Volume 9 (2008), Issue 4, Article 118, 20 pp.
MULTIPLICATION OF SUBHARMONIC FUNCTIONS
R. SUPPER
U NIVERSITÉ L OUIS PASTEUR
UFR DE M ATHÉMATIQUE ET I NFORMATIQUE , URA CNRS 001
7 RUE R ENÉ D ESCARTES
F–67 084 S TRASBOURG C EDEX , F RANCE
supper@math.u-strasbg.fr
Received 31 May, 2008; accepted 12 November, 2008
Communicated by S.S. Dragomir
A BSTRACT. 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.
Key words and phrases: Gamma and Beta functions, growth of subharmonic functions in the unit ball.
2000 Mathematics Subject Classification. 31B05, 33B15, 26D15, 30D45.
1. I NTRODUCTION
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
with dA(z) the normalized area measure element on D and ϕa (z) =
164-08
a−z
.
1−az
2
R. S UPPER
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.
2. N OTATIONS AND M AIN R ESULTS
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
N
X
hx, ai
hx, ai =
xj aj ,
Pa (x) =
a,
Qa (x) = x − Pa (x)
|a|2
j=1
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
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
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 2.1. 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 ,
then g ∈ Xλ+ N −1 in each of the six cases gathered in the following Table 2.1.
2
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 118, 20 pp.
http://jipam.vu.edu.au/
M ULTIPLICATION OF S UBHARMONIC F UNCTIONS
α
case
(ii)
β
N +1
2
γ
+β
β > − N 2+1
γ > max(α, −1 − β)
α=β+1
β > − N 4+3
γ > |1 + β|
α=
(i)
N +1
2
−γ
β ≥ −γ
(iv)
α=1
β≥0
(v)
α=1+β−γ
β > −1
β+1
2
β ≥ − 12
α=
(iii)
α=
(vi)
3
N +1
4
<γ<
N +1
2
γ>1
1+β
2
<γ<β+
N +3
4
γ > 1+β
2
Table 2.1: Six situations where Theorem 2.1 shows that g belongs to the set Xλ+ N −1 .
2
Theorem 2.2. Given λ ∈ R and g a subharmonic function defined on BN , satisfying:
∃C 00 > 0
MXλ+α+β+N (gu) ≤ C 00 MYα,β,γ (u)
then g ∈ SX λ+α+ N −1 provided that α ≥ 0, β ≥ − N 2+1 , γ >
2
∀u ∈ RSY α,β,γ ,
N −1
.
2
3. S OME P RELIMINARIES
Notation 3.1. Given a ∈ BN and R ∈]0, 1[, let B(a, Ra ) = {x ∈ BN : |x − a| < Ra } with
Ra = R
1 − |a|2
.
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)
∀u ∈ SY α,β,γ .
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
B(a, Ra ) ⊂ E(a, R) = {x ∈ BN : |Φa (x)| < R},
hence:
(3.1)
2 γ
2 α
Z
(1 − |x|2 )β u(x) dx
MYα,β,γ (u) ≥ (1 − R ) (1 − |a| )
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
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 118, 20 pp.
∀x ∈ B(a, Ra ).
http://jipam.vu.edu.au/
4
R. S UPPER
Let Cβ =
1−R β
1+R
if β ≥ 0 and Cβ = 2β if β < 0. Hence
2 γ
2 α+β
Z
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 SN in RN
R
(see [2, p. 29]) and Ra ≥ 1+R
(1 − |a|2 ). The subharmonicity of u now provides:
MYα,β,γ (u) ≥ Cβ (1 − R2 )γ (1 − |a|2 )α+β u(a) σN
≥ Cβ
(Ra )N
N
σN RN (1 − R)γ
(1 − |a|2 )α+β+N u(a).
N (1 + R)N −γ
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.
Proof. Since u ≥ 0, we have for any x ∈ BN :
(1 − |x|2 )λ+α+β+N g(x) u(x) ≤ MXλ (g) (1 − |x|2 )α+β+N u(x)
≤ MXλ (g) MXα+β+N (u)
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|
1
1 − hx, ai
1+R
<
<2
.
2
4
1 − |x|
1−R
Proof. Clearly hx, ai = ha + y, ai = |a|2 + hy, ai with |y| < Ra . From the Cauchy-Schwarz
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
R |a|
(1 − |a| ) 1 −
1 + R|a|
2
and 1 + R|a| < 2. The term on the right equals
2
(1 − |a| ) 1 +
with
R|a|
1+R|a|
= (1 − |a|2 )
R |a|
1 + R|a|
1
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].
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 118, 20 pp.
http://jipam.vu.edu.au/
M ULTIPLICATION OF S UBHARMONIC F UNCTIONS
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)
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
(cos θ)2x−1 (sin θ)2y−1 dθ =
=2
0
Γ(x) Γ(y)
Γ(x + y)
(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
1
P +Q
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
Lemma 3.5. Given A ≥ 0 and a ∈ BN , let u and fa denote the functions defined on BN by
1
1
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 ,
∂x2j
|a|2 A (A + 1)
≥0
(1 − hx, ai)A+2
∀x ∈ BN .
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
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 118, 20 pp.
http://jipam.vu.edu.au/
6
R. S UPPER
is subharmonic too. The computation
(∆fa )(x) ≥ fa (x)
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)
Then
√ Γ
IP (s, t, b1 , b2 ) = π
2k
− 1 X X Γ(j + 2k + P )
t b2
j
(b1 s)
.
Γ(P ) k∈N j∈N k! j! Γ N 2−1 + k
2
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
1−s b1
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
Γ N 2−2 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
√
πX
Γ(P )
k∈N
(t b2 )2k
Γ(2k + P )
.
Γ N 2−1 + k 22k k! (1 − s b1 )2k+P
The result follows from the expansion
X Γ(j + 2k + P )
Γ(2k + P )
=
(b1 s)j .
2k+P
(1 − s b1 )
j!
j∈N
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 118, 20 pp.
http://jipam.vu.edu.au/
M ULTIPLICATION OF S UBHARMONIC F UNCTIONS
7
4. P ROOF OF T HEOREM 2.1
The cases (i), (ii), (iii), (iv), (v) and (vi) of Theorem 2.1 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 ≤
K
(1 −
|a|2 )
A+P
2
−τ
A+P
.
2
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 =
A+P
,
2
IA,P,T (a, b) ≤
then
K
(1 − |b|2 )
A+P
2
∀a ∈ BN , ∀b ∈ BN ,
with
A+P −1
K=2
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 ,
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
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 118, 20 pp.
http://jipam.vu.edu.au/
8
R. S UPPER
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
N −1
k! j! `! Γ( 2 + k)
b2
2
2k
|a|` Jk,j,`
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:
IA,P,T (a, b)
2k
N −1
2h
X X
Γ(j + 2k + P ) Γ(2h − j + A) Γ h + 12 Γ(T + 1)
π 2
b2
j
=
(b1 )
|a|2h−j
N
Γ(P ) Γ(A)
2
k!
j!
(2h
−
j)!
Γ
k
+
h
+
+
T
+
1
2 j=0
2
(k,h)∈N
Taking [3, p. 40] into account:
N
2h
π 2 Γ(T + 1) X X (2h)! B(j + 2k + P, 2h − j + A)
(4.2) IA,P,T (a, b) =
Γ(P ) Γ(A)
22h+2k h! k! j! (2h − j)!
2 j=0
(k,h)∈N
×
Γ(2k + P + 2h + A) j 2k 2h−j
b b |a|
.
Γ(k + h + N2 + T + 1) 1 2
Let
L=
2P +A−1 Γ(T + 1) N −1
π 2 .
Γ(P ) Γ(A)
The duplication formula
√
1
π Γ(2z) = 2
Γ(z) Γ z +
2
(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
2z−1
since Γ increases on [1, +∞[ and
1≤k+h+
A+P +1
N
≤k+h+
+ T + 1.
2
2
This leads to:
IA,P,T (a, b)
2h
X X
(2h)! B(j + 2k + P, 2h − j + A) Γ k + h + A+P
2h−j
2
≤L
bj1 b2k
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
(k,h)∈N
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 118, 20 pp.
http://jipam.vu.edu.au/
M ULTIPLICATION OF S UBHARMONIC F UNCTIONS
The last sum turns into:
Z
2h
X
(2h)! bj |a|2h−j
1
ξ j+2k+P −1 (1 − ξ)2h−j+A−1 dξ
1
j! (2h − j)!
j=0
Z
1
=
0
Z
9
0
2h
X
(2h)! (b1 ξ)j [(1 − ξ) |a|]2h−j
j! (2h − j)!
j=0
!
ξ 2k+P −1 (1 − ξ)A−1 dξ
1
[b1 ξ + |a| (1 − ξ)]2h ξ 2k+P −1 (1 − ξ)A−1 dξ.
=
0
Hence the majorant of IA,P,T (a, b) becomes:
Z
L
0
1
X (b2 ξ)2k
k!
k∈N
Z
=L
0
1
A+P
)
2
X Γ(h + k +
h!
h∈N
X Γ(k +
A+P
) (b2
2
ξ)2k
k!
k∈N
!
ξ P −1 (1 − ξ)A−1 dξ
[b1 ξ + |a| (1 − ξ)]2h
1
1 − [b1 ξ + |a| (1 − ξ)]2
k+ A+P
2
ξ P −1 (1 − ξ)A−1 dξ
according to the expansion
X Γ(h + C)
Γ(C)
=
Xh
C
(1 − X)
h!
h∈N
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=
A+P
2
and
X=
(b2 ξ)2
1 − [b1 ξ + |a| (1 − ξ)]2
since |X| < 1, as
δ(ξ) := (b2 ξ)2 + [b1 ξ + |a| (1 − ξ)]2
= |b|2 ξ 2 + |a|2 (1 − ξ)2 + 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.
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) 2
2
0
(b2 ξ)2
(1
−
[b
ξ
+
|a|
(1
−
ξ)]
1
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
Now
1 − δ(ξ) ≥ 1 − [ξ |b| + |a| (1 − ξ)]2
≥ 1 − [ξ |b| + (1 − ξ)]2
= ξ(1 − |b|)[2 − ξ(1 − |b|)]
≥ ξ(1 − |b|2 )
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 118, 20 pp.
http://jipam.vu.edu.au/
10
R. S UPPER
since
[2 − ξ(1 − |b|)] − (1 + |b|) = (1 − ξ)(1 − |b|) ≥ 0.
Similarly,
1 − δ(ξ) ≥ (1 − ξ)(1 − |a|2 ).
Thus
1
A+P
2
1
≤
A+P
[1 − δ(ξ)]
[(1 − ξ)(1 − |a|2 )] 2 −τ [ξ(1 − |b|2 )]τ
since τ ≥ 0 and A+P
− τ ≥ 0. Finally:
2
Z 1
L · Γ A+P
A+P
2
ξ P −τ −1 (1 − ξ)A+τ − 2 −1 dξ.
IA,P,T (a, b) ≤
A+P
−τ
2
2
τ
(1 − |a| ) 2
(1 − |b| ) 0
This integral converges since P − τ > 0 and
A+P
A−P
A+τ −
=
+ 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
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 )
dr dθ1 ,
IA,P,T (a, b) =
(1 − |a| s)A
0
0
where
Z 2π
dθ2
JP (s, t, b1 , b2 ) =
= 2 IP (s, t, b1 , b2 )
(1 − s b1 − t b2 cos θ2 )P
0
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:
IA,P,T (a, b)
Z 2π Z 1
(1 − r2 )T r dr dθ
=
A
P
0 (1 − |a| s) (1 − s b1 − t b2 )
Z0 X
X (t b2 )n Γ(n + P )
Γ(` + A)
=
(|a| s)`
(1 − s2 − t2 )T ds dt
n+P
n! Γ(P ) (1 − s b1 )
B2 `∈N `! Γ(A)
n∈N
X Γ(` + A) |a|` (b2 )n Γ(j + n + P ) (b1 )j Z
s`+j tn (1 − s2 − t2 )T ds dt.
=
`!
Γ(A)
n!
Γ(P
)
j!
B2
3
(`,n,j)∈N
The last integral vanishes when n is odd or ` + j odd. Otherwise (n = 2k and ` + j = 2h), it
equals
Z
Γ h + 21 Γ k + 12 Γ(T + 1)
`+j n
2
2 T
2 s t (1 − s − t ) ds dt =
Γ(k + h + T + 2)
H
by Lemma 3.4 and turns into
n! (2h)! π Γ(T + 1)
h! k! Γ(k + h + T + 2)
22h+2k
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 118, 20 pp.
http://jipam.vu.edu.au/
M ULTIPLICATION OF S UBHARMONIC F UNCTIONS
11
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). For any
1
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.
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
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 =
A+P
.
2
4.1. Proof of Theorem 2.1 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
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
thus
A Z
1+R
(1 − |x|2 )λ+α+β+N fa (x) g(x)
Va g(a) ≤ 2
dx
N −1
1−R
(1 − |x|2 )λ+ 2
B(a,Ra )
A Z
1+R
dx
0
≤CK 2
λ+ N 2−1
1−R
B(a,Ra ) (1 − |x|2 )
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 118, 20 pp.
http://jipam.vu.edu.au/
12
R. S UPPER
from Corollary 4.2. Lemmas 1 and 5 of [5] provide
λ+ N2−1
1 − |x|2
≥ Cλ+ N −1
2
1 − |a|2
∀x ∈ B(a, Ra ),
with Cλ+ N −1 defined in the same pattern as Cβ in the proof of Proposition 3.1. Finally:
2
C 0K
Va g(a) ≤
Cλ+ N −1
2
1+R
2
1−R
A
Va
(1 − |a|2 )λ+
,
N −1
2
thus
C 0K
MXλ+ N −1 (g) ≤
Cλ+ N −1
2
2
1+R
2
1−R
2α
∀R ∈]0, 1[.
The majorant is an increasing function with respect to R. Letting R tend toward 0+ , we get:
MXλ+ N −1 (g) ≤
2
C 0 K 2α
2 .
Cλ+ N −1
2
4.2. Proof of Theorem 2.1 in the case (ii). Here we work with fa defined by:
fa (x) =
1
(1 − hx, ai)A
A = α + β + N.
where
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
= 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
τ −γ+β+
Hence, Jb (fa ) defined in Formula (4.3) now satisfies
(4.5)
Jb (fa ) ≤
K
(1 − |a|2 )
∀a ∈ BN , ∀b ∈ BN .
N −1
2
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 ) ≤
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 118, 20 pp.
C 0K
(1 − |a|2 )
N −1
2
∀a ∈ BN .
http://jipam.vu.edu.au/
M ULTIPLICATION OF S UBHARMONIC F UNCTIONS
13
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
dx
C 0K
1+R
≤
2
N −1
2 λ
1−R
(1 − |a|2 ) 2
B(a,Ra ) (1 − |x| )
and the last integral is majorized by
Proposition 3.1. Finally:
Va
Cλ (1−|a|2 )λ
MXλ+ N −1 (g) ≤
2
with Cλ defined similarly to Cβ in the proof of
C 0 K N +1+2β
2
.
Cλ
4.3. Proof of Theorem 2.1 in the case (iii) . Here fa is defined by:
1
fa (x) =
∀x ∈ BN ,
A
(1 − hx, ai) (1 − |x|2 )β+γ
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
N +1
N +1
0≤τ ≤
and 2γ −
< τ < 2γ.
2
2
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
A+β+γ =N +1−γ+β
N +1
=N +1+α−
+β
2
N −1
=α+β+N −
,
2
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
MXλ+ N −1 (g) ≤
2
C 0 K N +1−2γ
2
.
Cλ+ N −1
2
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 118, 20 pp.
http://jipam.vu.edu.au/
14
R. S UPPER
4.4. Proof of Theorem 2.1 in the case (iv). Here fa is defined by:
fa (x) =
1
(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
(1 − |x|2 )A+β fa (x) g(x) dx.
(4.8)
Va g(a) ≤ 2
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 )
and the proof ends as in the case (ii), here with:
MXλ+ N −1 (g) ≤
2
C 0 K N +1
2
.
Cλ
4.5. Proof of Theorem 2.1 in the case (v). Here fa is defined by:
fa (x) =
1
(1 − hx, ai)A (1 − |x|2 )γ
∀x ∈ BN ,
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 −
0≤τ ≤
N +1
+β
2
and
2γ −
N +1
− β < τ < 2γ
2
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
and the conclusion follows as in the previous case. Finally:
MXλ+ N −1 (g) ≤
2
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 118, 20 pp.
C 0 K N +1+2(β−γ)
2
.
Cλ
http://jipam.vu.edu.au/
M ULTIPLICATION OF S UBHARMONIC F UNCTIONS
15
4.6. Proof of Theorem 2.1 in the case (vi). Here fa is defined by:
fa (x) =
1
(1 − hx, ai)A (1 − |x|2 )α
with A = N + β > N 2−1 > 0, P = 2γ > 0, T =
The use of Theorem 4.1 is allowed since
β−1
2
∀x ∈ BN
+ γ > −1 (actually T + 1 =
β+1
2
+ γ > 0).
A + P = N + 1 + β − 1 + 2γ = N + 1 + 2T.
Now τ := α + γ = β+1
+ γ satisfies 0 ≤ τ ≤
2
N +β
γ − 2 < τ < 2γ is fulfilled too since
β+1
<γ
2
N +β
2
+ γ (because of γ > − β+1
). Moreover
2
and β + 1 + (N + β) = 1 + N + 2β > 0.
In addition,
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
5. T HE S ITUATION WITH R ADIAL S UBHARMONIC F UNCTIONS
5.1. 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
(1 − |b|2 )P/2
(equality holds when P = N + 1 + 2T ) with
IP,T (b) ≤
K0 =
∀b ∈ BN ,
Γ(T + 1) N
π2.
Γ P +1
2
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 = rN −1 (sin θ1 )N −2 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
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 118, 20 pp.
http://jipam.vu.edu.au/
16
R. S UPPER
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
Now [2, p. 29] and [3, p. 40] lead to:
√
Γ(T + 1) N −1 X |b|2k π Γ(2k + P )
2
.
π
IP,T (b) =
Γ(P )
22k k! Γ k + N2 + T + 1
k∈N
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
with
Γ(T + 1) N −1 P −1
P
K =
π 2 2 Γ
.
Γ(P )
2
Another application of the duplication formula provides the final expression of K 0 .
0
Proof in the case N = 2. Now
Z
2π
Z
1
IP,T (b) =
0
0
(1 − r2 )T r
dr dθ.
(1 − |b| r cos θ)P
Then
X Γ(n + P )
n
Z
1
n+1
2 T
Z
2π
n
|b|
r
(1 − r ) dr
(cos θ) dθ .
n!
Γ(P
)
0
0
n∈N
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.
IP,T (b) =
Corollary 5.2. Given α ≥ 0, β ≥ − N 2+1 and γ > N 2−1 , let A = N 2+1 + β and u defined on BN
by:
1
u(x) =
∀x ∈ BN .
(1 − |x|2 )A
Then u ∈ RSY α,β,γ and MYα,β,γ (u) ≤ K 0 where K 0 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,
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 118, 20 pp.
http://jipam.vu.edu.au/
M ULTIPLICATION OF S UBHARMONIC F UNCTIONS
17
Proposition 5.1 provides:
Jb (u) ≤ (1 − |b|2 )α+γ
K0
≤ K0
2
P/2
(1 − |b| )
since α ≥ 0. The conclusion proceeds from
MYα,β,γ (u) = sup Jb (u).
b∈BN
5.2. Proof of Theorem 2.2. Let A and u be defined as in Corollary 5.2. With R and Va as in
the proof of Theorem 2.1:
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:
A=
N −1
2
N +1
N −1
+β =β+N −
.
2
2
This leads to:
00
Va g(a) ≤ C K
Z
0
dx
B(a,Ra )
≤
(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 obtain finally:
2
MXλ+α+ N −1 (g) ≤
2
C 00 K 0
Cλ+α+ N −1
.
2
6. A NNEX : T HE S ETS SX λ
AND
SY α,β,γ
FOR SOME
S PECIAL 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).
Proof. For any a ∈ BN and x ∈ BN , the following holds:
(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
MYα,β,γ (u) ≤ 2−2γ MYα+γ,β+γ,0 (u).
Also, ha, xi < |a| and ha, xi < |x|, thus
(1 − ha, xi)(s−1)γ ≥ (1 − |a|)(s−1)γ
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 118, 20 pp.
and (1 − ha, xi)(−s−1)γ ≥ (1 − |x|)(−s−1)γ
http://jipam.vu.edu.au/
18
R. S UPPER
since (s − 1)γ ≥ 0 and (−s − 1)γ ≥ 0. Moreover
1 − |a| =
1 − |a|2
1 − |a|2
≥
1 + |a|
2
and 1 − |x| ≥
1 − |x|2
,
2
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 .
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
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 2.1, we have α = β + 1 > 1−N
. It is thus possible to choose β in such a
4
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 2.1 we have several examples of such situations: see Subsections 4.1 to 4.6 for examples
of non-vanishing subharmonic functions belonging to such sets SY α,β,γ .
Proposition 6.3. Let γ ∈ R and (α, β) ∈ R2 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
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 118, 20 pp.
http://jipam.vu.edu.au/
M ULTIPLICATION OF S UBHARMONIC F UNCTIONS
19
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 )
Z
2 β
≥ KR,γ Cβ (1 − |a| )
u(x) dx
B(a,Ra )
2 β
≥ KR,γ Cβ (1 − |a| ) Va u(a)
where the volume Va of B(a, Ra ) satisfies:
N
σN
R
Va ≥
(1 − |a|2 )N
N 1+R
(see the end of the proof of Proposition 3.1). Thus
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}.
Proof. Since 1 − hx, ai ∈]0, 2[, we have (1 − ha, xi)−2γ ≥ 2−2γ , ∀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)
σN
N
2 −α−γ
since u is subharmonic and the volume of B(ξ, r) is
0 ≤ u(ξ) ≤ κξ (1 − |a| )
rN . Finally, with ξ fixed, we have:
∀a ∈ BN ,
the constant κξ > 0 being independent of a. Hence u(ξ) = 0, letting |a| → 1− .
R EFERENCES
[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.
[3] R. REMMERT, Classical Topics in Complex Function Theory, Graduate Texts in Mathematics, 172.
Springer-Verlag, New York, 1998.
[4] W. RUDIN, Function theory in the unit ball of Cn , Fundamental Principles of Mathematical Science,
241, Springer-Verlag, New York-Berlin, 1980.
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 118, 20 pp.
http://jipam.vu.edu.au/
20
R. S UPPER
[5] R. SUPPER, Bloch and gap subharmonic functions, Real Analysis Exchange, 28(2) (2003), 395–414.
[6] R. SUPPER, Subharmonic functions with a Bloch type growth, Integral Transforms and Special
Functions, 16(7) (2005), 587–596.
[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.
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 118, 20 pp.
http://jipam.vu.edu.au/