Volume 8 (2007), Issue 1, Article 26, 17 pp.
ON THE ITERATED GREEN FUNCTIONS ON A BOUNDED DOMAIN AND
THEIR RELATED KATO CLASS OF POTENTIALS
HABIB MÂAGLI AND NOUREDDINE ZEDDINI
D ÉPARTEMENT DE M ATHÉMATIQUES ,
FACULTÉ DES S CIENCES DE T UNIS ,
C AMPUS U NIVERSITAIRE ,
2092 T UNIS , T UNISIA .
habib.maagli@fst.rnu.tn
noureddine.zeddini@ipein.rnu.tn
Received 30 May, 2006; accepted 18 February, 2007
Communicated by S.S. Dragomir
A BSTRACT. We use the results of Zhang [15, 16] and Davies [7] on the behavior of the heat
kernel p(t, x, y) on a bounded C 1,1 domain D to find again the result of Grunau-Sweers [9]
concerning the estimates of the iterated Greens functions Gm,n (D). Next, we use these estimates
to characterize, by means of p(t, x, y), the Kato class Km,n (D) and we give new examples of
functions belonging to this class.
Key words and phrases: Green function, Gauss semigroup, Kato class.
2000 Mathematics Subject Classification. 34B27.
1. I NTRODUCTION
Let D be a bounded C 1,1 domain in Rn , n ≥ 3 and p (t, x, y) be the density of the Gauss
semigroup on D. Combining the results of Zhang [15], [16] and those of Davies or DaviesSimon [7], [8] a qualitatively sharp understanding of the boundary behaviour of p (t, x, y) is
given as follows: There exist positive constants c1 , c2 and λ0 depending only on D such that for
all t > 0 and x , y ∈ D,
(1.1)
|x−y|2
δ(x)
δ(y)
c1 e−λ0 t−c2 t
√
√
∧1
∧1
n
t2
t∧1
t∧1
2
−λ0 t− |x−y|
c2 t
δ(y)
e
δ(x)
√
,
∧1
∧1
≤ p (t, x, y) ≤ √
n
c1 t 2
t∧1
t∧1
where δ(x) denotes the Euclidean distance from x to the boundary of D.
154-06
2
H ABIB M ÂAGLI AND N OUREDDINE Z EDDINI
Let G(x, y) be the Green’s function of the laplacien ∆ in D with a Dirichlet condition on
∂D. Then G is given by
Z ∞
(1.2)
G(x, y) =
p (t, x, y) dt, for x , y ∈ D.
0
For a positive integer m, we denote by Gm,n the Green’s function of the operator u 7→ (−∆)m u
on D with Navier boundary conditions ∆j u = 0 on ∂D for 0 ≤ j ≤ m − 1. Then G1,n = G
and Gm,n satisfies for m ≥ 2
Z Z
Gm,n (x, y) =
G(x, z)Gm−1,n (z, y) dz.
D
D
Using the Fubini theorem and the Chapman-Kolmogorov identity, we show by induction that
for each m ≥ 1 and x , y ∈ D we have
Z ∞
1
(1.3)
Gm,n (x, y) =
tm−1 p (t, x, y) dt.
(m − 1)! 0
In this paper we will use (1.1) and (1.3) to find again the result of Grunau and Sweers in [9]
concerning the sharp estimates of Gm,n . More precisely we will give another proof for the case
n ≥ 3 of the following theorem.
Theorem 1.1 ( see [9]). On D2 we have

δ(x)δ(y)
1

min 1 , |x−y|2
if n > 2m,

|x−y|n−2m






Log 1 + δ(x)δ(y)
if n = 2m,

|x−y|2

√

p
δ(x)δ(y)
Gm,n (x, y) ∼ Hm,n (x, y) =
δ(x)δ(y) min 1 , |x−y|
if n = 2m − 1,





1

if n = 2m − 2,

 δ(x)δ(y) Log 2 + |x−y|2 +δ(x)δ(y)



δ(x)δ(y)
if n < 2m − 2,
where the symbol ∼ is defined in the notations below.
As a second step we will also use (1.1) and (1.3) to give new contributions in the case n > 2m
to the study of the Kato class Km,n (D) defined in [11] for m = 1 and in [2] for m ≥ 2 as
follows.
Definition 1.1. A Borel measurable function q in D belongs to the Kato class Km,n (D) if q
satisfies the following condition
Z
δ(y)
(1.4)
lim sup
Gm,n (x, y)|q(y)| dy = 0.
α→0 x∈D D∩(|x−y|≤α) δ(x)
We note that in the case m = 1, the class K1,n (D) properly contains the classical Kato
class Kn (D) introduced in [1] as the natural class of singular functions which replaces the Lp Lebesgue spaces in order that the weak solutions of the Shrödinger equation are continuous and
satisfy a Harnack principle. More precisely, it is shown in [11] that the function ρα (y) = δα1(y)
belongs to K1,n (D) if and only if α < 2 but for 1 ≤ α < 2, ρα ∈
/ Kn (D).
Our second contribution here is to exploit estimates of Theorem 1.1 on the one hand, to give
new examples of functions belonging to the class Km,n (D) and to characterize this class by
means of the density of the Gauss semigroup in D on the other hand. In particular we will
prove the following results for the unit ball.
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 26, 17 pp.
http://jipam.vu.edu.au/
I TERATED G REEN F UNCTIONS
3
Proposition 1.2. For λ , µ ∈ R and y ∈ B(0, 1) we put
1
h
iµ .
ρλ, µ (y) =
2
(1 − |y|)λ Log( 1−|y|
)
For m ≥ 2 we have
ρλ, µ ∈ Km,n (B(0, 1)) if and only if
λ < 3 or (λ = 3 and µ > 1).
Theorem 1.3. Let n > 2m and q be a Borel measurable function in D. Then the following
assertions are equivalent:
1) q ∈ Km,n
(B(0, 1))
R t R δ(y) m−1
2) limt→0 supx∈B 0 B δ(x)
s
p(s, x, y)|q(y)| dyds = 0
We also note that in the case m = 1, similar characterizations have been obtained by Aizenman and Simon in [1] for the Kato class Kn (Rn ) and by Bachar and Mâagli in [4] for the half
space Rn+ , where they introduce a new Kato class that properly contains the classical one. This
was extended for m ≥ 2 by Mâagli and Zribi [12] to the class Km,n (Rn ) and by Bachar [3] to
the class Km,n (Rn+ ). The density of the Gauss semigroup in the case of Rn and Rn+ are explicitly
known, but this is not the case for a bounded C 1,1 domain even if D is an open ball .
In order to simplify our statements, we define some convenient notations.
Notations.
i) For x, y ∈ D, we denote by δ(x) the Euclidean distance from x to the boundary of D,
[x, y]2 = |x − y|2 + δ(x)δ(y) and d is the diameter of D.
ii) For a, b ∈ R, we denote by a ∧ b = min(a, b) and a ∨ b = max(a, b).
iii) Let f and g be two nonnegative functions on a set S.
We say that f g, if there exists c > 0 such that
f (x) ≤ c g(x)
for all x ∈ S.
We say that f ∼ g, if there exists C > 0 such that
1
g(x) ≤ f (x) ≤ Cg(x) for all x ∈ S.
C
The following properties will be used several times
iv) For a, b ≥ 0, we have
ab
ab
(1.5)
≤ min(a, b) ≤ 2
a+b
a+b
(1.6)
(a + b)p ∼ ap + bp for p ∈ R+ .
(1.7)
min(1, a) min(1, b) ≤ min(1, a b) ≤ min(1, a) max(1, b)
a
≤ Log(1 + a)
1+a
v) Let η, ν > 0 and 0 < γ ≤ 1. Then we have
(1.8)
(1.9)
Log(1 + t) tγ , for t ≥ 0.
(1.10)
Log(1 + η t) ∼ Log(1 + ν t) , for t ≥ 0.
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 26, 17 pp.
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4
H ABIB M ÂAGLI AND N OUREDDINE Z EDDINI
Finally we note that since for each a ≥ b ≥ 0 and c > 0 we have
(a + 1)(b + 1) −c (b−a)2
a+b
2
e
= 1+
e−c (b−a)
1 + ab
1 + ab
2a + ξ
2
= 1+
e−c ξ
1 + a(a + ξ)
2
≤ (2 + ξ)e−c ξ ≤ C.
Then, using (1.5) we deduce that for each x , y ∈ D and 0 < t ≤ 1 we have
|δ(x)−δ(y)|2
δ(x)δ(y)
δ(x)
δ(y)
t
min
, 1 ≤ C min √ , 1 min √ , 1 ec
t
t
t
|x−y|2
δ(x)
δ(y)
≤ C min √ , 1 min √ , 1 ec t .
t
t
So, using this fact, (1.7) and the fact that D is bounded we deduce that estimates (1.1) can be
written as follows:
There exist positive constants c, C and λ such that
1
h 1 (t, x, y) ≤ p (t, x, y) ≤ C hc,λ (t, x, y),
C c,λ
(1.11)
where
(1.12)

2
 min δ(x)δ(y) , 1 e−c |x−y|
t
,
n
t
t2
hc,λ (t, x, y) :=

δ(x)δ(y)e−λt ,
if 0 < t ≤ 1
if t > 1.
Throughout the paper, the letter C will denote a generic positive constant which may vary from
line to line.
2. P ROOF OF T HEOREM 1.1
First we need the following lemma.
Lemma 2.1. For each x, y ∈ D we have
a) For n ≥ 2m
δ(x)δ(y)
δ(x)δ(y) ≤ min 1 ,
|x − y|2
dn−2m+2
.
|x − y|n−2m
b)
δ(x)δ(y)
δ(x)δ(y) ≤ d min 1 ,
|x − y|2
2
δ(x)δ(y)
≤ 2d Log 1 +
|x − y|2
2
.
Now we will give the proof of Theorem 1.1. More precisely, using (1.3) and (1.11) we will
prove that for each c > 0, we have
Z ∞
tm−1 hc,λ (t, x, y)dt ∼ Hm,n (x, y) .
0
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 26, 17 pp.
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I TERATED G REEN F UNCTIONS
5
Without loss of generality we will assume that λ = 1 , c = 1 and denote by h1,1 (t, x, y) =
h(t, x, y). Hence, using a change of variable, we obtain
Z
0
∞
− |x−y|2
e t
δ(x)δ(y)
dt
tm−1 h(t, x, y)dt = C δ(x)δ(y) +
tm−1 min
,1
n
t
t2
0
Z ∞
n
δ(x)δ(y)
−m−1
2m−n
= C δ(x)δ(y) + |x − y|
min
r, 1 e− r dr.
r2
2
|x
−
y|
2
|x−y|
Z
1
n
Since we will sometimes omit e−r and we need to integrate the functions r → r 2 −m−1 and r →
n
r 2 −m near zero or near ∞, we will discuss the following cases
Case 1. n > 2m. Using (1.7) we obtain
min
δ(x)δ(y)
δ(x)δ(y)
, 1 min(r, 1) ≤ min
r, 1
|x − y|2
|x − y|2
δ(x)δ(y)
, 1 max(r , 1)).
≤ min
|x − y|2
Hence the lower bound follows from the fact that
Z
∞
min(1, r)r
n
−m−1
2
−r
e
Z
∞
|x−y|2
n
min(1, r)r 2 −m−1 e− r dr = C
dr ≥
d2
and the upper bound follows from Lemma 2.1.
Case 2. n = 2m. In this case
Z
∞
m−1
t
Z
∞
h(t, x, y)dt = C δ(x)δ(y) +
min
|x−y|2
0
δ(x)δ(y) 1
,
|x − y|2 r
e− r dr.
So using (1.5) and the fact that
|x − y|2 + (2d2 + 1)δ(x)δ(y)
≥ |x − y|2 + δ(x)δ(y) ,
1 + δ(x)δ(y)
we obtain
Z
∞
m−1
t
0
Z
2d2 +1
δ(x)δ(y)
dr
2
|x−y|2 |x − y| + rδ(x)δ(y)
|x − y|2 + (2d2 + 1)δ(x)δ(y)
= C Log
|x − y|2 (1 + δ(x)δ(y))
δ(x)δ(y)
≥ C Log 1 +
.
|x − y|2
h(t, x, y)dt ≥
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 26, 17 pp.
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6
H ABIB M ÂAGLI AND N OUREDDINE Z EDDINI
To prove the upper inequality we use (1.5), (1.11) and (1.10) to obtain
Z ∞
δ(x)δ(y) 1 − r
min
,
e dr
|x − y|2 r
|x−y|2
Z ∞
δ(x)δ(y)
e− r dr
≤C
2
δ(x)δ(y)r
+
|x
−
y|
2
|x−y|
Z
Z d2 +1
δ(x)δ(y)
δ(x)δ(y) ∞ − r
e dr
≤C
dr + C
2
[x, y]2 1+d2
|x−y|2 δ(x)δ(y)r + |x − y|
δ(x)δ(y)
|x − y|2 + (d2 + 1)δ(x)δ(y)
= C Log
+C
2
|x − y| (1 + δ(x)δ(y))
[x, y]2
(d2 + 1)δ(x)δ(y)
δ(x)δ(y)
+C
≤ C Log 1 +
2
|x − y| (1 + δ(x)δ(y))
|x, y|2 + δ(x)δ(y)
δ(x)δ(y)
≤ C Log 1 +
.
|x − y|2
Hence the result follows from Lemma 2.1.
Case 3. n = 2m − 1. In this case
Z ∞
tm−1 h(t, x, y) dt
0
Z
∞
− 12
δ(x)δ(y) 1
,
|x − y|2 r
= Cδ(x)δ(y) + |x − y|
r min
e− r dr
2
|x−y|
Z ∞
δ(x)δ(y)
δ(x)δ(y)
− 12 − r
≤C
d+
r e dr = C
.
|x − y|
|x − y|
0
On the other hand, an integration by parts shows that
Z ∞
δ(x)δ(y) 1 − r
− 12
|x − y|
r min
,
e dr
|x − y|2 r
|x−y|2
Z ∞
Z 2 |x−y|2
δ(x)δ(y) d δ(x)δ(y) − 1
− 32 − r
r
e dr
≤C
r 2 dr + |x − y|
|x−y|2
|x − y| 0
d2 δ(x)δ(y)
h
i∞
p
1
≤ C δ(x)δ(y) + |x − y| −2r− 2 e− r
|x−y|2
d2 δ(x)δ(y)
≤C
p
δ(x)δ(y).
Hence
Z
∞
m−1
t
h(t, x, y)dt ≤ C min
0
p
δ(x)δ(y)
δ(x)δ(y) ,
|x − y|
.
For the lower inequality we discuss two subcases
• If δ(x)δ(y) ≤ |x − y|2 . Then from (1.7) we have
Z ∞
Z ∞
3
δ(x)δ(y)
m−1
t
h(t, x, y)dt ≥ |x − y| min 1,
r− 2 e− r dr
2
|x − y|
0
1+d2
δ(x)δ(y)
=C
.
|x − y|
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 26, 17 pp.
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I TERATED G REEN F UNCTIONS
7
• If |x − y|2 ≤ δ(x)δ(y). Then
∞
Z
m−1
t
4d2 |x−y|2
(δ(x)δ(y))
Z
h(t, x, y)dt ≥ |x − y|
|x−y|2
0
δ(x)δ(y)
≥C
|x − y|
Z
δ(x)δ(y)
≥C
|x − y|
Z
δ(x)δ(y) 1
∧
|x − y|2
r
4d2 |x−y|2
(δ(x)δ(y))
1
r− 2 e− r dr
1
r− 2 e− r dr
|x−y|2
4d2 |x−y|2
(δ(x)δ(y))
1
r− 2 dr
|x−y|2
"
#
2d
≥ C δ(x)δ(y) p
−1
δ(x)δ(y)
h
i
p
p
≥ C δ(x)δ(y) 2d − δ(x)δ(y)
p
≥ C δ(x)δ(y).
Case 4. n = 2m − 2. In this case, we use (1.5) to deduce that
Z
∞
tm−1 h(t, x, y)dt
0
2
Z
∞
= Cδ(x)δ(y) + |x − y|
Z
|x−y|2
∞
∼ δ(x)δ(y) + δ(x)δ(y)
|x−y|2
δ(x)δ(y)
1
∧ 2
2
r|x − y|
r
e− r dr.
1
δ(x)δ(y)
−
r δ(x)δ(y)r + |x − y|2
e− r dr.
To prove the upper estimates we remark first that
1
δ(x)δ(y) ≤ C δ(x)δ(y) Log 2 +
[x, y]2
and we discuss the following subcases
1
• If 12 ≤ δ(x)δ(y) 1 + [x,y]
2 . Then 1 +
Z
∞
|x−y|2
1
δ(x)δ(y)
−
r δ(x)δ(y)r + |x − y|2
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 26, 17 pp.
−r
e
1
δ(x)δ(y)
Z
≤4+
∞
2
.
[x,y]2
So
1
δ(x)δ(y)
dr ≤
−
r δ(x)δ(y)r + |x − y|2
|x−y|2
1
= Log 1 +
δ(x)δ(y)
1
≤ Log 2 + Log 2 +
[x, y]2
1
.
≤ C Log 2 +
[x, y]2
dr
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8
H ABIB M ÂAGLI AND N OUREDDINE Z EDDINI
• If δ(x)δ(y) 1 +
1
[x,y]2
≤
1
.
2
Then δ(x)δ(y) ([x, y]2 + 1) ≤
1
2
[x, y]2 , which implies
that δ(x)δ(y) ≤ |x − y|2 and consequently [x, y]2 ≤ 2|x − y|2 . Hence
∞
Z
|x−y|2
δ(x)δ(y)
1
−
r δ(x)δ(y)r + |x − y|2
e− r dr
∞
r
≤ C Log (1 + δ(x)δ(y)) e
+C
Log
e− r dr
2
δ(x)δ(y)r
+
|x
−
y|
2
|x−y|
Z ∞
r
2
−|x−y|2
≤ C Log(1 + d ) e
e− r dr
+C
Log
2
δ(x)δ(y)r + |x − y|
|x−y|2
Z ∞
2
(1 + d )r
≤C
Log
e− r dr
2
δ(x)δ(y)r
+
|x
−
y|
2
|x−y|
Z ∞
(1 + d2 )r
≤C
Log
e− r dr
2 (1 + δ(x)δ(y))
|x
−
y|
2
|x−y|
Z ∞
1 + d2
1
≤ C Log
+C
Log
r e− r dr
|x − y|2
1 + δ(x)δ(y)
|x−y|2
Z ∞
1 + d2
+C
Log(r)e− r dr
≤ C Log
2
|x − y|
|x−y|2
Z ∞
2
1+d
≤ C Log
+C
Log(r)e− r dr
|x − y|2
1
2
1
1+d
≤ C + C Log
≤ C Log 2 +
.
|x − y|2
[x, y]2
− |x−y|2
Z
Hence
Z
∞
m−1
t
0
1
h(t, x, y)dt ≤ C δ(x)δ(y) Log 2 +
[x, y]2
.
Next we prove the lower estimates.
Z
∞
m−1
t
0
∞
1
δ(x)δ(y)
h(t, x, y)dt ∼ δ(x)δ(y) + δ(x)δ(y)
−
e− r dr
2
r
δ(x)δ(y)r
+
|x
−
y|
2
|x−y|
Z 2d2 1
δ(x)δ(y)
≥ Cδ(x)δ(y) + Cδ(x)δ(y)
−
dr
r δ(x)δ(y)r + |x − y|2
|x−y|2
2d2 (1 + δ(x)δ(y))
= Cδ(x)δ(y) + Cδ(x)δ(y) Log
.
|x − y|2 + 2d2 δ(x)δ(y)
Z
2
Let α > 1 such that α 2d2d2 +1 > 2[x, y]2 + 1; ∀x, y ∈ D. Then we have
2α d2 (1 + δ(x)δ(y))
2α d2
1
≥
≥2+
.
2
2
2
2
|x − y| + 2d δ(x)δ(y)
(1 + 2d )[x, y]
[x, y]2
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 26, 17 pp.
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I TERATED G REEN F UNCTIONS
9
Hence
Z
∞
tm−1 h(t, x, y)dt
2d2 (1 + δ(x)δ(y))
≥ Cδ(x)δ(y) Log α + Log
|x − y|2 + 2d2 δ(x)δ(y)
2α d2 (1 + δ(x)δ(y))
≥ Cδ(x)δ(y) Log
|x − y|2 + 2d2 δ(x)δ(y)
1
≥ Cδ(x)δ(y) Log 2 +
.
[x, y]2
0
Case 5. n < 2m − 2. In this case we need only to prove the upper inequality.
Z ∞
n
δ(x)δ(y) 1 − r
2m−n
−m
|x − y|
r2
min
,
e dr
|x − y|2 r
|x−y|2
Z d2 |x−y|2
Z ∞
δ(x)δ(y)
n
n
−m
2m−n
2m−n−2
≤ δ(x)δ(y)|x − y|
r 2 dr + |x − y|
r 2 −m−1 dr
2
|x−y|
|x−y|2
d2 δ(x)δ(y)
m−1− 2
m− n2
δ(x)δ(y)
2
δ(x)δ(y)
+
d2
2m − n
d2
m−1− n2
2
2
δ(x)δ(y)
≤
δ(x)δ(y) + 2
δ(x)δ(y)
2m − n − 2
d (2m − n)
d2
≤ C δ(x)δ(y).
"
2
≤
δ(x)δ(y) 1 −
2m − n − 2
#
n
This completes the proof of the theorem.
Now using estimates of Theorem 1.1 and similar arguments as in the proof of Corollary 2.5
in [5], we obtain the following.
Corollary 2.2. Let r0 > 0. For each x, y ∈ D such that |x − y| ≥ r0 , we have
(2.1)
Gm,n (x, y) δ(x)δ(y)
.
r0n+2−2m
Moreover, on D2 the following estimates hold
(
δ(x)∧δ(y)
|x−y|n+1−2m
δ(x)δ(y) Gm,n (x, y) ,
δ(x) ∧ δ(y) ,
for n ≥ 2m
for n ≤ 2m − 1.
3. T HE K ATO C LASS Km,n (D)
To give new examples of functions belonging to this class we need the following lemma
Lemma 3.1. For λ, µ ∈ R and x ∈ D, let ρλ, µ (x) =
ρλ, µ ∈ L1 (D) if and only if
1
µ
2d
δ λ (x)[Log( δ(x)
)]
. Then
λ < 1 or (λ = 1 and µ > 1).
Proof. Since for λ < 0 the function ρλ, µ is continuous and bounded in D we need only to prove
the result for λ ≥ 0.
Since D is a bounded C 1,1 domain and the function t 7→ tλ Log(1 2d ) µ is decreasing near 0 for
[
t ]
λ > 0, then the proof of the lemma on page 726 in [10] can be adapted.
Proposition 3.2. Let m ≥ 2 and p ∈ [1, ∞]. Then ρλ , µ (·)Lp (D) ⊂ Km,n (D), provided that:
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10
H ABIB M ÂAGLI AND N OUREDDINE Z EDDINI
i) For n ≥ 2m − 1, we have λ < 2 +
ii) For n = 2m − 2, we have λ < 2 +
iii) For n < 2m − 2, we have λ < 3 −
2(m−1)
n
− p1 and 2(m−1)
n
n−1
n
− p1 and n−1
< p.
n
1
.
p
< p.
Proof. Let h ∈ Lp (D) and q ∈ [1, ∞] such that p1 + 1q = 1. For x ∈ D and α ∈ (0, 1), we put
Z
δ(y)
I = I(x, α) :=
Gm,n (x, y)ρλ , µ (y)h(y) dy.
B(x,α)∩D δ(x)
Taking account of Theorem 1.1, we will discuss the following cases:
Case 1. n ≥ 2m − 1. In this case we have
Z
h(y)
dy
h
I
n−2(m−1)
λ−2
B(x,α)∩D |x − y|
δ(y)
Log
2d
δ(y)
iµ .
It follows from the Hölder inequality that

 1q
Z
1
dy
h
iqµ  .
I khkp 
(n−2(m−1))q
(λ−2)q
2d
B(x,α)∩D |x − y|
δ(y)
Log δ(y)
n
n
Since λ < 2 + 2(m−1)
− p1 and 2(m−1)
< p, then λ − 2 < 1q − n−2(m−1)
and q < n−2(m−1)
. Hence
n
n
1
n
we can choose q 0 > max 1, 1−(λ−2)q
so that q q 0 < n−2(m−1)
and (λ − 2)q < 1 − q10 := 1r .
We apply the Hölder inequality again and Lemma 3.1 to deduce that

 qr1
Z
dy
0
h
iqrµ  αn−(n−2m+2)qq .
I khkp 
2d
D δ(y)(λ−2)qr Log
δ(y)
Hence sup I(x, α) → 0 as α → 0.
x∈D
n
Case 2. n = 2m − 2. Assume that λ < 2 + n−1
− p1 and n−1
< p, then λ − 2 <
n
q < n.
Using (1.9) , (1.6) and the Hölder inequality we obtain
Z
1
h(y)
h
iµ dy
I
1+
[x, y] δ(y)λ−2 Log 2d
B(x,α)∩D
δ(y)
Z
1
h(y)
h
iµ dy
λ−2
2d
B(x,α)∩D |x − y| δ(y)
Log δ(y)

 1q
Z
1
1
h
iqµ dy  .
khkp 
q
(λ−2)q
|x
−
y|
B(x,α)∩D
δ(y)
Log 2d
1
q
−
1
n
and
δ(y)
0
Let us choose q 0 > 1 and r = q0q−1 such that qq 0 < n and (λ − 2)qr < 1. Then, using the Hölder
inequality again and Lemma 3.1 we obtain

 qr1
Z
dy
0
h
iqrµ  αn−qq .
I khkp 
(λ−2)qr
2d
D δ(y)
Log δ(y)
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I TERATED G REEN F UNCTIONS
11
Hence sup I(x, α) → 0 as α → 0.
x∈D
Case 3. n < 2m − 2. Using Theorem 1.1 and the Hölder inequality we obtain
Z
h(y)
h
iµ dy
I
2d
B(x,α)∩D δ(y)λ−2 Log
δ(y)

 1q
Z
1
h
iqµ dy  .
khkp 
(λ−2)q
2d
B(x,α)∩D δ(y)
Log δ(y)
As in the preceding cases we choose q 0 > 1 so that (λ − 2)qq 0 < 1 to deduce from the Hölder
inequality and Lemma 3.1 that sup I(x, α) → 0 as α → 0.
x∈D
This completes the proof of the proposition.
Next, we will prove Proposition 1.2. So we need the following results
Lemma 3.3 (see [5]). Let x, y ∈ D. Then the following properties are satisfied:
1) If δ(x)δ(y) ≤ |x − y|2 then
√
1+ 5
max(δ(x) , δ(y)) ≤
|x − y|.
2
2) If |x − y|2 ≤ δ(x)δ(y) then
√
√
(3 − 5)
(3 + 5)
δ(x) ≤ δ(y) ≤
δ(x).
2
2
Lemma 3.4. Let q ∈ Km,n (D). Then the function : x → δ 2 (x)q(x) is in L1 (D).
Proof. Let q ∈ Km,n (D). Then by (1.4), there exists α > 0 such that for all x ∈ D we have
Z
δ(y)
Gm,n (x, y)|q(y)| dy ≤ 1.
(|x−y|≤α)∩D δ(x)
S
Let x1 , x2 , . . . , xp ∈ D such that D ⊂ pi=1 B(xi , α). Then by Corollary 2.2, there exists
C > 0 such that ∀ y ∈ B(xi , α) ∩ D we have
δ 2 (y) ≤ C
δ(y)
Gm,n (xi , y).
δ(x)
Hence
Z
2
δ (y)|q(y)|dy ≤ C
D
p Z
X
i=1
B(xi ,α)∩D
δ(y)
Gm,n (xi , y)|q(y)| dy
δ(xi )
≤ C p < ∞.
Proof of Proposition 1.2. It follows from Lemmas 3.1 and 3.4 that a necessary condition for
ρλ , µ to belong to Km,n (B) is that λ < 3 or (λ = 3 and µ > 1). Let us prove that this condition
is sufficient.
For λ ≤ 2 the results follow from Proposition 3.2 by taking p = ∞. Hence we need only to
prove the results for 2 < λ < 3 or (λ = 3 and µ > 1).
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12
H ABIB M ÂAGLI AND N OUREDDINE Z EDDINI
µ
For x ∈ D and α ∈ 0, 4e− λ , we put
Z
I = I(x, α) :=
B(x,α)∩D
Z
=
B(x,α)∩D
δ(y)
Gm,n (x, y)ρλ , µ (y) dy
δ(x)
Gm,n (x, y)
h
iµ dy .
4
δ(x)δ(y)λ−1 Log δ(y)
Taking account of Theorem 1.1 we distinguish the following cases.
Case 1. n ≥ 2m − 1. Then we have
Z
1
1
h
iµ dy
I
n−2(m−1)
4
B(x,α)∩D1 |x − y|
(δ(y))λ−2 Log δ(y)
Z
1
1
h
iµ dy
+
n−2(m−1)
B(x,α)∩D2 |x − y|
(δ(y))λ−2 Log 4
δ(y)
= I1 + I2 ,
where
D1 = {x ∈ D : |x − y|2 ≤ δ(x)δ(y)}
and
D2 = {x ∈ D : δ(x)δ(y) ≤ |x − y|2 }.
• If y ∈ D1 , then from Lemma 3.3, we have δ(x) ∼ δ(y) and so |x − y| δ(y). Hence
Z
1
h
iµ dy
I1 C
B(x,α) |x − y|n−2m+λ Log
|x−y|
Z α 2m−(λ+1)
r
µ dr,
Log Cr
0
which tends to zero as α → 0.
• If y ∈ D2 , then using Lemma 3.3, we have max(δ(x), δ(y)) ≤
Z
I2 1
√
5
1−α( 1+2
)
tn−1
(1 − t)λ−2 Log
Z
4
1−t
µ
S n−1
√
1+ 5
|x
2
dσ(ω)
|x − tω|n−2(m−1)
− y|. Hence,
dt ,
where σ is the normalized measure on the unit sphere S n−1 of Rn .
Now by elementary calculus, we have
Z
dσ(ω)
1
t2(m−1)−n .
n−2(m−1)
n−2(m−1)
(|x| ∨ t)
S n−1 |x − tω|
So
Z
1
I2 1−α
√
1+ 5
2
t2m−3
(1 − t)λ−2 Log
4
1−t
µ dt,
which tends to zero as α tends to zero.
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I TERATED G REEN F UNCTIONS
13
Case 2. n = 2m − 2. In this case we have
Z
1
1
h
iµ dy
I
Log 2 +
2
[x, y] δ(y)λ−2 Log 4
B(x,α)∩D1
δ(y)
Z
1
1
h
iµ dy
+
Log 2 +
2
λ−2
[x, y] δ(y)
B(x,α)∩D2
Log 4
δ(y)
= I1 + I2 .
√
• If y ∈ D1 , it follows from the fact that Log(2 + t) ≤ t for t ≥ 2 that
Z
1
h
iµ dy
I1 4
B(x,α)∩D1 |x − y|(δ(y))λ−2 Log
δ(y)
Z
1
h
iµ dy
4
B(x,α)∩D1 |x − y|λ−1 Log
|x−y|
Z α
rn−λ
µ dr,
Log 4r
0
which tends to zero as α tends to zero.
• If y ∈ D2 , then
Z
1
1
h
iµ dy
I2 Log 2 +
|x − y|2 δ(y)λ−2 Log 4
B(x,α)∩D2
δ(y)
Z
1
h
iµ dy
λ−2
4
B(x,α)∩D2 |x − y|2 δ(y)
Log δ(y)
Z
Z 1
tn−1
dσ(ω)
µ
dt
√
4
2
λ−2 Log
n−1 |x − tω|
1−α( 1+2 5 ) (1 − t)
S
1−t
Z 1
n−1
t
1
µ
dt
√
4
λ−2
(|x| ∨ t)2
Log 1−t
1−α( 1+2 5 ) (1 − t)
Z 1
tn−3
µ dt,
√
4
λ−2 Log
1−α( 1+2 5 ) (1 − t)
1−t
which tends to zero as α tends to zero.
Case 3. n < 2m − 2. In this case
Z
I
B(x,α)∩D
1
h
(δ(y))λ−2 Log
4
δ(y)
iµ dy.
Hence the result follows from Lemma 3.3, using similar arguments as in the above cases.
In the sequel we aim at proving Theorem 1.3. Below we present some preliminary results
which we will need later.
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14
H ABIB M ÂAGLI AND N OUREDDINE Z EDDINI
Proposition 3.5.
a) For each t > 0 and all x, y ∈ D, we have
Z t
sm−1 p(s, x, y) ds Gm,n (x, y).
0
b) Let 0 < t ≤ 1 and x, y ∈ D. Then
t
Z
sm−1 p(s, x, y) ds ,
Gm,n (x, y) 0
provided that
√
i) n > 2m and |x − y| ≤ t; or
ii) n = 2m and [x, y]2 ≤ t ; or
iii) n = 2m − 1 and |x − y|2 + 2δ(x)δ(y) ≤ t.
Proof.
a) Follows from (1.3).
b) We deduce from (1.11) and (1.12) that
Z t
Z
m−1
2m−n
s
p(s, x, y) ds ∼ |x − y|
0
∞
|x−y|2
t
min
n
δ(x)δ(y)
r, 1 r 2 −m−1 e− r dr.
2
|x − y|
Next, we distinguish the following cases
i) n > 2m. In this case the result follows from (1.7) and Theorem 1.1.
ii) n = 2m. Using (1.5) we have
Z t
Z 2
δ(x)δ(y)
m−1
s
p(s, x, y) ds ≥ C
dr
|x−y|2 δ(x)δ(y)r + |x − y|2
0
t
[x, y]2 + δ(x)δ(y)
t
≥ C Log
·
.
δ(x)δ(y) + t
|x − y|2
t
Now since [x, y]2 ≤ t and the function t 7→ δ(x)δ(y)+t
is nondecreasing, then the result
follows from Theorem 1.1.
iii) n = 2m − 1. As in the proof of Theorem 1.1 we distinguish two cases
• If δ(x)δ(y) ≤ |x − y|2 . In this case the result follows from (1.7).
• If δ(x)δ(y) > |x − y|2 . Then
Z t
Z |x−y|2
δ(x)δ(y) δ(x)δ(y) − 1
m−1
s
p(s, x, y) ds ≥ C
r 2 dr.
2
|x − y| |x−y|
0
t
Since |x − y|2 + 2δ(x)δ(y) ≤ t, then
!2
|x − y|
|x − y|
1
|x − y|2
p
+ √
1− √
≤
.
δ(x)δ(y)
t
2
δ(x)δ(y)
Hence
Z
t
δ(x)δ(y) |x − y|
p
|x − y|
δ(x)δ(y)
p
= C δ(x)δ(y)
sm−1 p(s, x, y) ds ≥ C
0
≥ C Gm,n (x, y).
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I TERATED G REEN F UNCTIONS
15
Proposition 3.6. Let q ∈ Km,n (D). Then for each fixed α > 0, we have
Z
δ(y)
(3.1)
sup sup
p (t, x, y)|q(y)| dy := M (α) < ∞.
t≤1
x∈D (|x−y|>α)∩D δ(x)
Proof. Let 0 < t < 1, q ∈ Km,n (D) and 0 < α < 1. Then using (1.11) and (1.12) we have
Z
Z
|x−y|2
1
δ(y)
p (t, x, y)|q(y)| dy n +1
δ 2 (y)e− t |q(y)| dy
t2
(|x−y|>α)∩D δ(x)
(|x−y|>α)∩D
α2 Z
e− t
n +1
δ 2 (y)|q(y)| dy.
t2
D
Hence the result follows from Lemma 3.4.
Proof of Theorem 1.3. 2) ⇒ 1) Assume that
Z Z t
δ(y) m−1
lim sup
s
p(s, x, y)|q(y)| ds dy = 0.
t→0 x∈D D 0 δ(x)
Then by Proposition 3.5, there exists C > 0 such that for α > 0 we have
Z
Z Z α2
δ(y)
δ(y) m−1
Gm,n (x, y)|q(y)| dy ≤ C
s
p(s, x, y) dsdy,
δ(x)
(|x−y|≤α)∩D δ(x)
D 0
which shows that q satisfies (1.4).
1) ⇒ 2) Suppose that q ∈ Km,n (D) and let ε > 0. Then there exists 0 < α < 1 such that
Z
δ(y)
sup
Gm,n (x, y)|q(y)| dy ≤ ε.
x∈D (|x−y|≤α)∩D δ(x)
On the other hand, using Proposition 3.5 and (3.1), we have for 0 < t < 1
Z Z t
δ(y) m−1
s
p(s, x, y)|q(y)| dsdy
D 0 δ(x)
Z
Z t
δ(y) m−1
s
p(s, x, y)|q(y)| dsdy
(|x−y|≤α)∩D 0 δ(x)
Z
Z t
δ(y) m−1
+
s
p(s, x, y)|q(y)| dsdy
(|x−y|>α)∩D 0 δ(x)
Z
δ(y)
Gm,n (x, y)|q(y)| dy
(|x−y|≤α)∩D δ(x)
Z tZ
δ(y)
+
p(s, x, y)|q(y)| dsdy
0
(|x−y|>α)∩D δ(x)
ε + t M (α).
This achieves the proof.
Next we assume that m = 1 and we will give another characterization of the class K1,n (D).
Corollary 3.7. Let n ≥ 3 and q be a measurable function. For α > 0, put
Z Z ∞
δ(y)
Gα q(x) =
e−α s
p(s, x, y)|q(y)| dsdy, for x ∈ D
δ(x)
D 0
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16
H ABIB M ÂAGLI AND N OUREDDINE Z EDDINI
and
Z
1
α
Z
a(α) = sup
x∈D
0
D
δ(y)
p(s, x, y)|q(y)| dyds.
δ(x)
Then there exists C > 0 such that
1
a(α) ≤ kGα qk∞ ≤ C a(α) ,
e
where kGα qk∞ = sup |Gα q(x)|.
x∈D
In particular, we have
q ∈ K1,n (D) ⇐⇒ lim kGα qk∞ = 0.
α→∞
Proof. Let α > 0. Then using the Fubini theorem, we obtain for x ∈ D
Z t Z
Z ∞
δ(y)
−α t
Gα q(x) =
αe
p(s, x, y)|q(y)| dyds dt
0
0
D δ(x)
"Z t Z
#
Z ∞
α
δ(y)
=
e−t
p(s, x, y)|q(y)| dyds dt.
0
D δ(x)
0
Hence, 1e a(α) ≤ kGα qk∞ .
On the other hand if we denote by [t] the integer part of t, then we have


Z ∞
[t] Z k+1 Z
X
α
δ(y)
Gα q(x) ≤
e−t 
p(s, x, y)|q(y)| dyds dt
k
δ(x)
D
0
k=0 α


Z ∞
[t] Z 1 Z
X α
δ(y)
k
≤
e−t 
p s + , x, y |q(y)| dyds dt.
α
0
D δ(x)
k=0 0
Now, using the Chapmann-Kolmogorov identity and the Fubini theorem we obtain
Z 1Z
α
δ(y)
k
p s + , x, y |q(y)|dyds
α
0
D δ(x)
!
Z
Z 1Z
α
δ(y)
δ(z)
k
=
p(s, z, y)|q(y)|dyds
p
, x, z dz
δ(x)
α
0
D δ(z)
D
Z
δ(z)
k
≤ a(α)
p
, x, z dz.
α
D δ(x)
Since the first eigenfunction ϕ1 associated to −∆ satisfies ϕ1 (x) ∼ δ(x) and
Z
p(t, x, z)ϕ1 (z)dz = e−λ1 t ϕ1 (x) ≤ ϕ1 (x),
D
then
kGα qk∞ ≤ Ca(α).
So, the last assertion follows from Theorem 1.3.
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I TERATED G REEN F UNCTIONS
17
R EFERENCES
[1] M. AIZENMAN AND B. SIMON, Brownian motion an Harnack’s inequality for Shrödinger operators, Commun. Pure. Appl. Math., 35 (1982), 209–273.
[2] I. BACHAR, Estimates for the Green function and existence of positive solutions of polyharmonic nonlinear equations with Navier boundary conditions, New Trends in Potential Theory. Theta
(2005), 101–113.
[3] I. BACHAR, Estimates for the Green function and characterization of a certain Kato class by the
Gauss semigroup in the half space , J. Inequal. Pure & Appl. Math., 6(4) (2005), Art. 119. [ONLINE: http://jipam.vu.edu.au/article.php?sid=593].
[4] I. BACHAR AND H. MÂAGLI, Estimates on the Green function and existence of positive solutions
of nonlinear singular elliptic equations in the half space, Positivity, 9 (2005), 153–192.
[5] I. BACHAR, H. MÂAGLI, S. MASMOUDI AND M. ZRIBI, Estimates on the Green function and
singular solutions for polyharmonic nonlinear equations, Abstract and Applied Analysis, 12 (2003),
715–741.
[6] K.L. CHUNG AND Z. ZHAO, From Brownian Motion to Schrödinger’s Equation, Springer Verlag
(1995).
[7] E.B. DAVIES, The equivalence of certain heat kernel and Green function bounds, J. Func. Anal.,
71 (1987), 88–103.
[8] E.B. DAVIES AND B. SIMON, Ultracontractivity and the heat kernel for Shrödinger operators and
Dirichlet laplacians, J. Func. Anal., 59 (1984), 335–395.
[9] H.C. GRUNAU AND G. SWEERS, Sharp estimates for iterated Green functions, Proc. Royal Soc.
Edinburgh. Sec A, 132 (2002), 91–120.
[10] A.C. LAZER AND P.J. MCKENNA, On a singular nonlinear elliptic boundary value problem, Proc.
Amer. Math. Soc., 111(3) (1991), 721–730.
[11] H. MÂAGLI AND M. ZRIBI, On a new Kato class and singular solutions of a nonlinear elliptic
equation in bounded domain of Rn , Positivity, (2005), 667–686.
[12] H. MÂAGLI AND M. ZRIBI, Existence of positive solutions for some polyharmonic nonlinear
equations in Rn , Abstract and Applied Analysis, (2005).
[13] S.C. PORT
(1978).
AND
C.J. STONE, Brownian Motion and Classical Potential Theory, Academic Press
[14] J.A. VAN CASTEREN, Generators Strongly Continuous Semi-groups, Pitmann Advanced Publishing Program, Boston, (1985).
[15] Qi. S. ZHANG, The boundary behavior of heat kernels of Dirichlet laplacians, J. Diff. Equations,
182 (2002), 416–430.
[16] Qi. S. ZHANG, The global behavior of the heat kernels in exterior domains, J. Func. Anal., 200
(2003), 160–176.
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 26, 17 pp.
http://jipam.vu.edu.au/