ(n)

)−1+λ/n

is the best constant for the classical Hardy-Littlewood-Sobolev constant on Rn. Furthermore,the constant Cn,λ is sharp for the inequality (4.1) and there is no nonzero extremal functionfor the inequality (4.1).


Proof. We have, by (2.2),∫Bn

∫Bn


2

)λdVxdVy=

∫Bn

∫Bnf(x)

(2

1− |x|2

)n(2|x− a|√

(1− |a|2)(1− |x|2)

)−λg(y)

(2

1− |y|2

)ndxdy

=

∫Bn

∫Bnf(x)

(2

1− |x|2

)n−λ2

|x− y|−λg(y)

(2

1− |y|2

)n−λ2

dxdy.

Set f = f(x)(

21−|x|2

)n−λ2

and g = g(y)(

21−|y|2

)n−λ2. Then by the Hardy-Littlewood-Sobolev

inequality on Rn (see Lieb [39]),∣∣∣∣∣∣∣∫Bn

∫Bn


2

)λdVxdVy∣∣∣∣∣∣∣ =

∣∣∣∣∫Bn

∫Bnf(x)|x− y|−λg(y)dxdy

∣∣∣∣≤∫Bn

∫Bn|f(x)| · |x− y|−λ · |g(y)|dxdy

≤Cn,λ(∫

Bn|f(x)|

2n2n−λdx

) 2n−λ2n

·(∫

Bn|g(y)|

2n2n−λdy

) 2n−λ2n

=Cn,λ

(∫Bn|f |

2n2n−λdV

) 2n−λ2n

·(∫

Bn|g(y)|

2n2n−λdV

) 2n−λ2n

.

Furthermore, Cn,λ is sharp and there is no nonzero extreme function. The proof of Theorem4.1 is thereby completed. �

Before the proof of Theorem 1.1, we need the following Lemma.

Lemma 4.2. Let 0 < α < n, 0 < β < n and 0 < α + β < n. Then∫Bn

(sinh

ρ(x, y)

2

)α−n(cosh

ρ(x, y)

2

)−α−β (sinh

ρ(x, z)

2

)β−ndVx

=

{[(sinh

ρ

2

)α−n (cosh

ρ

2

)−α−β]∗(

sinhρ

2

)β−n}(Tz(y))

≤2nγn(α)γn(β)

γn(α + β)

(sinh

ρ(y, z)

2

)α+β−n(cosh

ρ(y, z)

2

)−α,

where

γn(α) = πn/22αΓ(α/2)

Γ(n−α2

), 0 < α < n.(4.3)


Proof. We firstly show

∫Bn

(sinh

ρ(x, y)

2

)α−n(cosh

ρ(x, y)

2

)−α−β (sinh

ρ(x, z)

2

)β−ndVx

=

{[(sinh

ρ

2

)α−n (cosh

ρ

2

)−α−β]∗(

sinhρ

2

)β−n}(Tz(y)).

(4.4)

In fact, using the following identity (see [40], (3.13)),

|Tz(Ty(x))| = |Tx(Ty(z))|, x, y, z ∈ Bn,

we have, by the Mobius shift invariance,

∫Bn

(sinh

ρ(x, y)

2

)α−n(cosh

ρ(x, y)

2

)−α−β (sinh

ρ(x, z)

2

)β−ndVx

=

∫Bn

(sinh

ρ(Ty(x), y)

2

)α−n(cosh

ρ(Ty(x), y)

2

)−α−β (sinh

ρ(Ty(x), z)

2

)β−ndVx

=

∫Bn

(sinh

ρ(Ty(Ty(x)))

2

)α−n(cosh

ρ(Ty(Ty(x)))

2

)−α−β (sinh

ρ(Tz(Ty(x)))

2

)β−ndVx

=

∫Bn

(sinh

ρ(x)

2

)α−n(cosh

ρ(x)

2

)−α−β (sinh

ρ(Tx(Tz(y)))

2

)β−ndVx

=

∫Bn

(sinh

ρ(x)

2

)α−n(cosh

ρ(x)

2

)−α−β (sinh

ρ(TTz(y)(x)

2

)β−ndVx

=

{[(sinh

ρ

2

)α−n (cosh

ρ

2

)−α−β]∗(

sinhρ

2

)β−n}(Tz(y)).

To finish the proof, it is enough to show{[(sinh

ρ

2

)α−n (cosh

ρ

2

)−α−β]∗(

sinhρ

2

)β−n}(y)

≤2nγn(α)γn(β)

γn(α + β)

(sinh

ρ(y)

2

)α+β−n(cosh

ρ(y)

2

)−α.

(4.5)

Notice that, for 0 < α < n, 0 < β < n and 0 < α + β < n, we have (see e.g. [54])∫Rn|x|α−n|y − x|β−ndx =

γn(α)γn(β)

γn(α + β)|y|α+β−n.(4.6)


Therefore, by (2.2) and (4.6),{[(sinh

ρ

2

)α−n (cosh

ρ

2

)−α−β]∗(

sinhρ

2

)β−n}(y)

=

∫Bn

(sinh

ρ(x)

2

)α−n(cosh

ρ(x)

2

)−α−β (sinh

ρ(Ty(x))

2

)β−n(2

1− |x|2

)ndx

=

∫Bn

(|x|√

1− |x|2

)α−n(1√

1− |x|2

)−α−β (|x− y|√

(1− |y|2)(1− |x|2)

)β−n(2

1− |x|2

)ndx

=2n

(1− |y|2)(β−n)/2

∫Bn|x|α−n|x− y|β−ndx

≤ 2n

(1− |y|2)(β−n)/2

∫Rn|x|α−n|x− y|β−ndx

=2n

(1− |y|2)(β−n)/2· γn(α)γn(β)

γn(α + β)|y|α+β−n

=2nγn(α)γn(β)

γn(α + β)

(sinh

ρ(y)

2

)α+β−n(cosh

ρ(y)

2

)−α.

Thus, the desired result follows. �

Next, we will estimate the kernel function Q−1k (ρ) for the 2k-th order operator Qk.

Lemma 4.3. Let k be a positive integer with 2 ≤ k < n2. The kernel Q−1

k (ρ) satisfies

Q−1k (ρ) .

(1

sinh ρ2

)n−2k

, ρ > 0.(4.7)

Proof. Recall that

Qk =

(−∆H −

(n− 1)2

4

)(P1 + 2) · · · · · (P1 + k(k − 1)).

We have, by (2.4),

Q−1k =

(−∆H −

(n− 1)2

4

)−1

∗ (P1 + 2)−1 ∗ · · · ∗ (P1 + k(k − 1))−1,

where, by Lemma 3.1, (P1 + i(i− 1))−1(2 ≤ i ≤ k) satisfies

(P1 + i(i− 1))−1 =

(i(i− 1)− n(n− 2)

4−∆H

)−1

=

((i− 1/2)2 − (n− 1)2

4−∆H

)−1

.

(1

sinh ρ2

)n−2(1

cosh ρ2

)2i

.

(4.8)


We shall prove (4.7) by induction. For k = 2, we have, by (4.8) and Lemma 4.2,

Q−12 =

(−∆H −

(n− 1)2

4

)−1

∗ (P1 + 2)−1

.

[(1

sinh ρ2

)n−21

cosh ρ2

]∗

[(1

sinh ρ2

)n−2(1

cosh ρ2

)4]

≤(

1

sinh ρ2

)n−2

∗

[(1

sinh ρ2

)n−2(1

cosh ρ2

)4]

.

(1

sinh ρ2

)n−4(1

cosh ρ2

)2

≤(

1

sinh ρ2

)n−4

.

Now suppose that equation (4.8) is valid for k = l, i.e. Q−1l (ρ) .

(1

sinh ρ2

)n−2l

. Then by

(4.8) and Lemma 4.2,

Q−1l+1(ρ) =Q−1

l (ρ) ∗ (P1 + (l + 1)l)

.

(1

sinh ρ2

)n−2l

∗

[(1

sinh ρ2

)n−2(1

cosh ρ2

)2l+2]

.

(1

sinh ρ2

)n−2l−2(1

cosh ρ2

)2

≤(

1

sinh ρ2

)n−2l−2

.

Thus, the proof of the lemma is completed. �

Proof of Theorem 1.1 We first prove, for some positive constant C > 0,

(4.9)

∫Bn

(Qku)udV ≥ C

(∫Bn|u|

2nn−2kdV

)n−2kn

, u ∈ C∞0 (Bn).

By Lemma 4.3 and Theorem 4.1, we have∫Bn|(Q−

12

k u)(x)|2dV =

∫Bnu(x)(Q−1

k u)(x)dV

.∫Bn

∫Bn

u(x)u(y)(2 sinh ρ(Ty(x))

2

)n−2kdVxdVy

.

(∫Bn|u(x)|

2nn+2kdV

)n+2kn

.

(4.10)


On the other hand,

∣∣∣∣∫Bnu(x)g(x)dV

∣∣∣∣2 =

∣∣∣∣∫Bn

(Q12k u)(x)(Q

− 12

k g)(x)dV

∣∣∣∣2≤∫Bn|(Q

12k u)(x)|2dV ·

∫Bn|(Q−

12

k g)(x)|2dV.(4.11)

Combing (4.10) and (4.11) yields

∣∣∣∣∫Bnu(x)g(x)dV

∣∣∣∣2 .∫Bn|(Q

12k u)(x)|2dV

(∫Bn|g(x)|

2nn+2kdV

)n+2kn

=

∫BnQku(x) · u(x)dV

(∫Bn|g(x)|

2nn+2kdV

)n+2kn

.

(4.12)

Taking g = |u|n+2kn−2k , we have, by (4.12),

(∫Bn|u(x)|

2nn−2kdV

)2

.∫BnQku(x) · u(x)dV

(∫Bn|u(x)|

2nn−2kdV

)n+2kn

.(4.13)

Therefore,

(∫Bn|u(x)|

2nn−2kdV

)n−2kn

.∫BnQku(x) · u(x)dV.(4.14)

Now we shall prove inequality (1.9). Notice that, by Plancherel formula, (2.6) and (1.3),

,

∫BnQku(x) · u(x)dV =Dn

∫ +∞

−∞

∫Sn−1

λ2

4

k∏i=2

(2i− 1)2 + λ2

4|u(λ, ζ)|2|c(λ)|−2dλdσ(ς)

≥k∏i=2

(2i− 1)2

4Dn

∫ +∞

−∞

∫Sn−1

λ2

4|u(λ, ζ)|2|c(λ)|−2dλdσ(ς)

=k∏i=2

(2i− 1)2

4

(∫Bn|∇Hu|2dV −

(n− 1)2

4

∫Bnu2dV

)

≥C(∫

Bn|u|pdV

) 2p

, 2 < p ≤ 2n

n− 2.

(4.15)


Therefore, for 2 < p ≤ 2nn−2k

, choose p ∈ (2, 2nn−2

] such that p < p. By Holder inequality and(4.15), ∫

Bn|u|pdV =

∫Bn|u|s|u|tdV

≤(∫

Bn|u|pdV

) sp(∫

Bn|u|

2nn−2kdV

) (n−2k)t2n

≤(∫

BnQku(x) · u(x)dV

) s2(∫

BnQku(x) · u(x)dV

) t2

=

(∫BnQku(x) · u(x)dV

) p2

,

(4.16)

where s = (1− n−2k2n

p)(1p− n−2k

2n)−1 and t = p− s. Thus, the proof of Theorem 1.1 is finished.

Having established Theorem 1.1, Theorem 1.2 follows immediately.

5. Proof of Theorem 1.3: Sharp higher order Hardy-Sobolev-Mazya’sinequalities

It has been shown on the Poincare ball model Bn by Liu (see [41], Theorem 2.3):

(5.1)

(1− |x|2

2

)k+n2

(−∆)k

[(1− |x|2

2

)k−n2

f

]= Pkf, f ∈ C∞0 (Bn), k ∈ N.

In terms of the Poincare half space model Hn, we can establish the following:

Lemma 5.1. Let k be a positive integer. There holds, for each f ∈ C∞0 (Hn),

(5.2) xn2

+k

1 (−∆)k(xk−n

21 f) = Pkf.

Proof. It is enough to show

(5.3) xn2

+k

1 ∆k(xk−n

21 f) =

k∏i=1

[∆H +

(n− 2i)(n+ 2i− 2)

4

]f.

We shall prove (5.3) by induction.

A simple calculation shows, for each α ∈ R and f ∈ C∞0 (Hn), there holds

xα+21 ∆(x−α1 f) =x2

1∆f − 2αx1∂f

∂x1

+ α(α + 1)f

= = [∆H + α(α + 1)]f + (n− 2− 2α)x1∂f

∂x1

.

(5.4)

If we choose α = n−22

in (5.4), then

(5.5) xn2

+1

1 ∆(x1−n

21 f) =

[∆H +

n(n− 2)

4

]f.


Now suppose that equation (5.3) is valid for k = l, i.e.

(5.6) xn2

+l

1 ∆l(xl−n

21 f) =

l∏i=1

[∆H +

(n− 2i)(n+ 2i− 2)

4

]f.

We note it is easy to check

(5.7) ∆l+1(x1g) = x1∆l+1g + 2(l + 1)∂

∂x1

∆lg, g ∈ C∞0 (Hn).

Therefore, by (5.6) and (5.7),

xn2

+l+1

1 ∆l+1(xl+1−n

21 f) = x

n2

+l+1

1 ∆l+1(x1 · xl−n

21 f)

=xn2

+l+1

1 x1∆l+1(xl−n

21 f) + 2(l + 1)x

n2

+l+1

1

∂

∂x1

∆l(xl−n

21 f)

=xn2

+l+2

1 ∆

{x−n

2−l

1

l∏i=1

[∆H +

(n− 2i)(n+ 2i− 2)

4

]f

}+

2(l + 1)xn2

+l+1

1

∂

∂x1

{x−n

2−l

1

l∏i=1

[∆H +

(n− 2i)(n+ 2i− 2)

4

]f

}.

(5.8)

By (5.4), we have

xn2

+l+2

1 ∆

{x−n

2−l

1

l∏i=1

[∆H +

(n− 2i)(n+ 2i− 2)

4

]f

}

=

[∆H +

(n+ 2l)(n+ 2l + 2)

4

] l∏i=1

[∆H +

(n− 2i)(n+ 2i− 2)

4

]f−

2(l + 2)x1∂

∂x1

l∏i=1

[∆H +

(n− 2i)(n+ 2i− 2)

4

]f.

(5.9)


Combining (5.8) and (5.9) yields

xn2

+l+1

1 ∆l+1(xl+1−n

21 f)

=

[∆H +

(n+ 2l)(n+ 2l + 2)

4

] l∏i=1

[∆H +

(n− 2i)(n+ 2i− 2)

4

]f−

2(l + 2)x1∂

∂x1

l∏i=1

[∆H +

(n− 2i)(n+ 2i− 2)

4

]f+

2(l + 1)xn2

+l+1

1

∂

∂x1

{x−n

2−l

1

l∏i=1

[∆H +

(n− 2i)(n+ 2i− 2)

4

]f

}

=

[∆H +

(n+ 2l)(n+ 2l + 2)

4

] l∏i=1

[∆H +

(n− 2i)(n+ 2i− 2)

4

]f−

2(l + 1)x1−n

2−l

1 · ∂xn2

+l

1

∂x1

·l∏

i=1

[∆H +

(n− 2i)(n+ 2i− 2)

4

]f.

(5.10)

To get the last equation in the above, we use the fact

x1∂

∂x1

l∏i=1

[∆H +

(n− 2i)(n+ 2i− 2)

4

]f

=x1∂

∂x1

{xn2

+l

1 x−n

2−l

1

l∏i=1

[∆H +

(n− 2i)(n+ 2i− 2)

4

]f

}

=xn2

+l+1

1

∂

∂x1

{x−n

2−l

1

l∏i=1

[∆H +

(n− 2i)(n+ 2i− 2)

4

]f

}+

x1−n

2−l

1

∂xn2

+l

1

∂x1

·l∏

i=1

[∆H +

(n− 2i)(n+ 2i− 2)

4

]f.

Therefore, by (5.10),

xn2

+l+1

1 ∆l+1(xl+1−n

21 f)

=

[∆H +

(n+ 2l)(n+ 2l + 2)

4

] l∏i=1

[∆H +

(n− 2i)(n+ 2i− 2)

4

]f−

2(l + 1)(n

2+ l) l∏i=1

[∆H +

(n− 2i)(n+ 2i− 2)

4

]f

=l+1∏i=1

[∆H +

(n− 2i)(n+ 2i− 2)

4

]f.

(5.11)


The proof of Lemma 5.1 is thus completed. �

Proof of Theorem 1.3 We claim that

(5.12)

∫Rn+|∇ku|2dx−

k∏i=1

(2i− 1)2

4

∫Rn+

u2

x2k1

dx =

∫Bn

(Pkv)vdV −k∏i=1

(2i− 1)2

4

∫Bnv2dV,

where v = xn2−k

1 u. In fact, by Lemma 5.1,∫Rn+|∇ku|2dx−

k∏i=1

(2i− 1)2

4

∫Rn+

u2

x2k1

dx

=

∫Rn+u · (−∆)kudx−

k∏i=1

(2i− 1)2

4

∫Rn+

u2

x2k1

dx

=

∫Hnxn2−k

1 u · Pk(xn2−k

1 u)dV −k∏i=1

(2i− 1)2

4

∫Hn

(xn2−k

1 u)2dV

=

∫Bn

(Pkv)vdV −k∏i=1

(2i− 1)2

4

∫Bnv2dV.

(5.13)

This proves the claim. Therefore, by Theorem 1.2,∫Rn+|∇ku|2dx−

k∏i=1

(2i− 1)2

4

∫Rn+

u2

x2k1

dx

=

∫Bn

(Pkv)vdV −k∏i=1

(2i− 1)2

4

∫Bnv2dV

≥C(∫

Hn|v|p dV

) 2p

= C

(∫Rn+xγ1 |u|pdx

) 2p

.

Similarly, using the identity (5.1), we obtain inequality (1.13). The proof of Theorem 1.3is thereby completed.

6. proof of Theorem 1.4: Fractional Poincare-Sobolev inequality

The main purpose of this section is two fold. We first establish the Poincare-Sobolevinequality of fractional order s with best Sobolev constant in dimensional 3. When s = 1,this is the result by Benguria, Frank and Loss (see [10]). Next, we will illustrate that Theorem6.2 below does not imply our sharp Hardy-Sobolev-Mazya inequalities.

We first show the following lemma, which is equivalent to Theorem 1.4.


Lemma 6.1. Let s2≤ s < 3

2. Then for f ∈ L

63+2s (B3), we have∫

B3

f(x)[(−∆H − 1)−sf

](x)dV ≤ 2−2sπ−

32

Γ(3−2s2

)

Γ(s)C3,3−2s

(∫B3

|f |6

3+2sdV

) 3+2s3

,(6.1)

where C3,3−2s is given by (4.2). Furthermore, the constant C3,3−2s is sharp and this constantis not attained in (6.1) for nonzero functions.

Proof. We have, for 0 < s < 3/2,

(−∆H − 1)−s =1

Γ(s)

∫ ∞0

ts−1et(∆H+1)dt

=1

Γ(s)· 2−3π−3/2 ρ

sinh ρ

∫ ∞0

t2s−5

2 e−ρ2

4t dt

=2−απ−3/2 Γ(3−2s2

)

Γ(s)

1

ρ2−2s sinh ρ

=2−απ−3/2 Γ(3−2s2

)

Γ(s)

(1

2 sinh ρ2

)3−2s

·Ψs(ρ),

where Ψs(ρ) =(

2 sinh ρ2

ρ

)2−2s1

cosh ρ2. It is easy to check, for s ≥ 1

2, the function Ψs(ρ) is

decreasing on (0,∞) andsup

ρ∈(0,∞)

Ψs(ρ) = 1.

Therefore, by Theorem 4.1,∫B3

f(x)[(−∆H − 1)−sf

](x)dV ≤2−απ−3/2 Γ(3−2s

2)

Γ(s)

∫B3

∫B3

f(x)f(y)(2 sinh ρ(Ty(x))

2

)3−2sdVxdVy

≤2−2sπ−32

Γ(3−2s2

)

Γ(s)C3,3−2s

(∫B3

|f |6

3+2sdV

) 3+2s3

.

�

Next we recall a result of Benguria, Frank and Loss (see [10], Corollary 3.1).

Theorem 6.2. Let n ≥ 2 and n − 1 ≤ α < n (resp. 0 < α < 1 if n = 1). The operator(−∆− 1

4x21)−

α2 is a bounded operator from Lp(Rn

+) to Lq(Rn+) for all 1 < p, q <∞ that satisfy

1q

= 1p− α

n, and its norm coincides with the one of (−∆)−

α2 : Lp(Rn)→ Lq(Rn).

Moreover, for such values of α we have(f, (−∆− 1

4x21

)−α2 f

)≤ 2−απ−

n2

Γ(n−α2

)

Γ(α2)Cn,n−α‖f‖p,(6.2)

where p = 2nn+α

and Cn,n−α is given by (4.2). Furthermore, the constant Cn,n−α is the sharpconstant and this constant is not attained in (6.2) for nonzero functions.


Remark 6.3. We note that the Hardy-Sobolev-Maz’ya inequality for higher order derivativesproved in our paper does not follow from the above result of [10]. This can be explainedthrough the Fourier analysis on hyperbolic spaces. For simplicity, we choosing n = 5 andα = 4 in (6.2) to illustrate this.

Following the proof in [10], we have the following sharp Sobolev type inequality∫R5+

∣∣∣∣∆u+u

4x21

∣∣∣∣2 dx ≥ S5,2

(∫R5+

|u|10dx

) 15

,(6.3)

where S5,2 is the best second order Sobolev constant on R5. Inequality (6.3) is equivalent to(see Lemma 6.4) ∫

H5

|∆Hf + 3f |2dV ≥ S5,2

(∫H5

|f |10dV

) 15

, f = x121 u.(6.4)

By the Plancherel formula (2.5),∫H5

|∆Hf + 3f |2dV =D5

∫ +∞

−∞

∫S4

(λ2 + 4)2

16|f(λ, ζ)|2|c(λ)|−2dλdσ(ς).

However, ∫H5

(P4f)fdV −2∏i=1

(2i− 1)2

4

∫H5

f 2dV

=D5

∫ +∞

−∞

∫S4

2∏i=1

λ2 + (2i− 1)2

4|f(λ, ζ)|2|c(λ)|−2dλdσ(ς)−

D5

2∏i=1

(2i− 1)2

4

∫ +∞

−∞

∫S4|f(λ, ζ)|2|c(λ)|−2dλdσ(ς)

=D5

∫ +∞

−∞

∫S4

λ4 + 10λ2

16|f(λ, ζ)|2|c(λ)|−2dλdσ(ς).

Since

infλ∈R

λ4+10λ2

16(λ2+4)2

16

= 0,

then one cannot find a positive constant C which is independent of f such that∫H5

(P4f)fdV −2∏i=1

(2i− 1)2

4

∫H5

f 2dV ≥ C

∫H5

|∆Hf + 3f |2dV.

Therefore, inequality (6.3) does not imply the following Hardy-Sobolev-Maz’ya inequalityon R5

+ ∫R5+

|∆u|2dx− 9

16

∫R5+

u2

x41

dx ≥ C

(∫R5+

|u|10dx

) 15

.


This shows that Theorem 6.2 does not lead to the fourth order Hardy-Sobolev-Maz’ya in-equality on the hyperbolic space in dimension 5.

We now complete the proof of the following lemma which was needed in the above illus-tration.

Lemma 6.4. Let u ∈ C∞0 (R5+) and f = x

121 u. Then∫

R5+

∣∣∣∣∆u+u

4x21

∣∣∣∣2 dx =

∫H5

|∆Hf + 3f |2dV.

Proof. We compute∫R5+

∣∣∣∣∆u+u

4x21

∣∣∣∣2 dx =

∫R5+

∣∣∣∣−∆u− u

4x21

∣∣∣∣2 dx=

∫R5+

(u(−∆)2u− u(−∆)u

2x21

+u2

16x41

)dx.

(6.5)

Since u = x− 1

21 f , we have, by (5.4) and (5.2),

(−∆)u =(−∆)(x− 1

21 f) = x

− 52

1

(−∆H −

3

4

)f − 2x

− 32

1

∂f

∂x1

;

(−∆)2u =(−∆)2(x− 1

21 f) = x

− 92

1

(−∆H −

15

4

)(−∆H −

7

4

)f.

Therefore, ∫R5+

u(−∆)2u =

∫R5+

x−51 f

(−∆H −

15

4

)(−∆H −

7

4

)fdx

=

∫H5

f

(−∆H −

15

4

)(−∆H −

7

4

)fdV ;

−∫R5+

u(−∆)u

2x21

dx =− 1

2

∫R5+

x−51 f

(−∆H −

3

4

)fdx+

∫R5+

x−41 f

∂f

∂x1

dx

=− 1

2

∫H5

f

(−∆H −

3

4

)fdV + 2

∫R5+

x−51 f 2dx

=− 1

2

∫H5

f

(−∆H −

3

4

)fdV + 2

∫H5

f 2dV ;∫R5+

u2

16x41

dx =

∫R5+

f 2

16x51

dx =1

16

∫H5

f 2dV.

(6.6)


Combining (6.5) and (6.6) yields∫R5+

∣∣∣∣−∆u− u

4x21

∣∣∣∣2 dx=

∫H5

f

(−∆H −

15

4

)(−∆H −

7

4

)fdV − 1

2

∫H5

f

(−∆H −

3

4

)fdV +

33

16

∫H5

f 2dx

=

∫H5

f (−∆H − 3)2 fdV =

∫H5

|∆Hf + 3f |2dV.

This completes the proof. �

7. An alternative proof of the sharp Sobolev inequalities on hyperbolicspaces

In this section, we shall give an alternative proof of the sharp constant for the Sobolevinequality in hyperbolic spaces via Theorem 4.1. This Sobolev inequality without subtractingthe Hardy term is weaker than the Poincare-Sobolev and Hardy-Sobolev-Mazya inequalitieswe established in Sections 4 and 5. The sharp Sobolev inequality on n-dimensional sphereSn was proved by Beckner [5]. The authors thank one of the referees who points out andinsists that Beckner’s theorem on the sphere can imply the main theorem on sharp Sobolevinequalities on hyperbolic spaces by Liu in [41] and thus can give a simpler proof of Liu’sresults [41]. In this section, we will provide a different proof of this sharp Sobolev inequalityon hyperbolic spaces using the Hardy-Littlewood-Sobolev inequality.

Before we begin the proof, we need the following Lemma.

Lemma 7.1. Let f ∈ C∞0 (Bn). Then there exists a function g ∈ C∞0 (Bn) such that Pkg = f .

Proof. Choose f ∈ C∞0 (Bn) such that

(−∆)kf =

(1− |x|2

2

)−k−n2

f.

Set g = (1−|x|22

)k−n2 f . Then g ∈ C∞0 (Bn). Furthermore, by (5.1),

Pkg = Pk

[(1− |x|2

2

)k−n2

f

]=

(1− |x|2

2

)k+n2

(−∆)kf = f.

�

It is known that the kernel (−∆)−k(1 ≤ k < n/2) is 1γn(2k)|x|n−2k , where

(7.1) γn(2k) =πn/222kΓ(k)

Γ(n2− k)

.

We have, for f ∈ C∞0 (Bn),

(7.2) f(0) =1

γn(2k)

∫Bn

(−∆)kf(x) · 1

|x|n−2kdx.


Replacing f by (1− |x|2)k−n2 f and using (5.1), we obtain

f(0) =1

γn(2k)

∫Bn

(−∆)k[(1− |x|2)k−n2 f ] · 1

|x|n−2kdx

=2k−

n2

γn(2k)

∫BnPkf(x) · 1

|x|n−2k

(1− |x|2

2

)−n2−k

dx

=1

γn(2k)

∫BnPkf(x) ·

(√1− |x|22|x|

)n−2k

dV

=1

γn(2k)

∫BnPkf(x) ·

(2 sinh

ρ(x)

2

)2k−n

dV.

(7.3)

Therefore, by the Mobius shift invariance, for f ∈ C∞0 (Bn) and y ∈ Bn,

f(y) =1

γn(2k)

∫BnPkf(x) ·

(2 sinh

ρ(Ty(x))

2

)2k−n

dVx.(7.4)

Theorem 7.2. Let 1 ≤ k < n/2. Then, for any f ∈ C∞0 (Bn),∫BnPkf(x) · f(x)dV ≥ Sn,k

(∫Bn|f(x)|

2nn−2kdV

)n−2kn

,

where Sn,k is the best k-th order Sobolev constant. Furthermore, the inequality is strict fornonzero f ’s.

Proof. Let g ∈ C∞0 (Bn) be such that Pkg = f . Then∫Bnu(x)g(x)dV =

∫Bn

(P−1k f)(x) · (Pkg)(x)dV

=

∫Bn

(P−1k f)(x) · f(x)dV

=

∫Bn|(P−1/2

k f)(x)|2dV.

(7.5)

By (7.4) and Theorem 1.2,∫Bnu(x)g(x)dV =

1

γn(2k)

∫Bn

∫Bn

f(x) · Pkg(2 sinh ρ(Ty(x))

2)n−2k

dVxdVy

=1

γn(2k)

∫Bn

∫Bn

f(x) · f(y)

(2 sinh ρ(Ty(x))

2)n−2k

dVxdVy

≤ 1

γn(2k)Cn,n−2k

(∫Bn|f(x)|

2nn+2kdV

)n+2kn

.

(7.6)


Combing (7.5) and (7.6) yields∫Bn|(P−1/2

k f)(x)|2dV ≤ 1

γn(2k)Cn,n−2k

(∫Bn|f(x)|

2nn+2kdV

)n+2kn

, f ∈ C∞0 (Bn).(7.7)

On the other hand, for h ∈ C∞0 (Bn),∣∣∣∣∫Bnh(x)f(x)dV

∣∣∣∣2 =

∣∣∣∣∫Bn

(P12k h)(x)(P

− 12

k f)(x)dV

∣∣∣∣2≤∫Bn|(P

12k h)(x)|2dV ·

∫Bn|(P−

12

k f)(x)|2dV.(7.8)

Combing (7.7) and (7.8) yields∣∣∣∣∫Bnh(x)f(x)dV

∣∣∣∣2 ≤ 1

γn(2k)Cn,n−2k

∫Bn|(P

12k h)(x)|2dV

(∫Bn|f(x)|

2nn+2kdV

)n+2kn

=1

γn(2k)Cn,n−2k

∫BnPkh(x) · h(x)dV

(∫Bn|f(x)|

2nn+2kdV

)n+2kn

.

(7.9)

Taking f = |h|n+2kn−2k , we have, by (7.9),

(∫Bn|h(x)|

2nn−2kdV

)2

≤ 1

γn(2k)Cn,n−2k

∫BnPkh(x) · h(x)dV

(∫Bn|h(x)|

2nn−2kdV

)n+2kn

.

(7.10)

Therefore, we have, for h ∈ C∞0 (Bn),

γn(2k)

Cn,n−2k

(∫Bn|h(x)|

2nn−2kdV

)n−2kn

≤∫BnPkh(x) · h(x)dV,(7.11)

where

γn(2k)

Cn,n−2k

=πn/222kΓ(k)

Γ(n/2− k)· Γ(n/2 + k)

πn/2−kΓ(k)

(Γ(n/2)

Γ(n)

)2k/n

= 22kπkΓ(n/2 + k)

n/2− k

(Γ(n/2)

Γ(n)

)2k/n

is the best k-th order Sobolev constant (see [39]). Moreover, by Theorem 4.1, the inequalityis strict for nonzero f ’s. �

8. Proof of Theorem 1.5: the best constants in Hardy-Sobolev-Maz’ya andSobolev inequalities coincide in dimension 5

In this section we shall show that the sharp constant of Hardy-Sobolev-Maz’ya inequalityfor n = 5 and k = 2 coincides with the corresponding Sobolev constant. We also give anevidence showing that such a coincidence in dimension 6 does not hold (see Remark 8.3 andLemma 8.4. As we have pointed out, we can continue to establish that the sharp constant ofHardy-Sobolev-Maz’ya inequality for special n and k = 2 coincides with the correspondingSobolev constant (see remarks after the statement of Theorem 1.5 and Conjecture 1.6 in theintroduction).


The proof of Theorem 1.5 depends on the following key lemma.

Lemma 8.1. Let n = 5. There holds

[(−4−∆H)(−3−∆H)]−1 =1

16π2· 1

2 sinh ρ2

· 1

cosh2 ρ2

≤ 1

γ5(4)· 1

2 sinh ρ2

, ρ > 0,

where γ5(4) = 116π2 is defined in (4.3).

Proof. By (3.9), we have, for n = 5,

(−∆H − 4)−1 =

∫ ∞0

e(−∆H−4)tdt =1

8π2· cosh ρ

sinh3 ρ.

On the other hand, choosing ν = 1 in (3.5), we have

(−3−∆H)−1 =1

(sinh ρ)3· (2π)−

52

Γ(52

+ 12)

212

+1Γ(12

+ 1)

∫ π

0

sin2 tdt

=1

8π2· 1

sinh3 ρ.

Therefore,

[(−4−∆H)(−3−∆H)]−1 =(−4−∆H)−1 − (−3−∆H)−1

=1

8π2· cosh ρ

sinh3 ρ− 1

8π2· 1

sinh3 ρ

=1

16π2· 1

2 sinh ρ2

· 1

cosh2 ρ2

The desired result follows. �

By Lemma 8.1 and Theorem 4.1, we have

Corollary 8.2. There holds, for each u ∈ C∞0 (B5),∫B5

((−4−∆H)(−3−∆H)u)udV ≥ S5,2

(∫Bn|u|10dV

) 15

,

where S5,2 = γ5(4)/C5,1 is the best second order Sobolev constant.

Proof of Theorem 1.5 By Corollary 8.2, it is enough to show∫B5

(P2u)udV − 9

16

∫B5

u2dV ≥∫B5

((−4−∆H)(−3−∆H)u)udV.


In fact, by the Plancherel formula (see (2.5)),∫B5

(P2u)udV − 9

16

∫B5

u2dV

=D5

∫ +∞

−∞

∫S4

[(λ2 + 1)(λ2 + 9)

16− 9

16

]|u(λ, ζ)|2|c(λ)|−2dλdσ(ς))

=D5

∫ +∞

−∞

∫S4

λ4 + 10λ2

16|u(λ, ζ)|2|c(λ)|−2dλdσ(ς))

≥D5

∫ +∞

−∞

∫S4

λ2(λ2 + 1)

16|u(λ, ζ)|2|c(λ)|−2dλdσ(ς))

=

∫B5

((−4−∆H)(−3−∆H)u)udV.

This completes the proof.

Remark 8.3. It seems that the sharp constant of Poincare-Sobolev inequalities in case n = 6and k = 2 is strictly less than Sobolev constant. In fact, we have the following:

Lemma 8.4. Let n = 6. There exists ε > 0 such that for 0 < ρ < ε,

(8.1) [(−25/4−∆H)(−4−∆H)]−1 >1

16π3· 1

(2 sinh ρ2)2

=1

γ6(4)· 1

(2 sinh ρ2)2.

Before the proof, we need some fact about the Legendre function of second type Qµν (z).

It is known that (see [23], page 773, 7.133(2))∫ ∞u

(x2 − 1)12λ(x− u)µ−1Q−λν (x)dx =Γ(µ)eµπi(u2 − 1)

12λ+ 1

2µQ−λ−µν (u),

| arg(u− 1)| < π, 0 < Reµ < 1 +Re(ν − λ).

(8.2)

Setting u = cosh ρ and using (8.2), we have∫ ∞ρ

(sinh r)λ+1

(cosh r − cosh ρ)1−µQ−λν (cosh r)dr =Γ(µ)eµπi(sinh ρ)λ+µQ−λ−µν (cosh ρ).(8.3)

Setting z = cosh ρ and using (3.4), we have

Qµν (cosh ρ) = ei(πµ)2−ν−1 Γ(ν + µ+ 1)

Γ(ν + 1)sinh−µ ρ

∫ π

0

(cosh ρ+ cos t)µ−ν−1(sin t)2ν+1dt.(8.4)

Proof of Lemma 8.4 We firstly show

(−25/4−∆H)−1 =3√

2

26π3(sinh ρ)−4

∫ π

0

(cosh ρ+ cos t)32dt.(8.5)

In fact, by (3.7),

(−25/4−∆H)−1 =

√π

(2π)72

∫ +∞

ρ


(− 1

sinh r

∂

∂r

)21

sinh rdr.(8.6)


By (8.4), we have(− 1

sinh r

∂

∂r

)21

sinh r=

2 cosh2 r + 1

sinh5 r=

2

π· 1

(sinh r)5

∫ π

0

(cosh r + cos t)2dt

=2

π· e−

7π2i2

32

Γ(3/2)

Γ(5)(sinh r)−

52Q

5/2−1/2(cosh r).

(8.7)

Therefore, by (8.6), (8.7) and (8.3), we have

(−25/4−∆H)−1 =

√π

(2π)72

2

π· e−

7π2i2

32

Γ(3/2)

Γ(5)

∫ +∞

ρ

(sinh r)−32

√cosh r − cosh ρ

Q5/2−1/2(cosh r)dr

=(2π)−3e−2πi(sinh ρ)−2Q2−1/2(cosh ρ).

(8.8)

Combining (8.4) and (8.8) yields (8.5).Next choosing ν = 3

2in (3.5), we have

(−4−∆H)−1 =(2π)−3Γ(4)

22Γ(2)(sinh ρ)−4

∫ π

0

(sin t)3dt =1

4π3(sinh ρ)−4.

Therefore,

[(−25/4−∆H)(−4−∆H)]−1 − 1

γ6(4)· 1

(2 sinh ρ2)2

=4

9[(−25/4−∆H)−1 − (−4−∆H)−1]− 1

16π3· 1

(2 sinh ρ2)2

=1

4π3(sinh ρ)−4

{4

9

[3√

2

16

∫ π

0

(cosh ρ+ cos t)32dt− 1

]− (sinh

ρ

2)2(cosh

ρ

2)4

}

=1

4π3(sinh ρ)−4

{4

9

[3√

2

16

∫ π

0

(1 + 2 sinh2 ρ

2+ cos t)

32dt− 1

]− (sinh

ρ

2)2(1 + 2 sinh2 ρ

2)2

}=

1

4π3(sinh ρ)−4ψ(sinh2 ρ

2),

where

ψ(r) =4

9

[3√

2

16

∫ π

0

(1 + 2r + cos t)32dt− 1

]− r(1 + r)2, r ≥ 0.

We compute

ψ′(r) =

√2

4

∫ π

0

(1 + 2r + cos t)12dt− (1 + r)(1 + 3r);

ψ′′(r) =

√2

4

∫ π

0

(1 + 2r + cos t)−12dt− 4− 6r.


It is easy to check that

ψ(0) = ψ′(0) = 0, limr→0+

ψ′′(r) = +∞.

Therefore, for small enough r > 0, we have ψ(r) > 0. Thus

[(−25/4−∆H)(−4−∆H)]−1 − 1

γ6(4)· 1

(2 sinh ρ2)2

=1

4π3(sinh ρ)−4ψ(sinh2 ρ

2) > 0, 0 < ρ < ε,

where ε > 0 is small enough.

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Department of Mathematics, University of Connecticut, Storrs, CT 06269, USAE-mail address: [email protected]

School of Mathematics and Statistics, Wuhan University, Wuhan, 430072, People’s Re-public of China

E-mail address: [email protected]

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