Journal of Inequalities in Pure and
Applied Mathematics
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
D ÉPARTEMENT DE M ATHÉMATIQUES
FACULTÉ DES S CIENCES DE T UNIS
C AMPUS U NIVERSITAIRE , 2092 T UNIS , T UNISIA .
Imed.Bachar@ipeit.rnu.tn
Received 06 July, 2005; accepted 11 August, 2005
Communicated by C. Bandle
A BSTRACT. We establish a 3G-theorem for the Green functions Gm,n of (−∆)m (m ≥ 1) on
Rn+ := {x = (x1 , . . . , xn ) ∈ Rn : xn > 0}, n ≥ 2m − 1, with Navier boundary conditions
∆j u |∂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 on Rn+ .
Key words and phrases: Green functions, 3G-theorem, Kato class.
2000 Mathematics Subject Classification. 34B27.
1. I NTRODUCTION
In [2], for n ≥ 3 and [3], for n = 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 space Rn+ := {x = (x1 , . . . , xn ) ∈ Rn , xn > 0}. In particular, they have
proved the following form of the 3G-Theorem:
Theorem 1.1. There exists a constant C > 0 such that for each x, y, z ∈ Rn+
G(x, z)G(z, y)
zn
zn
(1.1)
≤C
G(x, z) + G(y, z) .
G(x, y)
xn
yn
n
They then introduced a class of functions K1,n (R+ ) as follows:
Definition 1.1. A Borel measurable function q in Rn+ belongs to the class K1,n (Rn+ ) if q satisfies
the following condition
Z
yn
lim sup
G(x, y) |q(y)| dy = 0.
r→0 x∈Rn (|x−y|≤r)∩Rn xn
+
+
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
2
I MED BACHAR
They have studied the properties of functions belonging to this class.
In particular, in [2], the authors have showed the following characterization:
!
Z Z t
y
n
(1.2)
q ∈ K1,n (Rn+ ) ⇐⇒ lim sup
p(s, x, y) |q(y)| dsdy = 0,
t→0 x∈Rn Rn 0 xn
+
+
where p(s, x, y) is the density of the Gauss semigroup on Rn+ .
Note that similar characterizations have been already established in [1], (see also [5] and [8])
for the classical Kato class Kn (Rn ) defined as follows:
n
Definition
1.2. A Borel measurable function q in R+ (n ≥ 3) belongs to the Kato class
n
Kn R+ if q satisfies the following condition
!
Z
1
lim sup
|q(y)|dy = 0.
α→0 x∈Rn Rn ∩B(x,α) |x − y|n−2
+
+
For properties of functions in Kn (Rn+ ) we refer to [1], [5], [8], [10] and [11]).
Throughout this paper, we denote by Gm,n (x, y) the Green’s function of the operator u 7→
(−∆)m u on Rn+ with Navier boundary conditions ∆j u |∂Rn = 0, 0 ≤ j ≤ m − 1 for m ≥ 1 and
+
n ≥ max(3, 2m − 1).
The outline of the paper is as follows. In Section 2, we give explicitly the expression of
Gm,n (x, y) and we prove some inequalities on Gm,n (x, y) including a 3G-Theorem of the form
(1.1). In Section 3, we introduce a class of functions Km,n (Rn+ ) defined as follows:
Definition 1.3. A Borel measurable function q in Rn+ belongs to the class Km,n (Rn+ ) if q satisfies
Z
(1.3)
lim
r→0
sup
x∈Rn
+
(|x−y|≤r)∩Rn
+
yn
Gm,n (x, y) |q(y)| dy
xn
!
= 0.
We then study properties of functions belonging to this class. In particular, we prove the
following characterization for n > 2m:
!
Z Z t
yn m−1
n
(1.4)
q ∈ Km,n (R+ ) ⇔ lim sup
s
p(s, x, y) |q(y)| dsdy = 0,
t→0 x∈Rn Rn 0 xn
+
+
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 in Rn+ .
• s ∧ t = min(s, t) and s ∨ t = max(s, t) for s, t ∈ R.
• Let f and g be two nonnegative functions on a set S.
We say f g if there exists a constant c > 0, such that
f (x) ≤ cg(x) for all x ∈ S.
We say f ∼ g if
f g and g f.
• Let x, y ∈ Rn+ . Put y = (y1 , . . . , yn−1 , −yn ). Then we have
|x − y|2 = |x − y|2 + 4xn yn and |x − y|2 ≥ (xn + yn )2 ,
J. Inequal. Pure and Appl. Math., 6(4) Art. 119, 2005
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G REEN FUNCTION AND K ATO CLASS
3
which implies that
(1.5)
|x − y|2 ∼ |x − y|2 + xn yn
(1.6)
xn ∨ yn ≤ |x − y| .
The following properties will be used several times.
(i) For s, t ≥ 0, we have
st
s∧t∼
.
s+t
(ii) Let λ, µ > 0 and 0 < γ ≤ 1, then we have
(1.7)
(1.8)
1 − tλ ∼ 1 − tµ , for t ∈ [0, 1].
(1.9)
(1.10)
log(1 + λt) ∼ log(1 + µt), for t ≥ 0.
log 1 + tλ ∼ 1 ∧ tλ log (2 + t) , for t ≥ 0.
(1.11)
log(1 + t) tγ , for t ≥ 0.
(iii) Let a > 0, then we have
1 − e−a ∼ min(1, a).
(1.12)
2. I NEQUALITIES FOR
THE
G REEN ’ S F UNCTION
In the sequel for t > 0, x and y ∈ Rn+ , we denote by
!
!!
1
|x − y|2
|x − y|2
p(t, x, y) =
exp −
− exp −
n
4t
4t
(4πt) 2
!
x y 1
|x − y|2 n n
=
−
1 − exp −
,
n exp
2
4t
t
(4πt)
the density of the Gauss semigroup on Rn+ . Then the Green’s function of ∆ with the Dirichlet
condition on ∂Rn+ is given by
Z ∞
(2.1)
G(x, y) =
p(t, x, y)dt.
0
Let Gm,n (x, y) be the Green’s function of the operator u 7→ (−∆)m u on Rn+ with Navier
boundary conditions ∆j u |∂Rn = 0, 0 ≤ j ≤ m − 1.
+
Then Gm,n satisfies for m ≥ 2,
Z
Z
Gm,n (x, y) =
···
G(x, z1 )G(z1 , z2 ) · · · G(zm−1 , y)dz1 . . . dzm−1 .
Rn
+
Rn
+
Moreover, using the Fubini theorem, (2.1) and the Chapman-Kolmogorov identity we have
Gm,n (x, y)
Z
Z
=
···
Rn
+
G(x, z1 )G(z1 , z2 )dz1 )G(z2 , z3 ) · · · G(zm−1 , y)dz2 . . . dzm−1
Rn
+
Z
Z
Z
∞
Z
···
=
Rn
+
Z ∞
0
p(t1 + t2 , x, z2 )dt1 dt2 G(z2 , z3 ) · · · G(zm−1 , y)dz2 . . . dzm−1
Rn
Z +∞
0
0
p(t1 + t2 + · · · + tm , x, y)dt1 . . . dtm .
···
=
∞
0
J. Inequal. Pure and Appl. Math., 6(4) Art. 119, 2005
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4
I MED BACHAR
A simple computation shows that for each m ≥ 1 and x, y ∈ Rn+
Z ∞
1
(2.2)
Gm,n (x, y) =
sm−1 p(s, x, y)ds.
(m − 1)! 0
Next, we purpose to give an explicit expression for Gm,n .
Let δ > 0 and x, y ∈ Rn+ such that x 6= y. Put a = |x−y|
and b = |x−y|
. Then we have
2
2
Z δ
Z ∞
n−2m
2m−n
m−1
(2.3)
s
p(s, x, y)ds = αm,n |x − y|
r( 2 )−1 e−r dr
2
|x−y|
4δ
0
− |x−y|2m−n
∞
Z
ξ(
= βm,n
Z
!
∞
|x−y|2
4δ
n−2m
)−1
2
r
)−1
( n−2m
2
2
2
e−r dr
(e−a ξ − e−b ξ )dξ,
1
δ
where αm,n and βm,n are some positive constants.
Hence, using this fact and (2.2) it follows that
Z δ
lim
sm−1 p(s, x, y)ds = (m − 1)!Gm,n (x, y) < ∞ for x 6= y ⇐⇒ 2m − n < 2.
δ→∞
0
Moreover, we deduce from (2.3) by letting δ → ∞, the following explicit expression of Gm,n .
Proposition 2.1. For each x, y ∈ Rn+ , we have

1
1


a
−
, if n > 2m,
n−2m
n−2m
m,n

|x−y|
|x−y|


4xn yn
Gm,n (x, y) =
b
log
1
+
,
if n = 2m,
2
m,n

|x−y|



 c (|x−y| − |x − y| ),
if n = 2m − 1,
m,n
where am,n , bm,n and cm,n are some positive constants.
Corollary 2.2. For each x, y ∈ Rn+ , we have
(i) For n > 2m,
1
xn yn
Gm,n (x, y) ∼
n−2m
2 ∼
|x − y|
|x − y|
|x − y|n−2m
xn yn
1∧
|x − y|2
.
(ii) For n = 2m,
xn yn
Gm,n (x, y) ∼
log 2 +
|x − y|2
!
xn yn
|x − y|2
∼
log 1 +
.
|x − y|2
|x − y|2
xn yn
1∧
|x − y|2
(iii) For n = 2m − 1,
Gm,n (x, y) ∼
1
xn yn
∼ (xn yn ) 2
|x − y|
1∧
1
2
(xn yn )
|x − y|
!
.
Proof. The proof follows immediately from Proposition 2.1 and the statements (1.8), (1.5) and
(1.7) for n > 2m or n = 2m − 1 and using further (1.9) – (1.10) for n = 2m.
J. Inequal. Pure and Appl. Math., 6(4) Art. 119, 2005
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G REEN FUNCTION AND K ATO CLASS
5
Corollary 2.3. For each x, y ∈ Rn+ we have

1
,
if n > 2m,

|x−y|n−2m


yn
1 + Gm,n (x, y) if n = 2m,
Gm,n (x, y) 
xn


xn ∧ yn
if n = 2m − 1.
Remark 2.4. For each x, y ∈ Rn+ we have
yn
1
Gm,n (x, y) xn
|x − y|n−2m
1∧
yn
xn
2 !
, if n > 2m.
Indeed, from Corollary 2.3, we have
yn
1
Gm,n (x, y) .
xn
|x − y|n−2m
Interchanging the role of x and y, we get
yn
1
·
Gm,n (x, y) ,
xn |x − y|n−2m
which implies the result.
The next lemma is crucial in this work.
Lemma 2.5 (see [7]). Let x, y ∈ Rn+ . Then we have the following properties:
(1) If xn yn ≤ |x − y|2 , then
√
( 5 + 1)
|x − y|.
(xn ∨ yn ) ≤
2
(2) If |x − y|2 ≤ xn yn , then
√ !
√ !
3− 5
3+ 5
xn ≤ yn ≤
xn .
2
2
Corollary 2.6. For each x, y ∈ Rn+ , we have
Gm,n (x, y) (2.4)
(2.5)
(|x| +
xn yn
,
|x − y|n−2m+2
xn yn
n−2m+2
1)
(|y|
(2.6)
Gm,n (x, y),
+ 1)n−2m+2
xn ∧ yn
.
Gm,n (x, y) |x − y|n+1−2m
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
≤ log(1 + t) ≤ t,
1+t
for t ≥ 0.
To prove (2.6) we claim that
(2.7)
Gm,n (x, y) xn
.
|x − y|n+1−2m
Indeed, we have the following cases:
J. Inequal. Pure and Appl. Math., 6(4) Art. 119, 2005
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6
I MED BACHAR
Case 1. If n > 2m or n = 2m − 1, the inequality (2.7) follows from Corollary 2.2, (1.6) and
the fact that |x − y| ≥ |x − y| .
Case 2. If n = 2m, then we have the following subcases:
(1) If |x − y|2 ≤ xn yn , then by Lemma 2.5, we get xn ∼ yn .
Using this fact, Proposition 2.1, (1.9) and (1.11) we deduce that
cx2n
Gm,n (x, y) ∼ log 1 +
, (where c > 0),
|x − y|2
xn
.
|x − y|
(2) If xn yn ≤ |x − y|2 , then Lemma 2.5 gives that (xn ∨ yn ) |x − y|.
Hence from (2.4), we deduce that
xn yn
xn
Gm,n (x, y) .
2 |x − y|
|x − y|
This proves (2.7). Interchange the role of x and y, we obtain (2.6).
Proposition 2.7.
a) For each t > 0, and all x, y ∈ Rn+ , we have
Z t
sm−1 p(s, x, y)ds Gm,n (x, y).
b) Let t > 0 and x, y ∈
0
n
R+ .
Then
Z
Gm,n (x, y) t
sm−1 p(s, x, y)ds,
0
provided
√
i) n > 2m and |x − y| ≤ 2√t; or
ii) n = 2m and |x − y| ≤ 2 t; √or
iii) n = 2m − 1 and |x − y| ≤ 2 t.
Proof. Let t > 0 and x, y ∈ Rn+ . Then a) follows immediately from (2.2).
To prove b) we distinguish three cases.
i) For n > 2m, using (1.12) and the fact that for a, b ∈ (0, ∞) we have
(1 ∧ ab) ≥ (1 ∧ a)(1 ∧ b),
√
then there exists C > 0 such that for |x − y| ≤ 2 t,
!
Z t
Z t
2
1
|x
−
y|
xn yn exp
−
1
∧
ds
sm−1 p(s, x, y)ds ≥ C
n
+1−m
4s
s
0
0 s2
Z ∞
n
4rxn yn
C
−1−m
−r
≥
r2
e
1∧
dr
2
|x − y|n−2m |x−y|
|x − y|2
4t
Z ∞
n
C
xn yn
−1−m −r
2
≥
1∧
r
e (1 ∧ 4r)dr
n−2m
2
2
|x−y|
|x − y|
|x − y|
4t
Z ∞
n
C
xn yn
≥
1∧
r 2 −1−m e−r dr
n−2m
2
|x − y|
|x − y|
1
xn yn
C
≥
1∧
.
n−2m
|x − y|
|x − y|2
Hence the result follows from Corollary 2.2.
J. Inequal. Pure and Appl. Math., 6(4) Art. 119, 2005
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G REEN FUNCTION AND K ATO CLASS
7
√
ii) For n = 2m, by (2.3), there exists C > 0 such that for |x − y| ≤ 2 t
Z
t
sm−1 p(s, x, y)ds = αm,n
0
Z
|x−y|2
4t
|x−y|2
4t
| x−y|2
|x − y|2
e−r
dr ≥ C log
r
!
.
Hence the result follows from Proposition 2.1.
√
iii) Let n = 2m − 1, t > 0 and x, y ∈ Rn+ such that |x − y| ≤ 2 t.
|x−y|
√ and b = √ . Then using (2.3), we obtain
Put a = |x−y|
2 t
2 t
Z ∞
Z t
Z
√
−3
−r
m−1
I :=
s
p(s, x, y)ds = 2αm,n t a
r 2 e dr − b
a2
0
Now since for α > 0, we have
Z ∞
−3
r 2 e−r dr = 2
α2
2
e−α
−
α
∞
r
−3
2
−r
e dr .
b2
!
∞
Z
r
−1
2
e−r dr ,
α2
we deduce that
√
Z ∞
Z ∞
−1
−1
−r
−r
−a2
−b2
I = 4αm,n t (e
−e )+b
r 2 e dr − a
r 2 e dr
b2
a2
#
"
Z b2
Z ∞
√
−1
−1
1
r 2 e−r (r 2 − a)dr .
= 4αm,n t (b − a)
r 2 e−r dr +
a2
b2
Hence
√
Z
∞
I ≥ 4αm,n t(b − a)
r
−1
2
e−r dr.
b2
That is
∞
Z
I ≥ 2αm,n (|x − y| − |x − y|)
|x−y|2
4t
∞
Z
≥ 2αm,n (|x − y| − |x − y|)
r
r
−1
2
−1
2
e−r dr
e−r dr.
1
The result follows from Proposition 2.1.
Next we purpose to prove that Gm,n satisfies (1.1).
3G-Theorem. For x, y, z ∈ Rn+ , we have
Gm,n (x, z)Gm,n (z, y)
zn
zn
Gm,n (x, z) + Gm,n (y, z) .
Gm,n (x, y)
xn
yn
Proof. To prove the inequality, we denote by A(x, y) :=
quasi-metric, that is for each x, y, z ∈ Rn+ ,
(2.8)
xn y n
Gm,n (x,y)
and we claim that A is a
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], for n > 2m and as in [3], for n = 2m.
To prove (2.8), for n = 2m − 1, we derive from Corollary 2.2 that
1
A(x, y) ∼ (|x − y|2 ∨ xn yn ) 2 ∼ |x−y| .
J. Inequal. Pure and Appl. Math., 6(4) Art. 119, 2005
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8
I MED BACHAR
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. T HE C LASS Km,n (Rn+ )
Next we purpose to study and to characterize the class Km,n (Rn+ ) for n > 2m.
We recall that for 0 < α < n, we say that a Borel measurable function q in Rn+ belongs to
e α,n (Rn ) (see [6]) if q satisfies the following condition
the class K
+
Z
|q(y)|
(3.1)
lim sup
dy = 0.
r→0 x∈Rn (|x−y|≤r)∩Rn |x − y|n−α
+
+
The usual Kato class Kn (Rn+ ), corresponds to α = 2.
Remark 3.1. Let n > 2m. Using Corollary 2.3, the class Km,n (Rn+ ) obviously includes the
e 2m,n (Rn ). In particular, Kn (Rn ) ⊂ Km,n (Rn ).
class K
+
+
+
n
Example 3.1. Suppose that for p >
> 1, we have
2m
2p !
Z
yn
, 1 |q(y)|p dy < ∞,
M0 = sup
min
n
xn
x∈R+ (|x−y|≤1)∩Rn
+
then q ∈ Km,n (Rn+ ).
Indeed, let 0 < r < 1 and x ∈ Rn+ , then using Remark 2.4 and the Hölder inequality we get
Z
yn
Gm,n (x, y) |q(y)| dy
x
n
(|x−y|≤r)∩Rn
+
2 !
Z
yn
1
min
,1
|q(y)| dy
xn
|x − y|n−2m
(|x−y|≤r)∩Rn
+
! p1
2p !
Z
yn
min
, 1 |q(y)|p dy
n
x
n
(|x−y|≤r)∩R+
! p−1
Z
p
1
.
×
dy
p
(n−2m)
p−1
|x
−
y|
(|x−y|≤r)∩Rn
+
Hence
Z
1 2mp−n
yn
Gm,n (x, y)q(y)dy M0p r p → 0 as r → 0.
xn
x∈Rn
(|x−y|≤r)∩Rn
+
+
n
Proposition 3.2. Let p > max 2m
, 1 and f ∈ Lp Rn+ . Then
sup
y 7→
f (y)
∈ Km,n (Rn+ )
(|y| + 1)µ−λ ynλ
provided
i) n > 2m, λ ≤ 2 and λ < 2m − np and µ ≥ max(0, λ) or
ii) n = 2m and λ < min(2, 2m − np ) ≤ µ or
iii) n = 2m − 1, λ ≤ 2 and λ < 2m − np and µ ≥ max(1, λ).
J. Inequal. Pure and Appl. Math., 6(4) Art. 119, 2005
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G REEN FUNCTION AND K ATO CLASS
9
n
Proof. Let p > max 2m
, 1 and q ≥ 1 such that p1 + 1q = 1.
For f ∈ Lp (Rn+ ), x ∈ Rn+ and 0 < r < 1, put
Z
yn
|f (y)|
I = I(x, r) :=
Gm,n (x, y)
dy.
µ−λ y λ
x
(|y|
+
1)
n
B(x,r)∩Rn
n
+
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 and λ < 2m − np . Let µ ≥ max(0, λ) and put λ+ =
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
|f (y)|
Z
I
B(x,r)∩Rn
+
n
Z
n+λ+ −2m
|x − y|
dy kf kp
! 1q
1
B(x,r)∩Rn
+
(n+λ+ −2m)q
|x − y|
dy
+
r2m− p −λ , which tends to zero if r → 0.
Case 2. n = 2m. Assume that λ < min 2, 2m − np ≤ µ.
Using Proposition 2.1 and the Hölder inequality, we deduce that
! 1q
q q
4xn yn
1
yn
log 1 +
I kf kp
dy
2
(µ−λ)q y λq
x
|x
−
y|
(|y|
+
1)
n
(|x−y|≤r)∩Rn
n
+
Z
q q
1q
yn
4xn yn
1
log 1 +
dy
xn
|x − y|2
(|y| + 1)(µ−λ)q ynλq
(|x−y|≤r)∩D1
Z
q q
1q
yn
4xn yn
1
+
log 1 +
dy
xn
|x − y|2
(|y| + 1)(µ−λ)q ynλq
(|x−y|≤r)∩D2
= I1 + I2 ,
Z
where
D1 = {y ∈ Rn+ : xn yn ≤ |x − y|2 } and D2 = {y ∈ Rn+ : |x − y|2 ≤ xn yn }.
So, using that log(1 + t) ≤ t, for t ≥ 0 and Lemma 2.5, we obtain
(2−λ)q
Z
I1 (|x−y|≤r)∩D1
Z
yn
dy
|x − y|2q
1q
1
(|x−y|≤r)∩D1
! 1q
|x − y|λq
dy
n
r2m− p −λ , which converges to zero as r → 0.
On the other hand, from Lemma 2.5 and the fact that (|x| + 1) ∼ (|y| + 1), we obtain
1
I2 λ
xn (|x| + 1)(µ−λ)
Z
J. Inequal. Pure and Appl. Math., 6(4) Art. 119, 2005
(|x−y|≤r)∩D2
(cxn )2
log 1 +
|x − y|2
1q
q
dy
,
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10
I MED BACHAR
√
where c = 1 + 5. Let γ ∈ ] max(0, λ), min(2, 2m − np )[.
Since log(1 + t2 ) tγ , for t ≥ 0, then
Z
1q
xγ−λ
1
n
I2 dy
γq
(|x| + 1)µ−λ
(|x−y|≤r)∩D2 |x − y|
Z
1q
1
dy
γq
(|x−y|≤r)∩D2 |x − y|
n
r2m− p −γ , which converges to zero as r → 0.
Case 3. n = 2m − 1. Assume that λ ≤ 2 and λ < 2m − np . Let µ ≥ max(1, λ).
Using Corollary 2.2 and the Hölder inequality, we obtain

! 1q
! 1q 
Z
Z
(2−λ)q
(2−λ)q
yn
yn
I
dy
+
dy 
q
q
(µ−λ)q
(µ−λ)q
|x
−
y|
(|y|
+
1)
|x
−
y|
(|y|
+
1)
B(x,r)∩D1
B(x,r)∩D2
= I1 + I2 .
Now, if y ∈ D1 , then |x − y| ∼ |x − y| and so
! 1q
Z
n
1
I1 dy
r2m− p −λ , which tends to zero as r → 0.
(λ−1)q
B(x,r)∩D1 |x − y|
On the other hand, if y ∈ D2 , then |x − y|2 ∼ xn yn and by Lemma 2.5, we have further
xn ∼ yn . This implies that
Z
1q
x2−λ
n
I2 dy
xn (|x| + 1)µ−λ
B(x,r)∩D2
n
x1−λ
n
(r ∧ cxn ) q
µ−λ
(|x| + 1)
n
r q , which converges to zero as r → 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. Let q ∈ Km,n (Rn+ ), then for each compact L ⊆ Rn we have
Z
yn2
sup
|q(y)| dy < ∞.
1
+
x
y
x∈Rn
n
n
(x+L)∩Rn
+
+
Proof. Let q ∈ Km,n (Rn+ ), then by (1.3) there exists r > 0 such that
Z
yn
sup
Gm,n (x, y) |q(y)| dy ≤ 1.
xn
x∈Rn
(|x−y|≤r)∩Rn
+
+
Let a1 , a2 , . . . , ap ∈ Rn+ ∩ L such that Rn+ ∩ L ⊆ ∪ B(ai , r).
1≤i≤p
b
≤ 1 + |a − b| , then for each x, y, z ∈ Rn+ it follows
Since for a, b ∈ (0, ∞), we have 1+ab
that
1 + (xn + zn )yn
≤ [1 + zn (1 + |xn − yn |)].
1 + xn yn
Using this fact and Corollary 2.2, we obtain:
J. Inequal. Pure and Appl. Math., 6(4) Art. 119, 2005
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G REEN FUNCTION AND K ATO CLASS
11
For n > 2m,
yn2
[1 + zn (1 + |xn − yn |)]
|x + z − y|n−2m
1 + xn yn
1 + (xn + zn )yn
× |x + z − y|2 + 4(xn + zn )yn
For n = 2m, using further that
t
1+t
yn
Gm,n (x + z, y).
(xn + zn )
≤ log(1 + t), ∀t ≥ 0, we have
yn2
[1 + zn (1 + |xn − yn |)]
1 + xn yn
1 + (xn + zn )yn
× [|x + z − y|2 + 4(xn + zn )yn ]
yn
Gm,n (x + z, y).
(xn + zn )
For n = 2m − 1,
[1 + zn (1 + |xn − yn |)]
yn2
1 + xn yn
1 + (xn + zn )yn
1
× [|x + z − y|2 + 4(xn + zn )yn ] 2
yn
Gm,n (x + z, y).
(xn + zn )
Now, if z ∈ L and |x + z − y| ≤ r, then
yn2
yn
Gm,n (x + z, y).
1 + xn yn
(xn + zn )
Hence
Z
(x+L)∩Rn
+
yn2
|q(y)| dy
1 + xn yn
p Z
X
(|x+ai −y|≤r)∩Rn
+
i=1
So
Z
sup
x∈Rn
+
(x+L)∩Rn
+
yn
Gm,n (x + ai , y) |q(y)| dy p.
(xn + (ai )n )
yn2
|q(y)| dy < ∞.
1 + xn yn
Corollary 3.4. Let q ∈ Km,n (Rn+ ).Then we have for M > 0,
Z
yn2 |q(y)| dy < ∞.
(|y|≤M )∩Rn
+
Proposition 3.5. Let q ∈ Km,n (Rn+ ), then for each fixed α > 0, we have
Z
yn
(3.2)
sup sup
p(t, x, y) |q(y)| dy := M (α) < ∞.
x
t≤1 x∈Rn
n
(|x−y|>α)∩Rn
+
+
Proof. Let q ∈ Km,n (Rn+ ), 0 < t ≤ 1 and without loss of generality assume that 0 < α < 1.
Then by (1.12) and (1.7), it follows that
Z
yn
sup
p(t, x, y) |q(y)| dy
xn
x∈Rn
(|x−y|>α)∩Rn
+
+
!
Z
|x − y|2
1 − α2
yn2
exp −
|q(y)| dy.
n +1 e 8t sup
8
1 + xn yn
t2
x∈Rn
Rn
+
+
J. Inequal. Pure and Appl. Math., 6(4) Art. 119, 2005
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12
I MED BACHAR
To conclude, it is sufficient to prove that
!
Z
yn2
|x − y|2
sup
exp −
|q(y)| dy < ∞.
8
1
+
x
y
x∈Rn
n
n
Rn
+
+
Indeed, using Proposition 3.3, we have
Z
sup
x∈Rn
+
(x+B(0,1))∩Rn
+
yn2
f < ∞.
|q(y)| dy := M
1 + xn yn
Now since any ball B(0, k), of radius k ≥ 1 in Rn can be covered by An k n := α(n) balls
of radius 1, where An is a constant depending only on n (see [5, p. 67]), then there exists
a1 , a2 , . . . , aα(n) ∈ Rn+ such that
Rn+ ∩ B(0, k) ⊆
∪
1≤i≤α(n)
B(ai , 1).
Using the fact that for each x, y, z ∈ Rn+ ,
1 + (xn + zn )yn
≤ [1 + zn (1 + |xn − yn |)],
1 + xn yn
it follows that for all x ∈ Rn+ and k ≥ 1 :
Z
(x+B(0,k))∩Rn
+
α(n) Z
X
yn2
|q(y)| dy 1 + xn yn
i=1
B(x+ai ,1)∩Rn
+
α(n) Z
X
i=1
B(x+ai ,1)∩Rn
+
yn2
|q(y)| dy
1 + xn yn
yn2
|q(y)| dy
1 + (xn + (ai )n )yn
f.
An k n M
Hence for all x ∈ Rn+ , we have
!
Z
|x − y|2
yn2
exp −
|q(y)| dy
8
1 + xn yn
Rn
+
Z
∞
X
α2 k 2
yn2
exp −
|q(y)| dy
8
1 + xn yn
[kα≤|x−y|≤(k+1)α]∩Rn
+
k=0
∞
2 2
X
α
k
n
f (k + 1) exp −
An M
< ∞.
8
k=0
Thus
|x − y|2
sup
exp −
8
x∈Rn
Rn
+
+
Z
!
yn2
|q(y)| dy < ∞,
1 + xn yn
which completes the proof.
Theorem 3.6. Let n > 2m and q ∈ B(Rn+ ). Then the following assertions are equivalent:
(1) q ∈ Km,n (Rn+ )
R R t yn m−1
(2) lim sup Rn 0
s
p(s, x, y) |q(y)| dsdy = 0.
+
t→0 x∈Rn
xn
+
J. Inequal. Pure and Appl. Math., 6(4) Art. 119, 2005
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G REEN FUNCTION AND K ATO CLASS
13
Proof. 2) ⇒ 1) Assume that
Z
Z
t
lim sup
t→0 x∈Rn
+
Rn
+
0
yn m−1
s
p(s, x, y) |q(y)| dsdy = 0.
xn
Then by Proposition 2.7, there exists c > 0 such that for α > 0 we have
Z
Z Z α2
4 y
yn
n m−1
s
p(s, x, y) |q(y)| dsdy,
Gm,n (x, y) |q(y)| dy ≤ c
xn
xn
(|x−y|≤α)∩Rn
0
Rn
+
+
which shows that the function q satisfies (1.3).
Conversely suppose that q ∈ Km,n (Rn+ ). Let ε > 0, then there exists 0 < α < 1 such that
Z
yn
sup
Gm,n (x, y) |q(y)| dy ≤ ε.
xn
x∈Rn
(|x−y|≤α)∩Rn
+
+
On the other hand, using Proposition 2.7 and (3.2), we have for 0 < t < 1
Z Z t
yn m−1
s
p(s, x, y) |q(y)| dsdy
x
n
Rn
0
+
Z
Z t
yn m−1
s
p(s, x, y) |q(y)| dsdy
x
n
(|x−y|≤α)∩Rn
0
+
Z
Z t
yn m−1
+
s
p(s, x, y) |q(y)| dsdy
(|x−y|>α)∩Rn
0 xn
+
Z
yn
Gm,n (x, y) |q(y)| dy
x
n
(|x−y|≤α)∩Rn
+
Z tZ
yn
+
p(s, x, y) |q(y)| dyds
x
n
0
(|x−y|>α)∩Rn
+
ε + tM (α),
which implies that
Z
Z
lim sup
t→0 x∈Rn
+
Rn
+
0
t
yn m−1
s
p(s, x, y) |q(y)| dsdy = 0.
xn
Corollary 3.7. Let n > 2m and q ∈ B(Rn+ ). For α > 0 and x ∈ Rn+ , put
Z Z ∞
yn
Gα q(x) :=
e−αs sm−1 p(s, x, y) |q(y)| dsdy.
xn
Rn
0
+
Then
q ∈ Km,n (Rn+ ) ⇔ lim kGα qk∞ = 0,
α→+∞
where kGα qk∞ = sup |Gα q(x)| .
x∈Rn
+
Proof. (see [9]). Let q ∈ Km,n (Rn+ ), α > 0 and put
Z 1Z
α
yn m−1
a(α) = sup
s
p(s, x, y) |q(y)| dyds.
xn
x∈Rn
0
Rn
+
+
J. Inequal. Pure and Appl. Math., 6(4) Art. 119, 2005
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14
I MED BACHAR
Then we have
Z
∞
αe−αt
Gα q(x) =
"Z Z
t
0
Z
=
Rn
+
0
∞
−t
"Z
t
α
Z
e
0
0
Rn
+
#
yn m−1
s
p(s, x, y) |q(y)| dyds dt
xn
#
yn m−1
s
p(s, x, y) |q(y)| dyds dt.
xn
1
It follows that, a(α) ≤ kGα qk∞ .
e
On the other hand, for t > 0 and k ∈ N such that k ≤ t < k + 1, we have
#
"Z k+1 Z
m Z ∞
X
α
y
n
Gα q(x) ≤
e−t
sm−1 p(s, x, y) |q(y)| dyds dt
k
n xn
0
R
+
α
k=0
Z ∞
≤ a(α)
e−t (m + 1)dt
Z0 ∞
≤ a(α)
e−t (t + 1)dt = 2a(α),
0
1
which gives that a(α) ≤ kGα qk∞ ≤ 2a(α).
e
Hence the results follow from Theorem 3.6.
R EFERENCES
[1] M. AIZENMAN AND B. SIMON, Brownian motion and Harnack inequality for Schrödinger operators, Comm. Pure Appl. Math., XXXV (1982), 209–273.
[2] I. BACHAR AND H. MÂAGLI, Estimates on the Green’s function and existence of positive solutions of nonlinear singular elliptic equations in the half space, Positivity, 9(2) (2005), 153–192.
[3] I. BACHAR, H. MÂAGLI AND L. MÂATOUG, Positive solutions of nonlinear elliptic equations
in a half space in R2 , E.J.D.E, 2002 (2002), No. 41, 1–24.
[4] I. BACHAR, H. MÂAGLI AND M. ZRIBI, Estimates on the Green function and existence of positive solutions for some polyharmonic nonlinear equations in the half space, Manuscripta Math.,
113, (2004), 269–291.
[5] K.L. CHUNG AND Z. ZHAO, From Brownian Motion to Schrödinger’s Equation, Springer Verlag
(1995).
[6] E.B. DAVIES AND A.M. HINZ, Kato class potentials for higher order elliptic operators, J. London
Math. Soc., (2) 58 (1998) 669–678.
[7] H. MÂAGLI, Inequalities for the Riesz potentials, Archives of Inqualities and Applications, 1
(2003) 285–294.
[8] B. SIMON, Schrödinger semi-groups, Bull. Amer. Math. Soc., 7(3) (1982), 447–526.
[9] J.A. VAN CASTEREN, Generators Strongly Continuous Semi-groups, Pitman Advanced Publishing Program, Boston, (1985).
[10] Z. ZHAO, Subcriticality and gaugeability of the schrödinger operator, Trans. Amer. Math. Society,
334(1) (1992), 75–96.
[11] Z. ZHAO, On the existence of positive solutions of nonlinear elliptic equations. A probabilistic
potential theory approach, Duke Math. J., 69 (1993), 247–258.
J. Inequal. Pure and Appl. Math., 6(4) Art. 119, 2005
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