Rn : xn >0}, n ≥2m−1,with Navier boundary conditions ∆ju|∂Rn

http://jipam.vu.edu.au/

Volume 6, Issue 4, Article 119, 2005

ESTIMATES FOR THE GREEN FUNCTION AND CHARACTERIZATION OF A CERTAIN KATO CLASS BY THE GAUSS SEMIGROUP IN THE HALF SPACE

IMED BACHAR

DÉPARTEMENT DEMATHÉMATIQUES

FACULTÉ DESSCIENCES DETUNIS

CAMPUSUNIVERSITAIRE, 2092 TUNIS, TUNISIA. [email protected]

Received 06 July, 2005; accepted 11 August, 2005 Communicated by C. Bandle

ABSTRACT. We establish a3G-theorem for the Green functionsG_m,nof(−∆)^m(m ≥1)on Rⁿ+ := {x = (x₁, . . . , x_n) ∈ Rⁿ : x_n >0}, n ≥2m−1,with Navier boundary conditions

∆^ju|_∂Rn

+= 0,0≤j ≤m−1.

We exploit these results to define a certain Kato class of functions that we characterize by means of the Gauss semigroup onRⁿ+.

Key words and phrases: Green functions,3G-theorem, Kato class.

2000 Mathematics Subject Classification. 34B27.

1. INTRODUCTION

In [2], forn ≥ 3and [3], forn = 2, the authors have established interesting estimates for G(x, y),the Green function of the Laplace operator corresponding to zero Dirichlet boundary conditions in the half spaceRⁿ+ :={x = (x₁, . . . , x_n)∈ Rⁿ, x_n > 0}.In particular, they have proved the following form of the3G-Theorem:

Theorem 1.1. There exists a constantC >0such that for eachx, y, z ∈Rⁿ+

(1.1) G(x, z)G(z, y)

G(x, y) ≤C zn

x_nG(x, z) + zn

y_nG(y, z)

. They then introduced a class of functionsK_1,n(Rⁿ+)as follows:

Definition 1.1. A Borel measurable functionqinRⁿ+belongs to the classK_1,n(Rⁿ+)ifqsatisfies the following condition

limr→0 sup

x∈Rⁿ+

Z

(|x−y|≤r)∩Rⁿ+

y_n

x_nG(x, y)|q(y)|dy= 0.

ISSN (electronic): 1443-5756

c 2005 Victoria University. All rights reserved.

The author is especially thankful to Professor Habib Mâagli for valuable discussions.

He also thanks the referees for their careful reading of the paper.

204-05

They have studied the properties of functions belonging to this class.

In particular, in [2], the authors have showed the following characterization:

(1.2) q ∈K_1,n(Rⁿ+)⇐⇒lim

t→0 sup

x∈Rⁿ₊

Z

Rⁿ+

Z t 0

y_n

x_np(s, x, y)|q(y)|dsdy

!

= 0,

wherep(s, x, y)is the density of the Gauss semigroup onRⁿ+.

Note that similar characterizations have been already established in [1], (see also [5] and [8]) for the classical Kato classK_n(Rⁿ)defined as follows:

Definition 1.2. A Borel measurable function q in Rⁿ+ (n ≥ 3) belongs to the Kato class K_n Rⁿ+

ifqsatisfies the following condition

α→0lim sup

x∈Rⁿ₊

Z

Rⁿ+∩B(x,α)

1

|x−y|ⁿ⁻²|q(y)|dy

!

= 0.

For properties of functions inKn(Rⁿ+)we refer to [1], [5], [8], [10] and [11]).

Throughout this paper, we denote byGm,n(x, y) the Green’s function of the operatoru 7→

(−∆)^muonRⁿ+with Navier boundary conditions∆^ju|_∂

Rn +

= 0,0≤j ≤m−1form≥1and n≥max(3,2m−1).

The outline of the paper is as follows. In Section 2, we give explicitly the expression of G_m,n(x, y)and we prove some inequalities onG_m,n(x, y)including a3G-Theorem of the form (1.1). In Section 3, we introduce a class of functionsK_m,n(Rⁿ+)defined as follows:

Definition 1.3. A Borel measurable functionqinRⁿ+belongs to the classK_m,n(Rⁿ+)ifqsatisfies

(1.3) lim

r→0 sup

x∈Rⁿ+

Z

(|x−y|≤r)∩Rⁿ+

y_n xn

G_m,n(x, y)|q(y)|dy

!

= 0.

We then study properties of functions belonging to this class. In particular, we prove the following characterization forn >2m:

(1.4) q ∈K_m,n(Rⁿ+)⇔lim

t→0 sup

x∈Rⁿ+

Z

Rⁿ+

Z t 0

y_n xn

s^m−1p(s, x, y)|q(y)|dsdy

!

= 0,

which extends (1.2).

In order to simplify our statements, we define some convenient notations.

Notations:

• B(Rⁿ+)denotes the set of Borel measurable functions inRⁿ+.

• s∧t = min(s, t)ands∨t= max(s, t)fors, t∈R.

• Letf andg be two nonnegative functions on a setS.

We sayf g if there exists a constantc >0,such that f(x)≤cg(x)for allx∈S.

We sayf ∼g if

f gandg f.

• Letx, y ∈Rⁿ+.Puty= (y₁, . . . , yn−1,−y_n).Then we have

|x−y|² =|x−y|²+ 4x_ny_nand |x−y|² ≥(x_n+y_n)²,

which implies that

|x−y|² ∼ |x−y|²+x_ny_n (1.5)

x_n∨y_n ≤ |x−y|. (1.6)

The following properties will be used several times.

(i) Fors, t≥0,we have

(1.7) s∧t∼ st

s+t. (ii) Letλ, µ >0and0< γ≤1,then we have

1−t^λ ∼1−t^µ, fort∈[0,1].

(1.8)

log(1 +λt)∼log(1 +µt), fort≥0.

(1.9)

log 1 +t^λ

∼ 1∧t^λ

log (2 +t), fort ≥0.

(1.10)

log(1 +t)t^γ, fort≥0.

(1.11)

(iii) Leta >0,then we have

(1.12) 1−e^−a∼min(1, a).

2. INEQUALITIES FOR THEGREEN’SFUNCTION

In the sequel fort >0, xandy ∈Rⁿ+,we denote by p(t, x, y) = 1

(4πt)ⁿ² exp −|x−y|² 4t

!

−exp −|x−y|² 4t

!!

= 1

(4πt)ⁿ² exp −|x−y|² 4t

!

1−exp

−x_ny_n t

,

the density of the Gauss semigroup onRⁿ+.Then the Green’s function of∆with the Dirichlet condition on∂Rⁿ+is given by

(2.1) G(x, y) =

Z ∞ 0

p(t, x, y)dt.

Let G_m,n(x, y)be the Green’s function of the operator u 7→ (−∆)^mu on Rⁿ+ with Navier boundary conditions∆^ju|_∂Rn

+= 0,0≤j ≤m−1.

ThenG_m,n satisfies form ≥2, G_m,n(x, y) =

Z

Rⁿ+

· · · Z

Rⁿ+

G(x, z₁)G(z₁, z₂)· · ·G(zm−1, y)dz₁. . . dzm−1. Moreover, using the Fubini theorem, (2.1) and the Chapman-Kolmogorov identity we have

G_m,n(x, y)

= Z

Rⁿ+

· · · Z

Rⁿ+

G(x, z₁)G(z₁, z₂)dz₁)G(z₂, z₃)· · ·G(zm−1, y)dz₂. . . dzm−1

= Z

Rⁿ+

· · · Z

Rⁿ+

Z ∞ 0

Z ∞ 0

p(t₁+t₂, x, z₂)dt₁dt₂

G(z₂, z₃)· · ·G(zm−1, y)dz₂. . . dzm−1

= Z ∞

0

· · · Z ∞

0

p(t1 +t2+· · ·+tm, x, y)dt1. . . dtm.

A simple computation shows that for eachm≥1andx,y∈Rⁿ+

(2.2) G_m,n(x, y) = 1

(m−1)!

Z ∞ 0

s^m−1p(s, x, y)ds.

Next, we purpose to give an explicit expression forGm,n.

Letδ >0andx,y∈Rⁿ+such thatx6=y.Puta = ^|x−y|₂ andb= ^|x−y|₂ .Then we have Z δ

0

s^m−1p(s, x, y)ds =αm,n |x−y|^2m−n Z ∞

|x−y|2 4δ

r^{(n−2m2 )−1}e^−rdr (2.3)

− |x−y|^2m−n Z ∞

|x−y|2 4δ

r^{(n−2m2 )−1}e^−rdr

!

=β_m,n Z ∞

1 δ

ξ^{(n−2m2 )−1}(e^−a2ξ−e^−b2ξ)dξ, whereα_m,n andβ_m,n are some positive constants.

Hence, using this fact and (2.2) it follows that

δ→∞lim Z δ

0

s^m−1p(s, x, y)ds = (m−1)!G_m,n(x, y)<∞forx6=y⇐⇒2m−n <2.

Moreover, we deduce from (2.3) by lettingδ→ ∞,the following explicit expression ofG_m,n. Proposition 2.1. For eachx,y∈Rⁿ+,we have

G_m,n(x, y) =















a_m,n

1

|x−y|^n−2m − _|x−y|¹n−2m

, ifn >2m, b_m,nlog

1 + _|x−y|^4xnyn2

, ifn= 2m, c_m,n(|x−y| − |x−y|), ifn= 2m−1, wheream,n, bm,n andcm,nare some positive constants.

Corollary 2.2. For eachx,y∈Rⁿ+,we have (i) Forn >2m,

G_m,n(x, y)∼ x_ny_n

|x−y|^n−2m|x−y|² ∼ 1

|x−y|^n−2m

1∧ x_ny_n

|x−y|²

. (ii) Forn= 2m,

Gm,n(x, y)∼

1∧ xnyn

|x−y|²

log

2 + xnyn

|x−y|²

∼ xnyn

|x−y|² log 1 + |x−y|²

|x−y|²

! . (iii) Forn= 2m−1,

G_m,n(x, y)∼ x_ny_n

|x−y| ∼(x_ny_n)¹² 1∧ (x_ny_n)¹²

|x−y|

! .

Proof. The proof follows immediately from Proposition 2.1 and the statements (1.8), (1.5) and (1.7) forn >2morn = 2m−1and using further (1.9) – (1.10) forn = 2m.

Corollary 2.3. For eachx, y ∈Rⁿ+we have yn

x_nG_m,n(x, y)











1

|x−y|^n−2m, ifn >2m, 1 +G_m,n(x, y) ifn= 2m, x_n∧y_n ifn= 2m−1.

Remark 2.4. For eachx, y ∈Rⁿ+we have y_n

x_nG_m,n(x, y) 1

|x−y|^n−2m 1∧ y_n

x_n 2!

, if n > 2m.

Indeed, from Corollary 2.3, we have y_n

x_nG_m,n(x, y) 1

|x−y|^n−2m. Interchanging the role ofxandy,we get

G_m,n(x, y) y_n xn

· 1

|x−y|^n−2m, which implies the result.

The next lemma is crucial in this work.

Lemma 2.5 (see [7]). Letx, y ∈Rⁿ+.Then we have the following properties:

(1) Ifx_ny_n≤ |x−y|²,then

(x_n∨y_n)≤ (√ 5 + 1)

2 |x−y|.

(2) If|x−y|² ≤x_ny_n,then 3−√

5 2

!

x_n≤y_n≤ 3 +√ 5 2

! x_n. Corollary 2.6. For eachx,y∈Rⁿ+,we have

G_m,n(x, y) x_ny_n

|x−y|^n−2m+2, (2.4)

x_ny_n

(|x|+ 1)^n−2m+2(|y|+ 1)^n−2m+2 G_m,n(x, y), (2.5)

Gm,n(x, y) x_n∧y_n

|x−y|^n+1−2m. (2.6)

Proof. The assertions (2.4) and (2.5) follow from Corollary 2.2 and the fact that

|x−y| ≤ |x−y| ≤(|x|+ 1)(|y|+ 1)

and t

1 +t ≤log(1 +t)≤t, fort ≥0.

To prove (2.6) we claim that

(2.7) G_m,n(x, y) x_n

|x−y|^n+1−2m. Indeed, we have the following cases:

Case 1. Ifn > 2m orn = 2m−1,the inequality (2.7) follows from Corollary 2.2, (1.6) and the fact that|x−y| ≥ |x−y|.

Case 2. Ifn= 2m,then we have the following subcases:

(1) If|x−y|² ≤x_ny_n,then by Lemma 2.5, we getx_n∼y_n. Using this fact, Proposition 2.1, (1.9) and (1.11) we deduce that

G_m,n(x, y)∼log

1 + cx²_n

|x−y|²

, (wherec >0), xn

|x−y|.

(2) Ifxnyn≤ |x−y|²,then Lemma 2.5 gives that(xn∨yn) |x−y|.

Hence from (2.4), we deduce that

G_m,n(x, y) x_ny_n

|x−y|² x_n

|x−y|.

This proves (2.7). Interchange the role ofxandy,we obtain (2.6).

Proposition 2.7. a) For eacht >0,and allx, y∈Rⁿ+,we have Z t

0

s^m−1p(s, x, y)ds G_m,n(x, y).

b) Lett >0andx, y ∈Rⁿ+.Then G_m,n(x, y)

Z t 0

s^m−1p(s, x, y)ds, provided

i) n >2mand|x−y| ≤2√ t; or ii) n= 2mand|x−y| ≤2√

t; or iii) n= 2m−1and|x−y| ≤2√

t.

Proof. Lett >0andx, y∈Rⁿ+.Thena)follows immediately from (2.2).

To proveb)we distinguish three cases.

i)Forn > 2m,using (1.12) and the fact that fora, b∈(0,∞)we have (1∧ab)≥(1∧a)(1∧b),

then there existsC >0such that for|x−y| ≤2√ t, Z t

0

s^m−1p(s, x, y)ds≥C Z t

0

1

s^n2+1−m exp −|x−y|² 4s

!

1∧ x_ny_n s

ds

≥ C

|x−y|^n−2m Z ∞

|x−y|2 4t

r^n2−1−me^−r

1∧ 4rx_ny_n

|x−y|²

dr

≥ C

|x−y|^n−2m

1∧ x_ny_n

|x−y|² Z ∞

|x−y|2 4t

r^n2−1−me^−r(1∧4r)dr

≥ C

|x−y|^n−2m

1∧ x_ny_n

|x−y|² Z ∞

1

r^n2−1−me^−rdr

≥ C

|x−y|^n−2m

1∧ x_ny_n

|x−y|²

. Hence the result follows from Corollary 2.2.

ii)Forn = 2m,by (2.3), there existsC > 0such that for|x−y| ≤2√ t Z t

0

s^m−1p(s, x, y)ds =α_m,n

Z ^|x−y|

2 4t

|x−y|2 4t

e^−r

r dr ≥Clog |x−y|²

|x−y|²

! . Hence the result follows from Proposition 2.1.

iii)Letn= 2m−1, t >0andx, y∈Rⁿ+such that|x−y| ≤2√ t.

Puta = ^|x−y|

2√

t andb = ^|x−y|

2√

t .Then using (2.3), we obtain I :=

Z t 0

s^m−1p(s, x, y)ds= 2αm,n

√ t

a

Z ∞ a²

r⁻³² e^−rdr−b Z ∞

b²

r⁻³² e^−rdr

. Now since forα >0,we have

Z ∞ α²

r⁻³² e^−rdr = 2 e^−α2

α −

Z ∞ α²

r⁻¹² e^−rdr

! , we deduce that

I = 4α_m,n√ t

(e^−a2 −e^−b2) +b Z ∞

b²

r⁻¹² e^−rdr−a Z ∞

a²

r⁻¹² e^−rdr

= 4α_m,n√ t

"

(b−a) Z ∞

b²

r⁻¹² e^−rdr+ Z b²

a²

r⁻¹² e^−r(r¹² −a)dr

# . Hence

I ≥4α_m,n√

t(b−a) Z ∞

b²

r⁻¹² e^−rdr.

That is

I ≥2α_m,n(|x−y| − |x−y|) Z ∞

|x−y|2 4t

r⁻¹² e^−rdr

≥2α_m,n(|x−y| − |x−y|) Z ∞

1

r⁻¹² e^−rdr.

The result follows from Proposition 2.1.

Next we purpose to prove thatG_m,n satisfies (1.1).

3G-Theorem. Forx, y, z ∈Rⁿ+,we have G_m,n(x, z)G_m,n(z, y)

G_m,n(x, y) z_n

x_nG_m,n(x, z) + z_n

y_nG_m,n(y, z)

. Proof. To prove the inequality, we denote by A(x, y) := _G^xnyn

m,n(x,y) and we claim that A is a quasi-metric, that is for eachx, y, z ∈Rⁿ+,

(2.8) A(x, y)A(x, z) +A(y, z).

To this end, we observe that by using Corollary 2.2 and Lemma 2.5, the claim can be proved by similar arguments as in [2], forn >2mand as in [3], forn= 2m.

To prove (2.8), forn = 2m−1,we derive from Corollary 2.2 that A(x, y)∼(|x−y|²∨x_ny_n)¹² ∼ |x−y|.

Now since|x−z |≤|x−z|,we deduce that A(x, y) |x−z|+|z−y|

|x−z|+|z−y| (A(x, z) +A(y, z)).

3. THECLASSK_m,n(Rⁿ+)

Next we purpose to study and to characterize the classK_m,n(Rⁿ+)forn >2m.

We recall that for0 < α < n,we say that a Borel measurable functionq inRⁿ+ belongs to the classKe_α,n(Rⁿ+)(see [6]) ifqsatisfies the following condition

(3.1) lim

r→0 sup

x∈Rⁿ+

Z

(|x−y|≤r)∩Rⁿ+

|q(y)|

|x−y|^n−αdy = 0.

The usual Kato classK_n(Rⁿ+),corresponds toα= 2.

Remark 3.1. Let n > 2m. Using Corollary 2.3, the class K_m,n(Rⁿ+) obviously includes the classKe_2m,n(Rⁿ+).In particular,K_n(Rⁿ+)⊂K_m,n(Rⁿ+).

Example 3.1. Suppose that forp > n

2m >1,we have M₀ = sup

x∈Rⁿ+

Z

(|x−y|≤1)∩Rⁿ+

min yn

x_n 2p

,1

!

|q(y)|^pdy <∞, thenq ∈K_m,n(Rⁿ+).

Indeed, let0< r <1andx∈Rⁿ+,then using Remark 2.4 and the Hölder inequality we get Z

(|x−y|≤r)∩Rⁿ+

y_n

x_nG_m,n(x, y)|q(y)|dy

Z

(|x−y|≤r)∩Rⁿ+

min y_n

xn

2

,1

! 1

|x−y|^n−2m |q(y)|dy

Z

(|x−y|≤r)∩Rⁿ+

min yn

x_n 2p

,1

!

|q(y)|^pdy

!¹_p

× Z

(|x−y|≤r)∩Rⁿ+

1

|x−y|^{p−1p (n−2m)}dy

!^p−1_p . Hence

sup

x∈Rⁿ+

Z

(|x−y|≤r)∩Rⁿ₊

y_n

x_nGm,n(x, y)q(y)dyM

1 p

0 r

2mp−n

p →0asr→0.

Proposition 3.2. Letp >max _2mⁿ ,1

andf ∈L^p Rⁿ+

.Then y7→ f(y)

(|y|+ 1)^µ−λy_n^λ ∈K_m,n(Rⁿ+) provided

i) n >2m, λ≤2andλ <2m−ⁿ_p andµ≥max(0, λ) or ii) n = 2mandλ <min(2,2m− ⁿ_p)≤µ or

iii) n = 2m−1, λ≤2andλ <2m− ⁿ_p andµ≥max(1, λ).

Proof. Letp > max _2mⁿ ,1

andq ≥1such that ¹_p +¹_q = 1.

Forf ∈L^p(Rⁿ+), x∈Rⁿ+and0< r <1,put I =I(x, r) :=

Z

B(x,r)∩Rⁿ₊

y_n

x_nG_m,n(x, y) |f(y)|

(|y|+ 1)^µ−λy_n^λdy.

Note that if|x−y| ≤r,then(|x|+ 1)∼(|y|+ 1).So, we distinguish the following cases:

Case 1. n > 2m. Assume that λ ≤ 2 andλ < 2m− ⁿ_p.Let µ ≥ max(0, λ) and putλ⁺ = max(λ,0).Then using Corollary 2.2, (1.6) and the fact that |x−y| ≤ |x−y|,we deduce by the Hölder inequality that

I Z

B(x,r)∩Rⁿ+

|f(y)|

|x−y|^n+λ+−2mdy kfk_p Z

B(x,r)∩Rⁿ+

1

|x−y|^{(n+λ+−2m)q}dy

!¹_q

r^{2m−np−λ+}, which tends to zero ifr→0.

Case 2. n = 2m.Assume thatλ <min

2,2m− ⁿ_p

≤µ.

Using Proposition 2.1 and the Hölder inequality, we deduce that

I kfk_p Z

(|x−y|≤r)∩Rⁿ₊

y_n x_n

q log

1 + 4x_ny_n

|x−y|² q

1

(|y|+ 1)^(µ−λ)qy^λqn

dy

!¹_q

Z

(|x−y|≤r)∩D1

y_n x_n

q log

1 + 4x_ny_n

|x−y|² q

1

(|y|+ 1)^(µ−λ)qyn^λq

dy ¹_q

+ Z

(|x−y|≤r)∩D₂

y_n x_n

q log

1 + 4x_ny_n

|x−y|² q

1

(|y|+ 1)^(µ−λ)qy^λqn

dy ¹_q

=I₁+I₂, where

D₁ ={y∈Rⁿ+:x_ny_n ≤ |x−y|²} and D₂ ={y∈Rⁿ+ :|x−y|² ≤x_ny_n}.

So, using thatlog(1 +t)≤t, fort ≥0and Lemma 2.5, we obtain

I1 Z

(|x−y|≤r)∩D₁

y^(2−λ)qn

|x−y|^2qdy

!¹_q

Z

(|x−y|≤r)∩D₁ 1

|x−y|^λqdy ¹_q

r^2m−np−λ, which converges to zero asr→0.

On the other hand, from Lemma 2.5 and the fact that(|x|+ 1)∼(|y|+ 1), we obtain

I2 1

x^λ_n(|x|+ 1)^(µ−λ) Z

(|x−y|≤r)∩D₂

log

1 + (cx_n)²

|x−y|² q

dy ¹_q

,

wherec= 1 +√

5. Letγ ∈] max(0, λ),min(2,2m− ⁿ_p)[.

Sincelog(1 +t²)t^γ,fort≥0, then I₂ x^γ−λ_n

(|x|+ 1)^µ−λ Z

(|x−y|≤r)∩D2

1

|x−y|^γqdy ¹_q

Z

(|x−y|≤r)∩D2

1

|x−y|^γqdy ¹_q

r^2m−np−γ, which converges to zero asr→0.

Case 3. n = 2m−1.Assume thatλ≤2andλ <2m− ⁿ_p.Letµ≥max(1, λ).

Using Corollary 2.2 and the Hölder inequality, we obtain I



 Z

B(x,r)∩D1

yn^(2−λ)q

|x−y|^q(|y|+ 1)^(µ−λ)qdy

!¹_q +

Z

B(x,r)∩D2

yn^(2−λ)q

|x−y|^q(|y|+ 1)^(µ−λ)qdy

!¹_q



=I₁+I₂.

Now, ify∈D₁,then|x−y| ∼ |x−y|and so I₁

Z

B(x,r)∩D1

1

|x−y|^(λ−1)qdy

!¹_q

r^2m−np−λ, which tends to zero asr →0.

On the other hand, if y ∈ D2, then |x−y|² ∼ xnyn and by Lemma 2.5, we have further x_n∼y_n.This implies that

I₂ x^2−λ_n x_n(|x|+ 1)^µ−λ

Z

B(x,r)∩D2

dy ¹_q

x^1−λ_n

(|x|+ 1)^µ−λ (r∧cx_n)^nq

r^nq, which converges to zero asr →0.

The proof of the next results are similar to the case m = 1 and n ≥ 3, which has been considered in [2]. Since reference [2] is not available, I have chosen to reproduce it here.

Proposition 3.3. Letq∈K_m,n(Rⁿ+),then for each compactL⊆Rⁿwe have sup

x∈Rⁿ+

Z

(x+L)∩Rⁿ+

y_n²

1 +x_ny_n|q(y)|dy <∞.

Proof. Letq ∈K_m,n(Rⁿ+),then by (1.3) there existsr >0such that sup

x∈Rⁿ+

Z

(|x−y|≤r)∩Rⁿ₊

y_n

x_nGm,n(x, y)|q(y)|dy ≤1.

Leta₁, a₂, . . . , a_p ∈Rⁿ+∩Lsuch thatRⁿ+∩L⊆ ∪

1≤i≤pB(a_i, r).

Since fora, b ∈ (0,∞),we have _1+ab^b ≤ 1 +|a−b|,then for eachx, y, z ∈ Rⁿ+ it follows that 1 + (xn+zn)yn

1 +x_ny_n ≤[1 +z_n(1 +|x_n−y_n|)].

Using this fact and Corollary 2.2, we obtain:

Forn >2m, y²_n

1 +x_ny_n [1 +z_n(1 +|x_n−y_n|)]

1 + (x_n+z_n)y_n |x+z−y|^n−2m

×

|x+z−y|²+ 4(x_n+z_n)y_n y_n

(x_n+z_n)G_m,n(x+z, y).

Forn = 2m,using further that _1+t^t ≤log(1 +t),∀t ≥0,we have y²_n

1 +xnyn

[1 +z_n(1 +|x_n−y_n|)]

1 + (xn+zn)yn

×[|x+z−y|² + 4(x_n+z_n)y_n] yn

(x_n+z_n)G_m,n(x+z, y).

Forn = 2m−1, y²_n

1 +x_ny_n [1 +z_n(1 +|x_n−y_n|)]

1 + (x_n+z_n)y_n

×[|x+z−y|²+ 4(x_n+z_n)y_n]¹² y_n

(xn+zn)G_m,n(x+z, y).

Now, ifz ∈Land|x+z−y| ≤r,then y²_n

1 +x_ny_n y_n

(x_n+z_n)G_m,n(x+z, y).

Hence Z

(x+L)∩Rⁿ+

y_n²

1 +x_ny_n|q(y)|dy

p

X

i=1

Z

(|x+a_i−y|≤r)∩Rⁿ+

yn

(x_n+ (a_i)_n)G_m,n(x+a_i, y)|q(y)|dy p.

So

sup

x∈Rⁿ+

Z

(x+L)∩Rⁿ+

y_n² 1 +xnyn

|q(y)|dy <∞.

Corollary 3.4. Letq∈K_m,n(Rⁿ+).Then we have forM >0,

Z

(|y|≤M)∩Rⁿ+

y²_n|q(y)|dy <∞.

Proposition 3.5. Letq∈K_m,n(Rⁿ+),then for each fixedα >0,we have

(3.2) sup

t≤1

sup

x∈Rⁿ₊

Z

(|x−y|>α)∩Rⁿ+

y_n

x_np(t, x, y)|q(y)|dy :=M(α)<∞.

Proof. Letq ∈ K_m,n(Rⁿ+),0 < t ≤ 1and without loss of generality assume that0 < α < 1.

Then by (1.12) and (1.7), it follows that sup

x∈Rⁿ+

Z

(|x−y|>α)∩Rⁿ+

yn

x_np(t, x, y)|q(y)|dy

1

tⁿ²⁺¹e^−α

2 8t sup

x∈Rⁿ+

Z

Rⁿ₊

exp −|x−y|² 8

! y_n²

1 +x_ny_n|q(y)|dy.

To conclude, it is sufficient to prove that sup

x∈Rⁿ+

Z

Rⁿ₊

exp −|x−y|² 8

! y²_n

1 +x_ny_n|q(y)|dy <∞.

Indeed, using Proposition 3.3, we have sup

x∈Rⁿ₊

Z

(x+B(0,1))∩Rⁿ+

y_n²

1 +x_ny_n|q(y)|dy:=M <f ∞.

Now since any ball B(0, k), of radius k ≥ 1 in Rⁿ can be covered by A_nkⁿ := α(n) balls of radius 1, where A_n is a constant depending only on n (see [5, p. 67]), then there exists a₁, a₂, . . . , a_α(n) ∈Rⁿ+ such that

Rⁿ+∩B(0, k)⊆ ∪

1≤i≤α(n)B(a_i,1).

Using the fact that for eachx, y, z ∈Rⁿ+, 1 + (x_n+z_n)y_n

1 +x_ny_n ≤[1 +z_n(1 +|x_n−y_n|)], it follows that for allx∈Rⁿ+andk ≥1 :

Z

(x+B(0,k))∩Rⁿ+

y_n²

1 +x_ny_n|q(y)|dy

α(n)

X

i=1

Z

B(x+ai,1)∩Rⁿ+

y²_n

1 +x_ny_n|q(y)|dy

α(n)

X

i=1

Z

B(x+ai,1)∩Rⁿ₊

y_n²

1 + (x_n+ (a_i)_n)y_n|q(y)|dy A_nkⁿM .f

Hence for allx∈Rⁿ+,we have Z

Rⁿ+

exp −|x−y|² 8

! y_n² 1 +xnyn

|q(y)|dy

∞

X

k=0

exp

−α²k² 8

Z

[kα≤|x−y|≤(k+1)α]∩Rⁿ+

y_n²

1 +x_ny_n |q(y)|dy A_nMf

∞

X

k=0

(k+ 1)ⁿexp

−α²k² 8

<∞.

Thus

sup

x∈Rⁿ₊

Z

Rⁿ+

exp −|x−y|² 8

! y²_n

1 +x_ny_n|q(y)|dy <∞,

which completes the proof.

Theorem 3.6. Letn >2mandq∈ B(Rⁿ+). Then the following assertions are equivalent:

(1) q ∈K_m,n(Rⁿ+) (2) lim

t→0 sup

x∈Rⁿ+

R

Rⁿ+

Rt 0

y_n xn

s^m−1p(s, x, y)|q(y)|dsdy= 0.

Proof. 2)⇒1)Assume that limt→0 sup

x∈Rⁿ+

Z

Rⁿ+

Z t 0

y_n

x_ns^m−1p(s, x, y)|q(y)|dsdy = 0.

Then by Proposition 2.7, there existsc >0such that forα >0we have Z

(|x−y|≤α)∩Rⁿ₊

y_n

x_nGm,n(x, y)|q(y)|dy≤c Z

Rⁿ₊

Z ^α

2 4

0

y_n

x_ns^m−1p(s, x, y)|q(y)|dsdy, which shows that the functionqsatisfies (1.3).

Conversely suppose thatq∈K_m,n(Rⁿ+). Letε >0,then there exists0< α <1such that sup

x∈Rⁿ+

Z

(|x−y|≤α)∩Rⁿ₊

y_n

x_nG_m,n(x, y)|q(y)|dy≤ε.

On the other hand, using Proposition 2.7 and (3.2), we have for0< t <1 Z

Rⁿ+

Z t 0

y_n

x_ns^m−1p(s, x, y)|q(y)|dsdy

Z

(|x−y|≤α)∩Rⁿ+

Z t 0

y_n

x_ns^m−1p(s, x, y)|q(y)|dsdy +

Z

(|x−y|>α)∩Rⁿ+

Z t 0

y_n

x_ns^m−1p(s, x, y)|q(y)|dsdy

Z

(|x−y|≤α)∩Rⁿ₊

y_n xn

G_m,n(x, y)|q(y)|dy +

Z t 0

Z

(|x−y|>α)∩Rⁿ+

y_n xn

p(s, x, y)|q(y)|dyds ε+tM(α),

which implies that

limt→0 sup

x∈Rⁿ₊

Z

Rⁿ+

Z t 0

y_n

x_ns^m−1p(s, x, y)|q(y)|dsdy = 0.

Corollary 3.7. Letn >2mandq∈ B(Rⁿ+). Forα >0andx∈Rⁿ+,put

G_αq(x) :=

Z

Rⁿ+

Z ∞ 0

e^−αsy_n

x_ns^m−1p(s, x, y)|q(y)|dsdy.

Then

q ∈K_m,n(Rⁿ+)⇔ lim

α→+∞kG_αqk_∞ = 0, wherekG_αqk_∞= sup

x∈Rⁿ₊

|G_αq(x)|.

Proof. (see [9]). Letq ∈K_m,n(Rⁿ+),α >0and put a(α) = sup

x∈Rⁿ+

Z _α¹

0

Z

Rⁿ+

y_n xn

s^m−1p(s, x, y)|q(y)|dyds.

Then we have

Gαq(x) = Z ∞

0

αe^−αt

"

Z t 0

Z

Rⁿ₊

yn

x_ns^m−1p(s, x, y)|q(y)|dyds

# dt

= Z ∞

0

e^−t

"

Z _α^t

0

Z

Rⁿ+

y_n

x_ns^m−1p(s, x, y)|q(y)|dyds

# dt.

It follows that, 1

ea(α)≤ kG_αqk_∞.

On the other hand, fort >0andk ∈Nsuch thatk≤t < k+ 1,we have G_αq(x)≤

m

X

k=0

Z ∞ 0

e^−t

"

Z ^k+1_α

k α

Z

Rⁿ+

y_n

x_ns^m−1p(s, x, y)|q(y)|dyds

# dt

≤a(α) Z ∞

0

e^−t(m+ 1)dt

≤a(α) Z ∞

0

e^−t(t+ 1)dt = 2a(α), which gives that 1

ea(α)≤ kG_αqk_∞≤2a(α).

Hence the results follow from Theorem 3.6.

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