MULTIPLICATION OF SUBHARMONIC FUNCTIONS
R. SUPPER
Université Louis Pasteur
UFR de Mathématique et Informatique, URA CNRS 001
7 rue René Descartes
F–67 084 Strasbourg Cedex, France
EMail: supper@math.u-strasbg.fr
Multiplication of
Subharmonic Functions
R. Supper
vol. 9, iss. 4, art. 118, 2008
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Contents
Received:
31 May, 2008
Accepted:
12 November, 2008
Communicated by:
JJ
II
S.S. Dragomir
J
I
2000 AMS Sub. Class.:
31B05, 33B15, 26D15, 30D45.
Page 1 of 40
Key words:
Gamma and Beta functions, growth of subharmonic functions in the unit ball.
Abstract:
We study subharmonic functions in the unit ball of RN , with either a Bloch–
type growth or a growth described through integral conditions involving some
involutions of the ball. Considering mappings u 7→ gu between sets of functions
with a prescribed growth, we study how the choice of these sets is related to the
growth of the function g.
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Contents
1
Introduction
3
2
Notations and Main Results
4
3
Some Preliminaries
7
4
Proof of Theorem 2.2
4.1 Proof of Theorem 2.2 in the case (i)
4.2 Proof of Theorem 2.2 in the case (ii)
4.3 Proof of Theorem 2.2 in the case (iii)
4.4 Proof of Theorem 2.2 in the case (iv)
4.5 Proof of Theorem 2.2 in the case (v)
4.6 Proof of Theorem 2.2 in the case (vi)
5
6
Multiplication of
Subharmonic Functions
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14
22
24
25
27
28
29
The Situation with Radial Subharmonic Functions
5.1 The example of u : x 7→ (1 − |x|2 )−A with A ≥ 0 . . . . . . . . . .
5.2 Proof of Theorem 2.3 . . . . . . . . . . . . . . . . . . . . . . . . .
30
30
33
Page 2 of 40
Annex: The Sets SX λ and SY α,β,γ for some Special Values of λ, α, β, γ
35
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R. Supper
vol. 9, iss. 4, art. 118, 2008
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JJ
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1.
Introduction
This paper is devoted to functions u which are defined in the unit ball BN of RN
(relative to the Euclidean norm |·|), whose growth is described by the above boundedness on BN of x 7→ (1 − |x|2 )α v(x) for some parameter α. The function v may
denote merely u or some integral involving u and involutions Φx (precise definitions
and notations will be detailed in Section 2). In the first (resp. second) case, u is said
to belong to the set X (resp. Y). Given a function g defined on BN , we try to obtain
links between the growth of g and information on such mappings as
Y → X,
u 7→ gu.
This work is motivated by the situation known in the case of holomorphic functions
f in the unit disk D of C. Such a function is said to belong to the Bloch space Bλ if
||f ||Bλ := |f (0)| + sup(1 − |z|2 )λ |f 0 (z)| < +∞.
z∈D
It is said to belong to the space BM OAµ if
Z
2
2
||f ||BM OAµ := |f (0)| + sup (1 − |z|2 )2µ−2 |f 0 (z)|2 (1 − |ϕa (z)|2 ) dA(z) < +∞
a∈D
D
a−z
with dA(z) the normalized area measure element on D and ϕa (z) = 1−az
.
Given h a holomorphic function on D, the operator Ih : f 7→ Ih (f ) defined by:
Z z
(Ih (f ))(z) =
h(ζ) f 0 (ζ) dζ
∀z ∈ D
0
was studied for instance in [7] where it was proved that Ih : BM OAµ → Bλ is
bounded (with respect to the above norms) if and only if h ∈ Bλ−µ+1 (assuming
1 < µ < λ).
Since |f 0 |2 is subharmonic in the unit ball of R2 , the question naturally arose
whether some similar phenomena occur for subharmonic functions in BN for N ≥ 2.
Multiplication of
Subharmonic Functions
R. Supper
vol. 9, iss. 4, art. 118, 2008
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2.
Notations and Main Results
Let BN = {x ∈ RN : |x| < 1} with N ∈ N, N ≥ 2 and |·| the Euclidean norm in
RN . Given a ∈ BN , let Φa : BN → BN denote the involution defined by:
p
a − Pa (x) − 1 − |a|2 Qa (x)
Φa (x) =
∀x ∈ BN ,
1 − hx, ai
where
Multiplication of
Subharmonic Functions
R. Supper
hx, ai =
N
X
x j aj ,
Pa (x) =
j=1
hx, ai
a,
|a|2
vol. 9, iss. 4, art. 118, 2008
Qa (x) = x − Pa (x)
for all x = (x1 , x2 , . . . , xN ) ∈ RN and a = (a1 , a2 , . . . , aN ) ∈ RN , with Pa (x) = 0
if a = 0. We refer to [4, pp. 25–26] and [1, p. 115] for the main properties of the
map Φa (initially defined in the unit ball of CN ). For instance, we will make use of
the relation:
(1 − |a|2 ) (1 − |x|2 )
1 − |Φa (x)|2 =
.
(1 − hx, ai)2
In the following, α, β, γ and λ are given real numbers, with γ ≥ 0.
Definition 2.1. Let Xλ denote the set of all functions u : BN → [−∞, +∞[ satisfying:
MXλ (u) := sup (1 − |x|2 )λ u(x) < +∞.
x∈BN
Let Yα,β,γ denote the set of all measurable functions u : BN → [−∞, +∞[ satisfying:
Z
2 α
MYα,β,γ (u) := sup (1 − |a| )
(1 − |x|2 )β u(x) (1 − |Φa (x)|2 )γ dx < +∞.
a∈BN
BN
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The subset SX λ (resp. SY α,β,γ ) gathers all u ∈ Xλ (resp. u ∈ Yα,β,γ ) which
moreover are subharmonic and non–negative. The subset RSY α,β,γ gathers all u ∈
SY α,β,γ which moreover are radial.
Remark 1. When λ < 0 (resp. α + β < −N or α < −γ), the set SX λ (resp.
SY α,β,γ ) merely reduces to the single function u ≡ 0 (see Propositions 6.2, 6.3 and
6.4).
In Proposition 3.1 and Corollary 3.2, we will establish that SY α,β,γ ⊂ SX α+β+N
and that there exists a constant C > 0 such that
Multiplication of
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R. Supper
vol. 9, iss. 4, art. 118, 2008
MXλ+α+β+N (gu) ≤ C MXλ (g) MYα,β,γ (u)
for all u ∈ SY α,β,γ and all g ∈ Xλ with MXλ (g) ≥ 0. We will next study whether
some kind of a “converse” holds and obtain the following:
Theorem 2.2. Given λ ∈ R and g : BN → [0, +∞[ a subharmonic function satisfying:
∃C 0 > 0
MXλ+α+β+N (gu) ≤ C 0 MYα,β,γ (u)
∀u ∈ SY α,β,γ ,
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Page 5 of 40
then g ∈ Xλ+ N −1 in each of the six cases gathered in the following Table 1.
2
Theorem 2.3. Given λ ∈ R and g a subharmonic function defined on BN , satisfying:
00
00
∃C > 0
MXλ+α+β+N (gu) ≤ C MYα,β,γ (u)
∀u ∈ RSY α,β,γ ,
then g ∈ SX λ+α+ N −1 provided that α ≥ 0, β ≥ − N 2+1 , γ >
2
N −1
.
2
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α
case
(i)
(ii)
β
+β
β > − N2+1
γ > max(α, −1 − β)
α=β+1
− N4+3
γ > |1 + β|
α=
N +1
2
N +1
2
β>
−γ
β ≥ −γ
(iv)
α=1
β≥0
(v)
α=1+β−γ
β > −1
(vi)
β+1
2
− 12
(iii)
Multiplication of
Subharmonic Functions
γ
α=
α=
β≥
N +1
4
<γ<
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vol. 9, iss. 4, art. 118, 2008
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N +1
2
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γ>1
1+β
2
<γ<β+
N +3
4
γ > 1+β
2 JJ
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Table 1: Six situations where Theorem 2.2 shows that g belongs to the set Xλ+ N −1 .
2
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3.
Some Preliminaries
Notation 1. Given a ∈ BN and R ∈]0, 1[, let B(a, Ra ) = {x ∈ BN : |x − a| < Ra }
with
1 − |a|2
Ra = R
.
1 + R|a|
Proposition 3.1. There exists a C > 0 depending only on N , β, γ, such that:
MXα+β+N (u) ≤ C MYα,β,γ (u)
Multiplication of
Subharmonic Functions
∀u ∈ SY α,β,γ .
R. Supper
vol. 9, iss. 4, art. 118, 2008
Proof. Let some R ∈]0, 1[ be fixed in the following. Since u ≥ 0, we obtain for any
a ∈ BN :
Z
2 α
MYα,β,γ (u) ≥ (1 − |a| )
(1 − |x|2 )β u(x) (1 − |Φa (x)|2 )γ dx
ZBN
≥ (1 − |a|2 )α
(1 − |x|2 )β u(x) (1 − |Φa (x)|2 )γ dx.
B(a,Ra )
It follows from Lemma 1 of [6] that
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B(a, Ra ) ⊂ E(a, R) = {x ∈ BN : |Φa (x)| < R},
hence:
(3.1)
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MYα,β,γ (u) ≥ (1 − R2 )γ (1 − |a|2 )α
Z
(1 − |x|2 )β u(x) dx
B(a,Ra )
as γ ≥ 0. From Lemmas 1 and 5 of [5], it is known that
1−R
1 − |x|2
≤
≤2
1+R
1 − |a|2
∀x ∈ B(a, Ra ).
Close
Let Cβ =
1−R β
1+R
if β ≥ 0 and Cβ = 2β if β < 0. Hence
Z
2 γ
2 α+β
MYα,β,γ (u) ≥ Cβ (1 − R ) (1 − |a| )
u(x) dx.
B(a,Ra )
N
N/2
2π
The volume of B(a, Ra ) is σN (RNa ) with σN = Γ(N/2)
the area of the unit sphere
R
N
2
SN in R (see [2, p. 29]) and Ra ≥ 1+R (1 − |a| ). The subharmonicity of u now
provides:
2 γ
2 α+β
MYα,β,γ (u) ≥ Cβ (1 − R ) (1 − |a| )
≥ Cβ
(Ra )N
u(a) σN
N
σN RN (1 − R)γ
(1 − |a|2 )α+β+N u(a).
N (1 + R)N −γ
Multiplication of
Subharmonic Functions
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vol. 9, iss. 4, art. 118, 2008
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Corollary 3.2. Let g ∈ Xλ with MXλ (g) ≥ 0. Then:
MXλ+α+β+N (gu) ≤ C MXλ (g) MYα,β,γ (u)
∀u ∈ SY α,β,γ
where the constant C stems from Proposition 3.1.
(1 − |x| )
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2 α+β+N
g(x) u(x) ≤ MXλ (g) (1 − |x| )
≤ MXλ (g) MXα+β+N (u)
u(x)
because of MXλ (g) ≥ 0.
Lemma 3.3. Given a ∈ BN and R ∈]0, 1[, the following holds for any x ∈ B(a, Ra ):
1
1
1 − hx, ai
1 + 2R |a|
<
≤
≤
< 2 and
2
2
1 + R |a|
1 − |a|
1 + R |a|
II
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Proof. Since u ≥ 0, we have for any x ∈ BN :
2 λ+α+β+N
JJ
1
1 − hx, ai
1+R
<
<2
.
2
4
1 − |x|
1−R
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Proof. Clearly hx, ai = ha + y, ai = |a|2 + hy, ai with |y| < Ra . From the CauchySchwarz inequality, it follows that −Ra |a| ≤ hy, ai ≤ Ra |a|. Hence:
1 − |a|2 − R |a|
1 − |a|2
1 − |a|2
≤ 1 − hx, ai ≤ 1 − |a|2 + R |a|
.
1 + R|a|
1 + R|a|
The term on the left equals
(1 − |a|2 ) 1 −
R |a|
1 + R|a|
= (1 − |a|2 )
1
1 + R|a|
and 1 + R|a| < 2. The term on the right equals
R |a|
2
(1 − |a| ) 1 +
,
1 + R|a|
with
R|a|
1+R|a|
< 1. Now
1 − hx, ai
1 − hx, ai 1 − |a|2
=
1 − |x|2
1 − |a|2 1 − |x|2
and the last inequalities follow from Lemmas 1 and 5 of [5].
Lemma 3.4. Let H = {(s, t) ∈ R2 : t ≥ 0, s2 + t2 < 1} and P > −1, Q > −1,
T > −1. Then

0
if P is odd;

ZZ

sP tQ (1 − s2 − t2 )T ds dt =
P +1
Γ(T +1)
( Q+1

2 )
H
 Γ( 2 ) PΓ+Q
if P is even.
2 Γ( 2 +T +2)
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Proof. With polar coordinates s = r cos θ, t = r sin θ, this integral turns into I1 I2
with
Z 1
Z π
P +Q
2 T
I1 =
r
(1 − r ) r dr
and
I2 =
(cos θ)P (sin θ)Q dθ.
0
0
Keeping in mind the various expressions for the Beta function (see [3, pp. 67–68]):
Z 1
B(x, y) =
ξ x−1 (1 − ξ)y−1 dξ
0
Z π/2
Γ(x) Γ(y)
=2
(cos θ)2x−1 (sin θ)2y−1 dθ =
Γ(x + y)
0
(with x > 0 and y > 0), the change of variable ω = r2 leads to:
Z
1 1 P +Q
I1 =
ω 2 (1 − ω)T dω
2 0
+
1
Γ(T + 1)
Γ P +Q
P +Q
1
2
.
= B
+ 1, T + 1 =
P +Q
2
2
2Γ 2 + T + 2
When P is odd, I2 = 0 because cos(π − θ) = − cos(θ). However, when P is even:
Z π/2
I2 = 2
(cos θ)P (sin θ)Q dθ
0
Q+1
Γ P +1
Γ
P +1 Q+1
2
2
.
=B
,
=
2
2
Γ P +Q
+
1
2
Multiplication of
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Lemma 3.5. Given A ≥ 0 and a ∈ BN , let u and fa denote the functions defined
1
1
on BN by u(x) = (1−|x|
2 )A and fa (x) = (1−hx,ai)A ∀x ∈ BN . They are both subharmonic in BN .
Remark 2. u is radial, but not fa .
Proof. For u, the result of Lemma 3.5 has already been proved in Proposition 1 of
[5]. For any j ∈ {1, 2, . . . , N }, we now compute:
∂fa
(x) = aj A (1−hx, ai)−A−1
∂xj
so that:
(∆fa )(x) =
and
∂ 2 fa
(x) = (aj )2 A (A+1)(1−hx, ai)−A−2 ,
2
∂xj
|a|2 A (A + 1)
≥0
(1 − hx, ai)A+2
∀x ∈ BN .
Multiplication of
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vol. 9, iss. 4, art. 118, 2008
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Remark 3. Given A ≥ 0, A0 ≥ 0, the function fa defined on BN by
1
fa (x) =
A
(1 − hx, ai) (1 − |x|2 )A0
is subharmonic too. The computation
(∆fa )(x) ≥ fa (x)
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A |a|
2A0 |x|
−
1 − hx, ai 1 − |x|2
2
≥0
is left to the reader.
Proposition 3.6. Given N ∈ N, N > 3, (s, t, b1 , b2 ) ∈ R4 such that |s b1 |+|t b2 | < 1
and P > 0, let
Z π
(sin θ)N −3 dθ
IP (s, t, b1 , b2 ) =
.
P
0 (1 − s b1 − t b2 cos θ)
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Then
√ Γ
IP (s, t, b1 , b2 ) = π
2k
− 1 X X Γ(j + 2k + P )
t b2
j
(b1 s)
.
2
Γ(P ) k∈N j∈N k! j! Γ N 2−1 + k
N
2
Proof. As
t b2 cos θ t b2 ≤
1 − s b1 1 − s b1 < 1,
the following development is valid:
Z π
(sin θ)N −3 dθ
IP (s, t, b1 , b2 ) =
P
t b2 cos θ
0 (1 − s b )P
1 − 1−s b1
1
X Γ(n + P ) t b2 n Z π
1
=
(sin θ)N −3 (cos θ)n dθ.
(1 − s b1 )P n∈N n! Γ(P ) 1 − s b1
0
The last integral vanishes when n is odd. When n is even (n = 2k), then
Z π/2
N −2
1
N −3
2k
2
(sin θ)
(cos θ) dθ = B
,k +
2
2
0
N −2
Γ 2 Γ k + 12
=
Γ N 2−1 + k
√
Γ N 2−2 (2k)! π
=
Γ N 2−1 + k 22k k!
by [3, p. 40]. Hence:
IP (s, t, b1 , b2 ) =
Γ
N −2
2
√
Γ(P )
πX
k∈N
(t b2 )2k
Γ(2k + P )
.
Γ N 2−1 + k 22k k! (1 − s b1 )2k+P
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The result follows from the expansion
X Γ(j + 2k + P )
Γ(2k + P )
=
(b1 s)j .
(1 − s b1 )2k+P
j!
j∈N
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4.
Proof of Theorem 2.2
The cases (i), (ii), (iii), (iv), (v) and (vi) of Theorem 2.2 will be proved separately at
the end of this section.
Theorem 4.1. Given A > 0, P > 0, T > −1 and N ∈ N (N ≥ 2) such that
1 ≤ A + P ≤ N + 1 + 2T , let
Z
(1 − |x|2 )T
IA,P,T (a, b) =
dx
∀a ∈ BN , ∀b ∈ BN
A
P
BN (1 − hx, ai) (1 − hx, bi)
and τ a number satisfying both
IA,P,T (a, b) ≤
P −A
2
< τ < P and 0 ≤ τ ≤
A+P
2
−τ
IA,P,T (a, b) ≤
A+P
,
2
vol. 9, iss. 4, art. 118, 2008
Then
∀a ∈ BN , ∀b ∈ BN
(1 − |b|2 )τ
where the constant K is independent of a and b.
Example 4.1. If P > A and τ =
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K
(1 − |a|2 )
A+P
.
2
Multiplication of
Subharmonic Functions
then
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K
(1 − |b|2 )
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A+P
2
∀a ∈ BN , ∀b ∈ BN ,
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with
K = 2A+P −1 π
N −1
2
Γ(T + 1)
Γ
Γ(P )
P −A
2
.
Example 4.2. If P < A and τ = 0, then
IA,P,T (a, b) ≤
K
(1 − |a|2 )
A+P
2
∀a ∈ BN , ∀b ∈ BN ,
Close
with
A+P −1
K=2
π
N −1
2
Γ(T + 1)
Γ
Γ(A)
A−P
2
.
Proof. Up to a unitary transform, we assume a = (|a|, 0, 0, . . . , 0) and b = (b1 , b2 , 0, . . . , 0).
Proof of Theorem 4.1 in the case N > 3. Polar coordinates in RN provide the formulas: x1 = r cos θ1 with r = |x|, x2 = r sin θ1 cos θ2 (the formulas for x3 , . . . , xN
are available in [9, p. 15]) where θ1 , θ2 ,. . . , θN −2 ∈]0, π[ and θN −1 ∈]0, 2π[. The
volume element dx becomes rN −1 dr dσ (N ) where dσ (N ) denotes the area element
on SN , with
dσ (N ) = (sin θ1 )N −2 (sin θ2 )N −3 dθ1 dθ2 dσ (N −2)
(see [9, p. 15] for full details). Here θ2 ∈]0, π[ since N > 3. In the following, we
will write s = r cos θ1 and t = r sin θ1 , thus hx, bi = s b1 + t b2 cos θ2 and
Z πZ 1
(1 − r2 )T rN −1 (sin θ1 )N −2 IP (s, t, b1 , b2 )
(4.1) IA,P,T (a, b) = σN −2
dr dθ1
(1 − |a|s)A
0
0
with IP (s, t, b1 , b2 ) defined in the previous proposition. From [2, p. 29] we notice
that
N −1
N −2 √
σN −2 Γ
π = 2π 2 .
2
The expansion
X Γ(` + A)
1
=
(|a| s)`
(1 − |a| s)A
`!
Γ(A)
`∈N
leads to:
N −1
2π 2
IA,P,T (a, b) =
Γ(P ) Γ(A)
X
(k,j,`)∈N3
Γ(j + 2k + P ) Γ(` + A)
(b1 )j
k! j! `! Γ( N 2−1 + k)
b2
2
2k
|a|` Jk,j,`
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where
Z
π
1
Z
sj+` t2k (1 − r2 )T rN −1 (sin θ1 )N −2 dr dθ1
Jk,j,` =
Z0Z
=
0
sj+` t2k+N −2 (1 − s2 − t2 )T ds dt
H
with H as in Lemma 3.4. Now Jk,j,` = 0 unless j + ` = 2h (h ∈ N). Thus:
Multiplication of
Subharmonic Functions
N −1
π 2
IA,P,T (a, b) =
Γ(P ) Γ(A)
2k
2h
X X
Γ(j + 2k + P ) Γ(2h − j + A) Γ h + 21 Γ(T + 1)
b2
j
(b1 )
|a|2h−j
×
N
2
k! j! (2h − j)! Γ k + h + 2 + T + 1
(k,h)∈N2 j=0
Taking [3, p. 40] into account:
(4.2) IA,P,T (a, b) =
2h
π Γ(T + 1) X X (2h)! B(j + 2k + P, 2h − j + A)
Γ(P ) Γ(A)
22h+2k h! k! j! (2h − j)!
2 j=0
N
2
(k,h)∈N
×
Let
L=
Γ(2k + P + 2h + A) j 2k 2h−j
b b |a|
.
Γ(k + h + N2 + T + 1) 1 2
2z−1
π Γ(2z) = 2
1
Γ(z) Γ z +
2
vol. 9, iss. 4, art. 118, 2008
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2P +A−1 Γ(T + 1) N −1
π 2 .
Γ(P ) Γ(A)
The duplication formula
√
R. Supper
(see [3, p. 45]) is applied with 2z = 2k + P + 2h + A. Now
A+P +1
N
Γ(k + h +
)≤Γ k+h+
+T +1
2
2
since Γ increases on [1, +∞[ and
1≤k+h+
N
A+P +1
≤k+h+
+ T + 1.
2
2
This leads to:
Multiplication of
Subharmonic Functions
R. Supper
vol. 9, iss. 4, art. 118, 2008
IA,P,T (a, b)
2h
X X
(2h)! B(j + 2k + P, 2h − j + A) Γ k + h + A+P
2h−j
2
bj1 b2k
≤L
2 |a|
h!
k!
j!
(2h
−
j)!
(k,h)∈N2 j=0
2h
X Γ k + h + A+P
X
(2h)!
2k
2
=L
b2
bj1 |a|2h−j B(j + 2k + P, 2h − j + A).
h! k!
j! (2h − j)!
2
j=0
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(k,h)∈N
Page 17 of 40
The last sum turns into:
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Z 1
2h
X
(2h)! bj1 |a|2h−j
ξ j+2k+P −1 (1 − ξ)2h−j+A−1 dξ
j!
(2h
−
j)!
0
j=0
!
Z 1 X
2h
(2h)! (b1 ξ)j [(1 − ξ) |a|]2h−j
=
ξ 2k+P −1 (1 − ξ)A−1 dξ
j!
(2h
−
j)!
0
j=0
Z 1
=
[b1 ξ + |a| (1 − ξ)]2h ξ 2k+P −1 (1 − ξ)A−1 dξ.
0
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Hence the majorant of IA,P,T (a, b) becomes:
Z
1
X (b2 ξ)2k
X Γ(h + k +
!
A+P
)
2
[b1 ξ + |a| (1 − ξ)]2h ξ P −1 (1−ξ)A−1 dξ
h!
0 k∈N
h∈N
k+ A+P
Z 1X
A+P
2
Γ(k + 2 ) (b2 ξ)2k
1
=L
ξ P −1 (1−ξ)A−1 dξ
2
k!
1 − [b1 ξ + |a| (1 − ξ)]
0 k∈N
L
k!
R. Supper
according to the expansion
vol. 9, iss. 4, art. 118, 2008
Γ(C)
=
(1 − X)C
X Γ(h + C)
h!
h∈N
Xh
Title Page
with |X| < 1 when C > 0 (see [8, p. 53]). Here X = [b1 ξ + |a| (1 − ξ)]2 belongs
to ] − 1, 1[ since b1 and |a| do and ξ ∈ [0, 1]. The same expansion now applies with
C=
Multiplication of
Subharmonic Functions
2
A+P
2
X=
and
(b2 ξ)
1 − [b1 ξ + |a| (1 − ξ)]2
since |X| < 1, as
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2
2
δ(ξ) := (b2 ξ) + [b1 ξ + |a| (1 − ξ)]
2
2
2
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2
= |b| ξ + |a| (1 − ξ) + 2ξ(1 − ξ) b1 |a|
≤ |b|2 ξ 2 + |a|2 (1 − ξ)2 + 2ξ(1 − ξ)|b| |a|
= [ξ |b| + |a| (1 − ξ)]2 < 1.
Close
Hence
IA,P,T (a, b)
Z 1
Γ A+P
ξ P −1 (1 − ξ)A−1 dξ
2
≤L
A+P
A+P
2
0
(b2 ξ)2
(1 − [b1 ξ + |a| (1 − ξ)]2 ) 2
1 − 1−[b1 ξ+|a|
(1−ξ)]2
Z 1
A+P
ξ P −1 (1 − ξ)A−1 dξ
=L· Γ
A+P .
2
0 (1 − [b1 ξ + |a| (1 − ξ)]2 − (b2 ξ)2 ) 2
Multiplication of
Subharmonic Functions
R. Supper
vol. 9, iss. 4, art. 118, 2008
Now
1 − δ(ξ) ≥ 1 − [ξ |b| + |a| (1 − ξ)]2
Title Page
≥ 1 − [ξ |b| + (1 − ξ)]2
= ξ(1 − |b|)[2 − ξ(1 − |b|)]
Contents
≥ ξ(1 − |b|2 )
since
[2 − ξ(1 − |b|)] − (1 + |b|) = (1 − ξ)(1 − |b|) ≥ 0.
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Similarly,
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1 − δ(ξ) ≥ (1 − ξ)(1 − |a|2 ).
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Thus
1
[1 − δ(ξ)]
since τ ≥ 0 and
A+P
2
IA,P,T (a, b) ≤
A+P
2
≤
1
[(1 − ξ)(1 −
|a|2 )]
− τ ≥ 0. Finally:
L · Γ A+P
2
(1 −
|a|2 )
A+P
2
−τ
(1 −
Z
|b|2 )τ
0
A+P
2
Close
−τ
[ξ(1 − |b|2 )]τ
1
ξ P −τ −1 (1 − ξ)A+τ −
A+P
2
−1
dξ.
This integral converges since P − τ > 0 and
A+τ −
A+P
A−P
=
+ τ > 0.
2
2
Now the result follows with
A+P
A−P
A−P
K =L·Γ
B P − τ,
+ τ = L Γ(P − τ ) Γ
+τ .
2
2
2
Multiplication of
Subharmonic Functions
R. Supper
Proof of Theorem 4.1 in the case N = 3. Here
Z πZ 1
(1 − r2 )T r2 (sin θ1 ) JP (s, t, b1 , b2 )
IA,P,T (a, b) =
dr dθ1 ,
(1 − |a| s)A
0
0
vol. 9, iss. 4, art. 118, 2008
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Contents
where
Z
JP (s, t, b1 , b2 ) =
0
2π
dθ2
= 2 IP (s, t, b1 , b2 )
(1 − s b1 − t b2 cos θ2 )P
with IP (s, t, b1 , b2 ) as in Proposition 3.6, with N = 3. Hence IA,P,T (a, b) has the
same expression as in Formula (4.1), with N = 3, since σ1 = 2. Thus the proof ends
in the same manner as that above in the case N > 3.
Proof of Theorem 4.1 in the case N = 2. Now x1 = s = r cos θ and x2 = t =
r sin θ:
Z 2π Z 1
(1 − r2 )T r dr dθ
IA,P,T (a, b) =
P
A
0
0 (1 − |a| s) (1 − s b1 − t b2 )
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Z
=
B2 `∈N
=
X (t b2 )n Γ(n + P )
(1 − s2 − t2 )T ds dt
n+P
`! Γ(A)
n!
Γ(P
)
(1
−
s
b
)
1
n∈N
Z
Γ(` + A) |a|` (b2 )n Γ(j + n + P ) (b1 )j
s`+j tn (1 − s2 − t2 )T ds dt.
`! Γ(A) n! Γ(P ) j!
B2
X Γ(` + A)
X
(`,n,j)∈N3
(|a| s)`
The last integral vanishes when n is odd or ` + j odd. Otherwise (n = 2k and
` + j = 2h), it equals
Z
1
1
Γ
h
+
Γ
k
+
Γ(T + 1)
2
2
2 s`+j tn (1 − s2 − t2 )T ds dt =
Γ(k + h + T + 2)
H
by Lemma 3.4 and turns into
n! (2h)! π Γ(T + 1)
2h+2k
2
h! k! Γ(k + h + T + 2)
according to [3, p. 40]. Thus IA,P,T (a, b) is again recognized as Formula (4.2) now
with N = 2 and the proof ends as for the case N > 3.
We now present an example of a family of functions {fa }a which is uniformly
bounded above in Yα,β,γ :
Corollary 4.2. Given β > − N 2+1 (N ≥ 2) let α = N 2+1 +β and γ > max(α, −1−β).
1
For any a ∈ BN let fa denote the function defined by: fa (x) = (1−hx,ai)
2α , ∀x ∈ BN .
Then fa ∈ Yα,β,γ , ∀a ∈ BN . Moreover, there exists K > 0 such that MYα,β,γ (fa ) ≤
K, ∀a ∈ BN .
Remark 4. This constant K is the same as that in the previous theorem, with A = 2α,
P = 2γ and T = β + γ.
Multiplication of
Subharmonic Functions
R. Supper
vol. 9, iss. 4, art. 118, 2008
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Proof. With the above choices for parameters A, P , T , we actually have: P > A >
0, T > −1 and
A + P = 2α + 2γ = N + 1 + 2β + 2γ = N + 1 + 2T > 1.
The conditions 0 ≤ τ ≤ α + γ together with γ − α < τ < 2γ reduce to: γ − α <
τ ≤ α + γ. Let
Z
2 α
(4.3)
Jb (fa ) = (1 − |b| )
(1 − |x|2 )β fa (x) (1 − |Φb (x)|2 )γ dx.
BN
Multiplication of
Subharmonic Functions
R. Supper
vol. 9, iss. 4, art. 118, 2008
Now
(1 − |x|2 )β+γ
dx
N +1+2β (1 − hx, bi)2γ
BN (1 − hx, ai)
∀a ∈ BN , ∀b ∈ BN
2 α+γ
Z
Jb (fa ) = (1 − |b| )
≤K
according to Theorem 4.1 applied with τ = α + γ =
4.1.
A+P
.
2
Proof of Theorem 2.2 in the case (i)
Given R ∈]0, 1[, the subharmonicity of g provides for any a ∈ BN the majoration:
Z
1
g(a) ≤
g(x) dx
Va B(a,Ra )
with Va the volume of B(a, Ra ). From Lemma 3.3, it is clear that:
A
1 + R 1 − |x|2
1≤ 2
∀x ∈ B(a, Ra )
1 − R 1 − hx, ai
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with A = 2α > 0. Now g(x) ≥ 0, ∀x ∈ BN . With fa as in Corollary 4.2, this leads
to:
A Z
1+R
Va g(a) ≤ 2
(1 − |x|2 )A fa (x) g(x) dx.
1−R
B(a,Ra )
Now
A=α+β+
N −1
N +1
=α+β+N −
,
2
2
Multiplication of
Subharmonic Functions
thus
R. Supper
A Z
(1 − |x|2 )λ+α+β+N fa (x) g(x)
1+R
dx
N −1
1−R
(1 − |x|2 )λ+ 2
B(a,Ra )
A Z
dx
1+R
0
≤CK 2
λ+ N 2−1
1−R
B(a,Ra ) (1 − |x|2 )
Va g(a) ≤
from Corollary 4.2. Lemmas 1 and 5 of [5] provide
1 − |x|2
1 − |a|2
λ+ N2−1
≥ Cλ+ N −1
2
∀x ∈ B(a, Ra ),
with Cλ+ N −1 defined in the same pattern as Cβ in the proof of Proposition 3.1.
2
Finally:
A
Va
C 0K
1+R
2
Va g(a) ≤
N −1 ,
1−R
Cλ+ N −1
(1 − |a|2 )λ+ 2
2
thus
vol. 9, iss. 4, art. 118, 2008
2
C 0K
MXλ+ N −1 (g) ≤
Cλ+ N −1
2
2
1+R
2
1−R
2α
∀R ∈]0, 1[.
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The majorant is an increasing function with respect to R. Letting R tend toward 0+ ,
we get:
C 0 K 2α
MXλ+ N −1 (g) ≤
2 .
Cλ+ N −1
2
2
4.2.
Proof of Theorem 2.2 in the case (ii)
Multiplication of
Subharmonic Functions
Here we work with fa defined by:
1
fa (x) =
(1 − hx, ai)A
R. Supper
where
A = α + β + N.
Theorem 4.1 applies once again, with A = N + 1 + 2β > N 2−1 > 0, P = 2γ > 0
and T = β + γ > −1 (because γ > −1 − β). Condition A + P = N + 1 + 2T is
fulfilled too. Moreover τ := α + γ = β + γ + 1 satisfies both 0 ≤ τ ≤ β + γ + N 2+1
(obviously 0 < β + γ + 1 and 1 < N 2+1 ) and γ − β − N 2+1 < τ < 2γ:
N +3
N +1
τ −γ+β+
= 2β +
> 0 and 2γ − τ = γ − 1 − β > 0.
2
2
With such a choice for τ we have
A+P
N +1
N −1
−τ =
−1=
,
2
2
2
thus
K
(4.4)
IA,P,T (a, b) ≤
∀a ∈ BN , ∀b ∈ BN .
N +1
(1 − |a|2 ) 2 −1 (1 − |b|2 )α+γ
Hence, Jb (fa ) defined in Formula (4.3) now satisfies
(4.5)
Jb (fa ) ≤
K
(1 − |a|2 )
N −1
2
∀a ∈ BN , ∀b ∈ BN .
vol. 9, iss. 4, art. 118, 2008
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In other words,
(4.6)
MYα,β,γ (fa ) ≤
K
(1 − |a|2 )
N −1
2
∀a ∈ BN .
This implies:
(4.7)
MXλ+α+β+N (g fa ) ≤
C 0K
(1 − |a|2 )
N −1
2
∀a ∈ BN .
R. Supper
With R and Va as in the previous proof, we obtain here:
A Z
1+R
(1 − |x|2 )λ+α+β+N fa (x) g(x)
Va g(a) ≤ 2
dx
1−R
(1 − |x|2 )λ
B(a,Ra )
A Z
C 0K
1+R
dx
≤
2
N −1
2 λ
1−R
(1 − |a|2 ) 2
B(a,Ra ) (1 − |x| )
Va
and the last integral is majorized by Cλ (1−|a|
2 )λ with Cλ defined similarly to Cβ in
the proof of Proposition 3.1. Finally:
MXλ+ N −1 (g) ≤
2
C 0 K N +1+2β
2
.
Cλ
1
(1 −
− |x|2 )β+γ
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Here fa is defined by:
hx, ai)A (1
vol. 9, iss. 4, art. 118, 2008
Go Back
4.3. Proof of Theorem 2.2 in the case (iii)
fa (x) =
Multiplication of
Subharmonic Functions
∀x ∈ BN ,
where A = N + 1 − 2γ > 0. Theorem 4.1 is applied with P = 2γ > 0 and
T = 0 > −1. Thus
A + P = N + 1 = N + 1 + 2T.
We have to choose τ satisfying both
0≤τ ≤
N +1
2
and 2γ −
N +1
< τ < 2γ.
2
Multiplication of
Subharmonic Functions
Now
N +1
A+P
τ :=
=
=α+γ
2
2
fulfills the last condition since:
N +1
N +1
N +1
2γ − τ = 2 γ −
> 0 and τ − 2γ +
=2
− γ > 0.
4
2
2
Formula (4.3) implies Jb (fa ) ≤ K for all a ∈ BN and all b ∈ BN . Thus MYα,β,γ (fa ) ≤
K, ∀a ∈ BN . As before,
A Z
1+R
(1 − |x|2 )A+β+γ g(x)
Va g(a) ≤ 2
dx.
A
2 β+γ
1−R
B(a,Ra ) (1 − hx, ai) (1 − |x| )
Now
R. Supper
vol. 9, iss. 4, art. 118, 2008
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A+β+γ =N +1−γ+β
N +1
=N +1+α−
+β
2
N −1
=α+β+N −
,
2
Close
whence
A Z
1+R
(1 − |x|2 )α+β+N fa (x) g(x)
Va g(a) ≤ 2
dx
N −1
1−R
(1 − |x|2 ) 2
B(a,Ra )
A Z
1+R
dx
0
≤CK 2
λ+ N 2−1
1−R
B(a,Ra ) (1 − |x|2 )
and the proof ends as in the case (i). Here
C 0 K N +1−2γ
MXλ+ N −1 (g) ≤
2
.
Cλ+ N −1
2
Multiplication of
Subharmonic Functions
R. Supper
vol. 9, iss. 4, art. 118, 2008
2
Title Page
4.4.
Proof of Theorem 2.2 in the case (iv)
Contents
Here fa is defined by:
1
fa (x) =
(1 − hx, ai)A (1 − |x|2 )β
∀x ∈ BN ,
where A = N + 1, T = γ and P = 2γ thus A + P = N + 1 + 2T , allowing us
to use Theorem 4.1, with τ = α + γ = 1 + γ (since 0 ≤ τ ≤ N 2+1 + γ and
γ − N 2+1 < τ < 2γ). Hence Inequalities (4.4), (4.5), (4.6) and (4.7) follow. Now
A Z
1+R
(4.8)
Va g(a) ≤ 2
(1 − |x|2 )A+β fa (x) g(x) dx.
1−R
B(a,Ra )
Since A + β = α + β + N , this turns into:
A Z
MXλ+α+β+N (g fa )
1+R
Va g(a) ≤ 2
dx
1−R
(1 − |x|2 )λ
B(a,Ra )
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and the proof ends as in the case (ii), here with:
MXλ+ N −1 (g) ≤
2
4.5.
C 0 K N +1
2
.
Cλ
Proof of Theorem 2.2 in the case (v)
Multiplication of
Subharmonic Functions
Here fa is defined by:
fa (x) =
1
(1 − hx, ai)A (1 − |x|2 )γ
∀x ∈ BN ,
R. Supper
vol. 9, iss. 4, art. 118, 2008
where
N +3
N −1
=
> 0.
2
2
With P = 2γ > 0 and T = β, the condition A + P = N + 1 + 2T of Theorem 4.1
is fulfilled. Moreover τ := α + γ = 1 + β satisfies
A = N + 1 + 2(β − γ) > N + 1 −
N +1
0≤τ ≤
+β
2
N +1
and 2γ −
− β < τ < 2γ
2
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since:
2γ − τ = 2γ − (1 + β) > 0 and τ − 2γ +
N +1
N +3
+ β = −2γ +
+ 2β > 0.
2
2
Again
A+P
N +1
N −1
−τ =
−1=
2
2
2
and inequalities (4.4) to (4.7) follow. Formula (4.8) still holds with (1 − |x|2 )A+γ
instead of (1 − |x|2 )A+β . Here
A + γ = N + 1 + 2β − γ = N + α + β
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and the conclusion follows as in the previous case. Finally:
MXλ+ N −1 (g) ≤
2
4.6.
C 0 K N +1+2(β−γ)
2
.
Cλ
Proof of Theorem 2.2 in the case (vi)
Here fa is defined by:
fa (x) =
Multiplication of
Subharmonic Functions
1
(1 − hx, ai)A (1 − |x|2 )α
∀x ∈ BN
R. Supper
vol. 9, iss. 4, art. 118, 2008
β−1
2
with A = N + β > N 2−1 > 0, P = 2γ > 0, T =
+ γ > −1 (actually
β+1
T + 1 = 2 + γ > 0). The use of Theorem 4.1 is allowed since
Title Page
A + P = N + 1 + β − 1 + 2γ = N + 1 + 2T.
Contents
Now τ := α + γ = β+1
+ γ satisfies 0 ≤ τ ≤ N 2+β + γ (because of γ > − β+1
).
2
2
N +β
Moreover γ − 2 < τ < 2γ is fulfilled too since
β+1
<γ
2
and
β + 1 + (N + β) = 1 + N + 2β > 0.
In addition,
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A+P
N +β β+1
N −1
−τ =
−
=
.
2
2
2
2
Again it induces Formula (4.6). With (1 − |x|2 )A+β replaced by (1 − |x|2 )A+α ,
inequality (4.8) remains valid. Since A + α = N + α + β, the conclusion is once
again obtained in a similar way as in the cases (iv) and (v), here with
MXλ+ N −1 (g) ≤
2
C 0 K N +β
2
.
Cλ
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5.
5.1.
The Situation with Radial Subharmonic Functions
The example of u : x 7→ (1 − |x|2 )−A with A ≥ 0
Proposition 5.1. Given P ≥ 1, T > −1 and N ∈ N (N ≥ 2) such that P ≤
N + 1 + 2T , let
Z
(1 − |x|2 )T
IP,T (b) =
dx
∀b ∈ BN .
P
BN (1 − hx, bi)
Then
K0
IP,T (b) ≤
(1 − |b|2 )P/2
(equality holds when P = N + 1 + 2T ) with
Multiplication of
Subharmonic Functions
R. Supper
vol. 9, iss. 4, art. 118, 2008
∀b ∈ BN ,
Title Page
Contents
Γ(T + 1) N
π2.
K =
Γ P +1
2
0
Proof. Letting A → 0+ in Theorem 4.1, the majorization of IP,T (b) is an immediate
result, since K (as a function of A) tends towards K 0 : see Example 4.1. Nonetheless,
we still have to show that equality holds in the case P = N + 1 + 2T .
Proof in the case N ≥ 3. Up to a unitary transform, we may assume b = (|b|, 0, 0, . . . , 0),
so that hx, bi = |b| x1 = |b| r cos θ1 with θ1 ∈]0, π[ (we will have θ1 ∈]0, 2π[ in the
case N = 2). Now
dx = r
N −1
N −2
(sin θ1 )
dr dθ1 dσ
(N −1)
,
with the same notations as in the proof of Theorem 4.1. Here:
Z πZ 1
(1 − r2 )T rN −1 (sin θ1 )N −2
IP,T (b) = σN −1
dr dθ1 .
(1 − |b| r cos θ1 )P
0
0
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Then
(5.1)
IP,T (b) = σN −1
X Γ(n + P )
n∈N
n! Γ(P )
n
ZZ
|b|
sn tN −2 (1 − s2 − t2 )T ds dt
H
with s = r cos θ1 and t = r sin θ1 . This integral vanishes for odd n. If n = 2k, its
value is given by Lemma 3.4. Thus
σN −1 Γ N 2−1 Γ(T + 1) X |b|2k Γ k + 12 Γ(2k + P )
.
IP,T (b) =
N
2 Γ(P )
(2k)!
Γ
k
+
+
T
+
1
2
k∈N
Multiplication of
Subharmonic Functions
R. Supper
vol. 9, iss. 4, art. 118, 2008
Now [2, p. 29] and [3, p. 40] lead to:
√
Γ(T + 1) N −1 X |b|2k π Γ(2k + P )
.
IP,T (b) =
π 2
2k k! Γ k + N + T + 1
Γ(P )
2
2
k∈N
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Through the duplication formula ([3, p. 45]), it follows that:
Γ(T + 1) N −1 X |b|2k 22k+P −1 Γ k + P2 Γ k +
IP,T (b) =
π 2
2k k! Γ k + N + T + 1
Γ(P )
2
2
k∈N
XΓ k+P
2
|b|2k
= K0
P
k!
Γ
2
k∈N
P +1
2
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with
K0 =
Γ(T + 1) N −1
π 2
Γ(P )
P
P −1
2 Γ
.
2
Another application of the duplication formula provides the final expression of K 0 .
Proof in the case N = 2. Now
Z
IP,T (b) =
0
2π
1
Z
0
(1 − r2 )T r
dr dθ.
(1 − |b| r cos θ)P
Then
IP,T (b) =
X Γ(n + P )
n∈N
n! Γ(P )
n
Z
|b|
1
r
n+1
Z
2 T
2π
(1 − r ) dr
0
n
(cos θ) dθ .
Multiplication of
Subharmonic Functions
0
R. Supper
Rπ
The last integral equals 2 0 (cos θ)n dθ for any n. As σ1 = 2, here we recognize the
same expression as in formula (5.1), replacing N by 2. Hence the same conclusion.
vol. 9, iss. 4, art. 118, 2008
Title Page
Corollary 5.2. Given α ≥ 0, β ≥ − N 2+1 and γ > N 2−1 , let A =
defined on BN by:
1
u(x) =
∀x ∈ BN .
(1 − |x|2 )A
0
N +1
2
+ β and u
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0
Then u ∈ RSY α,β,γ and MYα,β,γ (u) ≤ K where K stems from Proposition 5.1
(with P = 2γ > 1 and T = β + γ − A = γ − N 2+1 > −1).
Proof. The subharmonicity of u follows from Lemma 3.5 since A ≥ 0. Let Jb (u) be
defined similarly as in formula (4.3). Then
Z
(1 − |x|2 )β+γ−A
2 α+γ
Jb (u) = (1 − |b| )
dx.
(1 − hx, bi)P
BN
As
N + 1 + 2T = N + 1 + 2γ − (N + 1) = P,
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Proposition 5.1 provides:
K0
≤ K0
2
P/2
(1 − |b| )
Jb (u) ≤ (1 − |b|2 )α+γ
since α ≥ 0. The conclusion proceeds from
MYα,β,γ (u) = sup Jb (u).
Multiplication of
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b∈BN
R. Supper
vol. 9, iss. 4, art. 118, 2008
5.2.
Proof of Theorem 2.3
Let A and u be defined as in Corollary 5.2. With R and Va as in the proof of Theorem 2.2:
Z
Va g(a) ≤
(1 − |x|2 )A u(x) g(x) dx
B(a,Ra )
(1 − |x|2 )λ+α+β+N u(x) g(x) dx
Z
=
(1 − |x|2 )λ+α+
B(a,Ra )
since:
N −1
2
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N +1
N −1
A=
+β =β+N −
.
2
2
This leads to:
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Va g(a) ≤ C 00 K 0
Z
B(a,Ra )
≤
dx
(1 − |x|2 )λ+α+
1
N −1
2
C 00 K 0 Va
,
Cλ+α+ N −1 (1 − |a|2 )λ+α+ N2−1
2
with Cλ+α+ N −1 defined in the same way as Cβ in the proof of Proposition 3.1. We
2
obtain finally:
C 00 K 0
MXλ+α+ N −1 (g) ≤
.
Cλ+α+ N −1
2
2
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vol. 9, iss. 4, art. 118, 2008
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6.
Annex: The Sets SX λ and SY α,β,γ for some Special Values of
λ, α, β, γ
Throughout the paper, it was assumed that γ ≥ 0. When γ ≤ 0, the set SY α,β,γ is
related to other sets of the same kind by:
Proposition 6.1. Given α ∈ R, β ∈ R and γ ≤ 0, then
+
+
+
Yα+γ,β+γ,0
⊂ Yα,β,γ
⊂ Yα+sγ,β−sγ,0
∀s ∈ [−1, 1],
+
where Yα,β,γ
denotes the subset of Yα,β,γ consisting of all non-negative u ∈ Yα,β,γ
(not necessarily subharmonic).
Multiplication of
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R. Supper
vol. 9, iss. 4, art. 118, 2008
Title Page
Proof. For any a ∈ BN and x ∈ BN , the following holds:
Contents
(6.1) (1 − |a|2 )α (1 − |x|2 )β (1 − |Φa (x)|2 )γ
= (1 − |a|2 )α+γ (1 − |x|2 )β+γ (1 − ha, xi)−2γ .
Since ha, xi ∈]−1, 1[ through the Cauchy-Schwarz inequality, we have (1−ha, xi)−2γ ≤
2−2γ as −2γ ≥ 0. If u ∈ Yα+γ,β+γ,0 and u(x) ≥ 0, ∀x ∈ BN , then u ∈ Yα,β,γ with
(1 − ha, xi)(−s−1)γ ≥ (1 − |x|)(−s−1)γ
since (s − 1)γ ≥ 0 and (−s − 1)γ ≥ 0. Moreover
1 − |a| =
1 − |a|2
1 − |a|2
≥
1 + |a|
2
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Also, ha, xi < |a| and ha, xi < |x|, thus
and
II
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MYα,β,γ (u) ≤ 2−2γ MYα+γ,β+γ,0 (u).
(1 − ha, xi)(s−1)γ ≥ (1 − |a|)(s−1)γ
JJ
and 1 − |x| ≥
1 − |x|2
,
2
Close
thus
−2γ
(1 − ha, xi)
2 (s−1)γ
≥ (1 − |a| )
2 (−s−1)γ
(1 − |x| )
−2γ
1
.
2
Finally
(1 − |a|2 )α (1 − |x|2 )β (1 − |Φa (x)|2 )γ ≥ 22γ (1 − |a|2 )α+sγ (1 − |x|2 )β−sγ .
Any non-negative u ∈ Yα,β,γ then belongs to Yα+sγ,β−sγ,0 with
MYα+sγ,β−sγ,0 (u) ≤ 2−2γ MYα,β,γ (u).
Remark 5. Even with γ ≤ 0, Proposition 3.1 still holds, since
−γ −γ
1 − ha, xi
1 − ha, xi
2 γ
(1 − |Φa (x)| ) =
1 − |x|2
1 − |a|2
−γ −γ
1
1
= 23γ ∀x ∈ B(a, Ra )
≥
2
4
according to Lemma 3.3. For the proof of Proposition 3.1 in the case γ ≤ 0, it is
enough to replace (1 − R2 )γ in formula (3.1) by 23γ .
Multiplication of
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vol. 9, iss. 4, art. 118, 2008
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Proposition 6.2. If λ < 0, then the set SX λ contains only the function u ≡ 0 on
BN .
Proof. Given u ∈ SX λ and ξ ∈ BN , let r ∈]|ξ|, 1[. Then
u(ξ) ≤ max u(x) = max u(x)
|x|≤r
|x|=r
Close
according to the maximum principle (see [2, pp. 48–49]). Thus
0 ≤ u(ξ) ≤ MXλ (u) (1 − r2 )−λ
which tends towards 0 as r → 1− (since −λ > 0). Finally u(ξ) = 0.
Remark 6. When α < 0, it is not compulsory that SY α,β,γ = {0}. For instance,
with α, β, γ as in case (ii) of Theorem 2.2, we have α = β + 1 > 1−N
. It is
4
thus possible to choose β in such a way that α < 0. In Subsection 4.2 we have
an example of function fa ∈ SY α,β,γ (with a fixed in BN ) and this function is not
vanishing. Similarly β < 0 does not imply SY α,β,γ = {0}. In Table 1 we have
several examples of such situations: see Subsections 4.1 to 4.6 for examples of nonvanishing subharmonic functions belonging to such sets SY α,β,γ .
Multiplication of
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vol. 9, iss. 4, art. 118, 2008
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2
Proposition 6.3. Let γ ∈ R and (α, β) ∈ R such that α+β < −N , then SY α,β,γ =
{0}.
Proof. Given R ∈]0, 1[, let KR,γ = (1 − R2 )γ if γ ≥ 0, or KR,γ = 23γ if γ ≤ 0.
Then: (1 − |Φa (x)|2 )γ ≥ KR,γ , ∀a ∈ BN , ∀x ∈ B(a, Ra ) according to Remark
5 (also remember that |Φa | < R on B(a, Ra ), see [6]). With Cβ as in the proof
of Proposition 3.1, the following inequalities hold for any u ∈ SY α,β,γ and any
a ∈ BN . The second inequality is based upon u ≥ 0 and the last one makes use of
the suharmonicity of u.
Z
2 −α
(1 − |a| ) MYα,β,γ (u) ≥
(1 − |x|2 )β u(x) (1 − |Φa (x)|2 )γ dx
BN
Z
≥ KR,γ
(1 − |x|2 )β u(x) dx
B(a,Ra )
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2 β
Z
≥ KR,γ Cβ (1 − |a| )
u(x) dx
B(a,Ra )
≥ KR,γ Cβ (1 − |a|2 )β Va u(a)
where the volume Va of B(a, Ra ) satisfies:
N
σN
R
Va ≥
(1 − |a|2 )N
N 1+R
Multiplication of
Subharmonic Functions
R. Supper
(see the end of the proof of Proposition 3.1). Thus
vol. 9, iss. 4, art. 118, 2008
u(a) ≤ κ (1 − |a|2 )−α−β−N
∀a ∈ BN ,
the constant κ > 0 being independent of a.
Given ξ ∈ BN , the maximum principle now provides for any r ∈]|ξ|, 1[:
0 ≤ u(ξ) ≤ max u(x) = max u(x) ≤ κ (1 − r2 )−α−β−N
|x|≤r
|x|=r
−
which tends towards 0 as r → 1 , since −α − β − N > 0. Hence u(ξ) = 0.
Proposition 6.4. Given γ ≥ 0, α < −γ and β ∈ R, then SY α,β,γ = {0}.
−2γ
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−2γ
Proof. Since 1−hx, ai ∈]0, 2[, we have (1−ha, xi)
≥ 2 , ∀x ∈ BN , ∀a ∈ BN .
Given u ∈ SY α,β,γ , ξ ∈ BN and r ∈]0, 1 − |ξ|[, the formula (6.1) leads to:
Z
2 α+γ −2γ
MYα,β,γ (u) ≥ (1 − |a| )
2
(1 − |x|2 )β+γ u(x) dx
∀a ∈ BN
B(ξ,r)
since u ≥ 0 on BN ⊃ B(ξ, r). Now |x| ≤ |ξ| + r, ∀x ∈ B(ξ, r). Let Lξ =
[1 − (|ξ| + r)2 ]β+γ if β + γ ≥ 0, or Lξ = 1 if β + γ ≤ 0. Then
Z
σN N
2 −α−γ 2γ
(1 − |a| )
2 MYα,β,γ (u) ≥ Lξ
u(x) dx ≥ Lξ
r u(ξ)
∀a ∈ BN
N
B(ξ,r)
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since u is subharmonic and the volume of B(ξ, r) is
have:
0 ≤ u(ξ) ≤ κξ (1 − |a|2 )−α−γ
σN
N
rN . Finally, with ξ fixed, we
∀a ∈ BN ,
the constant κξ > 0 being independent of a. Hence u(ξ) = 0, letting |a| → 1− .
Multiplication of
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vol. 9, iss. 4, art. 118, 2008
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References
[1] A.B. ALEKSANDROV, Function theory in the ball, in: Several Complex Variables, Part II, Encyclopedia of Mathematical Sciences, Volume 8, Springer Verlag, 1994.
[2] W.K. HAYMAN AND P.B. KENNEDY, Subharmonic Functions, Vol. I, London
Mathematical Society Monographs, No. 9. Academic Press, London–New York,
1976.
Multiplication of
Subharmonic Functions
R. Supper
[3] R. REMMERT, Classical Topics in Complex Function Theory, Graduate Texts
in Mathematics, 172. Springer-Verlag, New York, 1998.
vol. 9, iss. 4, art. 118, 2008
[4] W. RUDIN, Function theory in the unit ball of Cn , Fundamental Principles of
Mathematical Science, 241, Springer-Verlag, New York-Berlin, 1980.
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[5] R. SUPPER, Bloch and gap subharmonic functions, Real Analysis Exchange, 28(2) (2003), 395–414.
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II
[6] R. SUPPER, Subharmonic functions with a Bloch type growth, Integral Transforms and Special Functions, 16(7) (2005), 587–596.
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[7] R. YONEDA, Pointwise multipliers from BMOAα to BMOAβ , Complex Variables, Theory and Applications, 49(14) (2004), 1045–1061.
[8] K.H. ZHU, Operator Theory in Function Spaces, Monographs and Textbooks in
Pure and Applied Mathematics, 139. Marcel Dekker, Inc., New York, 1990.
[9] C. ZUILY, Distributions et équations aux dérivées partielles, Collection Méthodes, Hermann, Paris, 1986.
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