MULTIPLICATION OF SUBHARMONIC FUNCTIONS

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Multiplication of Subharmonic Functions

R. Supper vol. 9, iss. 4, art. 118, 2008

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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

Received: 31 May, 2008

Accepted: 12 November, 2008

Communicated by: S.S. Dragomir

2000 AMS Sub. Class.: 31B05, 33B15, 26D15, 30D45.

Key words: Gamma and Beta functions, growth of subharmonic functions in the unit ball.

Abstract: We study subharmonic functions in the unit ball of R^N, with either a Bloch–

type growth or a growth described through integral conditions involving some involutions of the ball. Considering mappingsu7→gubetween sets of functions with a prescribed growth, we study how the choice of these sets is related to the growth of the functiong.

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1. Introduction

This paper is devoted to functions u which are defined in the unit ball B_N of R^N (relative to the Euclidean norm|·|), whose growth is described by the above bound- edness onB_N ofx 7→ (1− |x|²)^αv(x)for some parameterα. The functionv may denote merelyuor some integral involvinguand involutionsΦ_x(precise definitions and notations will be detailed in Section2). In the first (resp. second) case,uis said to belong to the setX (resp.Y). Given a functiong defined onB_N, we try to obtain links between the growth ofg and information on such mappings as

Y → X, u7→gu.

This work is motivated by the situation known in the case of holomorphic functions f in the unit diskDofC. Such a function is said to belong to the Bloch spaceB_λ if

||f||B_λ :=|f(0)|+ sup

z∈D

(1− |z|²)^λ|f⁰(z)|<+∞.

It is said to belong to the spaceBM OA_µif

||f||²_{BM OA}_µ :=|f(0)|²+ sup

a∈D

Z

D

(1− |z|²)^2µ−2|f⁰(z)|²(1− |ϕ_a(z)|²)dA(z)<+∞

withdA(z)the normalized area measure element onDandϕ_a(z) = _1−az^a−z.

Givenha holomorphic function onD, the operatorI_h :f 7→I_h(f)defined by:

(I_h(f))(z) = Z z

0

h(ζ)f⁰(ζ)dζ ∀z ∈D

was studied for instance in [7] where it was proved that I_h : BM OA_µ → B_λ is bounded (with respect to the above norms) if and only if h ∈ Bλ−µ+1 (assuming 1< µ < λ).

Since |f⁰|² is subharmonic in the unit ball of R², the question naturally arose whether some similar phenomena occur for subharmonic functions inB_N forN ≥2.

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2. Notations and Main Results

LetB_N ={x ∈R^N : |x|< 1}withN ∈ N,N ≥2and|·|the Euclidean norm in R^N. Givena ∈B_N, letΦ_a :B_N →B_N denote the involution defined by:

Φ_a(x) = a−Pa(x)−p

1− |a|²Qa(x)

1− hx, ai ∀x∈B_N, where

hx, ai=

N

X

j=1

x_ja_j, P_a(x) = hx, ai

|a|² a , Q_a(x) =x−P_a(x)

for allx= (x1, x2, . . . , xN)∈R^N anda= (a1, a2, . . . , aN) ∈R^N, withPa(x) = 0 ifa = 0. We refer to [4, pp. 25–26] and [1, p. 115] for the main properties of the mapΦ_a (initially defined in the unit ball ofC^N). For instance, we will make use of the relation:

1− |Φ_a(x)|² = (1− |a|²) (1− |x|²) (1− hx, ai)² . In the following,α,β,γ andλare given real numbers, withγ ≥0.

Definition 2.1. LetX_λ denote the set of all functionsu:B_N →[−∞,+∞[satisfy- ing:

MX_λ(u) := sup

x∈BN

(1− |x|²)^λu(x)<+∞.

LetYα,β,γ denote the set of all measurable functions u: BN → [−∞,+∞[satisfy- ing:

MY_α,β,γ(u) := sup

a∈B_N

(1− |a|²)^α Z

BN

(1− |x|²)^βu(x) (1− |Φ_a(x)|²)^γdx <+∞.

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The subset SX_λ (resp. SY_α,β,γ) gathers all u ∈ X_λ (resp. u ∈ Y_α,β,γ) which moreover are subharmonic and non–negative. The subsetRSY_α,β,γ gathers allu∈ 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 functionu ≡0(see Propositions6.2, 6.3and 6.4).

In Proposition3.1and Corollary3.2, we will establish that SY_α,β,γ ⊂ SX_α+β+N and that there exists a constantC >0such that

MX_λ+α+β+N(gu)≤C MX_λ(g)MY_α,β,γ(u)

for allu ∈ SYα,β,γ and all g ∈ Xλ withMX_λ(g) ≥ 0.We will next study whether some kind of a “converse” holds and obtain the following:

Theorem 2.2. Givenλ∈Randg :B_N →[0,+∞[a subharmonic function satisfy- ing:

∃C⁰ >0 MX_λ+α+β+N(gu)≤C⁰MY_α,β,γ(u) ∀u∈ SY_α,β,γ, theng ∈ X_λ+^N−1

2 in each of the six cases gathered in the following Table1.

Theorem 2.3. Givenλ∈Randga subharmonic function defined onBN, satisfying:

∃C⁰⁰ >0 M_X_λ+α+β+N(gu)≤C⁰⁰M_Y_α,β,γ(u) ∀u∈ RSY_α,β,γ, theng ∈ SX_λ+α+^N−1

2 provided thatα ≥0,β ≥ −^N+1₂ ,γ > ^N−1₂ .

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case α β γ

(i) α= ^N+1₂ +β β >−^N₂⁺¹ γ >max(α,−1−β)

(ii) α=β+ 1 β >−^N₄⁺³ γ >|1 +β|

(iii) α=^N₂⁺¹−γ β ≥ −γ ^N+1₄ < γ < ^N₂⁺¹

(iv) α= 1 β≥0 γ >1

(v) α= 1 +β−γ β >−1 ^1+β₂ < γ < β+^N₄⁺³

(vi) α=^β+1₂ β ≥ −¹₂ γ >

1+β 2

Table 1: Six situations where Theorem2.2shows thatgbelongs to the setX_λ+^N−1

2 .

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3. Some Preliminaries

Notation 1. Givena∈B_N andR ∈]0,1[, letB(a, R_a) ={x∈B_N : |x−a|< R_a} with

R_a =R 1− |a|² 1 +R|a|.

Proposition 3.1. There exists aC > 0depending only onN,β,γ, such that:

MX_α+β+N(u)≤C MY_α,β,γ(u) ∀u∈ SY_α,β,γ.

Proof. Let someR ∈]0,1[be fixed in the following. Sinceu≥0, we obtain for any a∈B_N:

MY_α,β,γ(u)≥(1− |a|²)^α Z

BN

(1− |x|²)^βu(x) (1− |Φ_a(x)|²)^γdx

≥(1− |a|²)^α Z

B(a,Ra)

(1− |x|²)^βu(x) (1− |Φ_a(x)|²)^γdx.

It follows from Lemma 1 of [6] that

B(a, R_a)⊂E(a, R) ={x∈B_N : |Φ_a(x)|< R}, hence:

(3.1) MY_α,β,γ(u)≥(1−R²)^γ(1− |a|²)^α Z

B(a,Ra)

(1− |x|²)^βu(x)dx asγ ≥0. From Lemmas 1 and 5 of [5], it is known that

1−R

1 +R ≤ 1− |x|²

1− |a|² ≤2 ∀x∈B(a, R_a).

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LetC_β = ^1−R_1+Rβ

ifβ ≥0andC_β = 2^β ifβ <0. Hence MY_α,β,γ(u)≥C_β(1−R²)^γ(1− |a|²)^α+β

Z

B(a,Ra)

u(x)dx.

The volume of B(a, Ra)is σN (Ra)^N

N with σN = _Γ(N/2)²^π^N/2 the area of the unit sphere S_N inR^N (see [2, p. 29]) andR_a ≥ _1+R^R (1− |a|²). The subharmonicity ofunow provides:

MY_α,β,γ(u)≥C_β(1−R²)^γ(1− |a|²)^α+βu(a)σ_N (R_a)^N N

≥C_β σ_N N

R^N(1−R)^γ

(1 +R)^N−γ (1− |a|²)^α+β+Nu(a).

Corollary 3.2. Letg ∈ X_λ withMX_λ(g)≥0. Then:

MX_λ+α+β+N(gu)≤C MX_λ(g)MY_α,β,γ(u) ∀u∈ SY_α,β,γ where the constantCstems from Proposition3.1.

Proof. Sinceu≥0, we have for anyx∈B_N:

(1− |x|²)^λ+α+β+Ng(x)u(x)≤MX_λ(g) (1− |x|²)^α+β+Nu(x)

≤MX_λ(g)MX_α+β+N(u) because ofMX_λ(g)≥0.

Lemma 3.3. Givena∈BNandR ∈]0,1[, the following holds for anyx∈B(a, Ra):

1

2 < 1

1 +R|a| ≤ 1− hx, ai

1− |a|² ≤ 1 + 2R|a|

1 +R|a| <2 and 1

4 < 1− hx, ai

1− |x|² <21 +R 1−R .

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Proof. Clearlyhx, ai=ha+y, ai=|a|²+hy, aiwith|y|< R_a. From the Cauchy- Schwarz inequality, it follows that−R_a|a| ≤ hy, ai ≤R_a|a|. Hence:

1− |a|²−R|a| 1− |a|²

1 +R|a| ≤1− hx, ai ≤1− |a|²+R|a| 1− |a|² 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|

and1 +R|a|<2. The term on the right equals (1− |a|²)

1 + R|a|

1 +R|a|

,

with _1+R|a|^R|a| <1. Now

1− hx, ai

1− |x|² = 1− hx, ai 1− |a|²

1− |a|² 1− |x|² and the last inequalities follow from Lemmas 1 and 5 of [5].

Lemma 3.4. LetH = {(s, t) ∈ R² :t ≥ 0, s²+t² <1}andP > −1, Q >−1, T >−1. Then

Z Z

H

s^Pt^Q(1−s²−t²)^Tds dt=







0 if P is odd;

Γ(^P+1₂ )^Γ(^Q+1₂ )^Γ(T⁺¹⁾

2 Γ(^P^+Q2 +T+2) if P is even.

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Proof. With polar coordinatess =r cosθ,t =r sinθ, this integral turns intoI₁I₂ with

I₁ = Z 1

0

r^P^+Q(1−r²)^T r dr and I₂ = Z π

0

(cosθ)^P (sinθ)^Qdθ.

Keeping in mind the various expressions for the Beta function (see [3, pp. 67–68]):

B(x, y) = Z 1

0

ξ^x−1(1−ξ)^y−1dξ

= 2 Z π/2

0

(cosθ)^2x−1(sinθ)^2y−1dθ = Γ(x) Γ(y) Γ(x+y)

(withx >0andy >0), the change of variableω=r² leads to:

I₁ = 1 2

Z 1 0

ω^P+Q² (1−ω)^Tdω

= 1 2B

P +Q

2 + 1, T + 1

= Γ ^P+Q₂ + 1

Γ(T + 1) 2 Γ ^P^+Q₂ +T + 2 .

WhenP is odd,I₂ = 0becausecos(π−θ) = −cos(θ). However, whenP is even:

I₂ = 2 Z π/2

0

(cosθ)^P(sinθ)^Qdθ

=B

P + 1

2 ,Q+ 1 2

= Γ ^P⁺¹₂

Γ ^Q+1₂ Γ ^P^+Q₂ + 1 .

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Lemma 3.5. GivenA ≥ 0anda ∈ B_N, let uand f_a denote the functions defined onB_N byu(x) = _(1−|x|¹2)^A andf_a(x) = _(1−hx,ai)¹ A ∀x∈ B_N. They are both subhar- monic inB_N.

Remark 2. uis radial, but notf_a.

Proof. For u, the result of Lemma3.5 has already been proved in Proposition 1 of [5]. For anyj ∈ {1,2, . . . , N}, we now compute:

∂fa

∂x_j(x) =ajA(1−hx, ai)^−A−1 and ∂²fa

∂x²_j (x) = (aj)²A(A+1)(1−hx, ai)^−A−2, so that:

(∆f_a)(x) = |a|²A(A+ 1)

(1− hx, ai)^A+2 ≥0 ∀x∈B_N.

Remark 3. GivenA≥0,A⁰ ≥0, the functionf_adefined onB_N by

f_a(x) = 1

(1− hx, ai)^A(1− |x|²)^A⁰ is subharmonic too. The computation

(∆f_a)(x)≥f_a(x)

A|a|

1− hx, ai − 2A⁰|x|

1− |x|² 2

≥0 is left to the reader.

Proposition 3.6. GivenN ∈N,N >3,(s, t, b₁, b₂)∈R⁴such that|s b₁|+|t b₂|<1 andP > 0, let

I_P(s, t, b₁, b₂) = Z π

0

(sinθ)^N−3dθ (1−s b1−t b2 cosθ)^P.

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Then

I_P(s, t, b₁, b₂) = √

πΓ ^N₂ −1 Γ(P)

X

k∈N

X

j∈N

Γ(j + 2k+P)

k!j! Γ ^N₂⁻¹ +k(b₁s)^j t b₂

2 2k

.

Proof. As

t b₂ cosθ 1−s b₁

≤

t b₂ 1−s b₁

<1,

the following development is valid:

I_P(s, t, b₁, b₂) = Z π

0

(sinθ)^N⁻³dθ

(1−s b₁)^P

1− ^{t b}_{1−s b}² ^cos^θ

1

P

= 1

(1−s b₁)^P X

n∈N

Γ(n+P) n! Γ(P)

t b₂ 1−s b₁

nZ π 0

(sinθ)^N⁻³(cosθ)ⁿdθ.

The last integral vanishes whennis odd. Whennis even (n = 2k), then 2

Z π/2 0

(sinθ)^N−3(cosθ)^2kdθ =B

N −2

2 , k+ 1 2

= Γ ^N−2₂

Γ k+ ¹₂ Γ ^N−1₂ +k

= Γ ^N−2₂

(2k)!√ π Γ ^N−1₂ +k

2^2kk!

by [3, p. 40]. Hence:

I_P(s, t, b₁, b₂) = Γ ^N−2₂ √ π Γ(P)

X

k∈N

Γ(2k+P) Γ ^N−1₂ +k

2^2kk!

(t b₂)^2k (1−s b₁)^2k+P.

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The result follows from the expansion Γ(2k+P)

(1−s b1)^2k+P =X

j∈N

Γ(j+ 2k+P)

j! (b₁s)^j.

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4. Proof of Theorem 2.2

The cases (i), (ii), (iii), (iv), (v) and (vi) of Theorem2.2will 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

I_A,P,T(a, b) = Z

BN

(1− |x|²)^T

(1− hx, ai)^A(1− hx, bi)^P dx ∀a∈B_N,∀b ∈B_N andτ a number satisfying both ^P^−A₂ < τ < P and0≤τ ≤ ^A+P₂ . Then

I_A,P,T(a, b)≤ K

(1− |a|²)^A+P² ^−τ(1− |b|²)^τ ∀a∈B_N,∀b∈B_N

where the constantK is independent ofaandb.

Example 4.1. IfP > Aandτ = ^A+P₂ , then I_A,P,T(a, b)≤ K

(1− |b|²)^A+P² ∀a ∈B_N,∀b∈B_N, with

K = 2^A+P⁻¹π^N−1² Γ(T + 1) Γ(P) Γ

P −A 2

.

Example 4.2. IfP < Aandτ = 0, then I_A,P,T(a, b)≤ K

(1− |a|²)^A+P² ∀a ∈B_N,∀b∈B_N,

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with

K = 2^A+P⁻¹π^N−1² Γ(T + 1) Γ(A) Γ

A−P 2

.

Proof. Up to a unitary transform, we assumea= (|a|,0,0, . . . ,0)andb= (b₁, b₂,0, . . . ,0).

Proof of Theorem4.1in the case N > 3. Polar coordinates inR^N provide the formulas:x₁ =r cosθ₁withr=|x|,x₂ =r sinθ₁ cosθ₂(the formulas forx₃, . . . , x_N are available in [9, p. 15]) where θ₁, θ₂,. . . , θN−2 ∈]0, π[ andθ_N−1 ∈]0,2π[. The volume element dx becomesr^N⁻¹dr dσ^(N) wheredσ^(N) denotes the area element onS_N, with

dσ^(N) = (sinθ₁)^N⁻²(sinθ₂)^N−3dθ₁dθ₂dσ^(N−2)

(see [9, p. 15] for full details). Hereθ₂ ∈]0, π[sinceN > 3. In the following, we will writes=r cosθ₁ andt =r sinθ₁, thushx, bi=s b₁+t b₂ cosθ₂and

(4.1) I_A,P,T(a, b) =σ_N−2 Z π

0

Z 1 0

(1−r²)^Tr^N⁻¹(sinθ₁)^N⁻²I_P(s, t, b₁, b₂) (1− |a|s)^A dr dθ₁ withI_P(s, t, b₁, b₂)defined in the previous proposition. From [2, p. 29] we notice that

σN−2Γ

N −2 2

√

π= 2π^N−1² .

The expansion

1

(1− |a|s)^A =X

`∈N

Γ(`+A)

`! Γ(A) (|a|s)^` leads to:

I_A,P,T(a, b) = 2π^N−1² Γ(P) Γ(A)

X

(k,j,`)∈N³

Γ(j+ 2k+P) Γ(`+A) k!j!`! Γ(^N₂⁻¹ +k) (b₁)^j

b₂ 2

2k

|a|^`J_k,j,`

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where

Jk,j,` = Z π

0

Z 1 0

s^j+`t^2k(1−r²)^Tr^N−1(sinθ1)^N−2dr dθ1

= Z Z

H

s^j+`t^2k+N⁻²(1−s²−t²)^Tds dt

withH as in Lemma3.4. NowJk,j,` = 0unlessj+`= 2h(h∈N). Thus:

IA,P,T(a, b) = π^N−1² Γ(P) Γ(A)

× X

(k,h)∈N² 2h

X

j=0

Γ(j+ 2k+P) Γ(2h−j +A) Γ h+¹₂

Γ(T + 1) k!j! (2h−j)! Γ k+h+^N₂ +T + 1 (b₁)^j

b₂ 2

2k

|a|^2h−j

Taking [3, p. 40] into account:

(4.2) I_A,P,T(a, b) = π^N² Γ(T + 1) Γ(P) Γ(A)

X

(k,h)∈N² 2h

X

j=0

(2h)!B(j+ 2k+P,2h−j+A) 2^2h+2kh!k!j! (2h−j)!

× Γ(2k+P + 2h+A)

Γ(k+h+^N₂ +T + 1) b^j₁b^2k₂ |a|^2h−j. Let

L= 2^P+A−1Γ(T + 1) Γ(P) Γ(A) π^N−1² . The duplication formula

√πΓ(2z) = 2^2z−1Γ(z) Γ

z+1 2

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(see [3, p. 45]) is applied with2z = 2k+P + 2h+A. Now Γ(k+h+A+P + 1

2 )≤Γ

k+h+ N

2 +T + 1

sinceΓincreases on[1,+∞[and

1≤k+h+ A+P + 1

2 ≤k+h+N

2 +T + 1.

This leads to:

I_A,P,T(a, b)

≤L X

(k,h)∈N² 2h

X

j=0

(2h)!B(j+ 2k+P,2h−j +A) Γ k+h+^A+P₂

h!k!j! (2h−j)! b^j₁b^2k₂ |a|^2h−j

=L X

(k,h)∈N²

Γ k+h+^A+P₂ h!k! b^2k₂

2h

X

j=0

(2h)!

j! (2h−j)!b^j₁|a|^2h−jB(j+ 2k+P,2h−j +A).

The last sum turns into:

2h

X

j=0

(2h)!b^j₁|a|^2h−j j! (2h−j)!

Z 1 0

ξ^j+2k+P⁻¹(1−ξ)^2h−j+A−1dξ

= Z 1

0 2h

X

j=0

(2h)! (b1ξ)^j[(1−ξ)|a|]^2h−j j! (2h−j)!

!

ξ^2k+P⁻¹(1−ξ)^A−1dξ

= Z 1

0

[b₁ξ+|a|(1−ξ)]^2hξ^2k+P⁻¹(1−ξ)^A−1dξ.

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Hence the majorant ofI_A,P,T(a, b)becomes:

L Z 1

0

X

k∈N

(b2ξ)^2k k!

X

h∈N

Γ(h+k+ ^A+P₂ )

h! [b₁ξ+|a|(1−ξ)]^2h

!

ξ^P⁻¹(1−ξ)^A−1dξ

=L Z 1

0

X

k∈N

Γ(k+^A+P₂ ) (b2ξ)^2k k!

1

1−[b₁ξ+|a|(1−ξ)]²

k+^A+P₂

ξ^P⁻¹(1−ξ)^A−1dξ

according to the expansion

Γ(C)

(1−X)^C =X

h∈N

Γ(h+C) h! X^h

with|X|< 1whenC > 0(see [8, p. 53]). HereX = [b₁ξ+|a|(1−ξ)]² belongs to]−1,1[sinceb₁and|a|do andξ ∈[0,1]. The same expansion now applies with

C = A+P

2 and X = (b₂ξ)²

1−[b₁ξ+|a|(1−ξ)]² since|X|<1, as

δ(ξ) := (b₂ξ)²+ [b₁ξ+|a|(1−ξ)]²

=|b|²ξ²+|a|²(1−ξ)² + 2ξ(1−ξ)b₁|a|

≤ |b|²ξ²+|a|²(1−ξ)²+ 2ξ(1−ξ)|b| |a|

= [ξ|b|+|a|(1−ξ)]² <1.

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Hence

IA,P,T(a, b)

≤L Z 1

0

Γ ^A+P₂

1− _1−[b ^(b²^ξ)²

1ξ+|a|(1−ξ)]²

^A+P₂

ξ^P−1(1−ξ)^A−1dξ (1−[b₁ξ+|a|(1−ξ)]²)^A+P²

=L· Γ

A+P 2

Z 1 0

ξ^P⁻¹(1−ξ)^A−1dξ

(1−[b₁ξ+|a|(1−ξ)]²−(b₂ξ)²)^A+P² .

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) since

[2−ξ(1− |b|)]−(1 +|b|) = (1−ξ)(1− |b|)≥0.

Similarly,

1−δ(ξ)≥(1−ξ)(1− |a|²).

Thus 1

[1−δ(ξ)]^A+P² ≤ 1

[(1−ξ)(1− |a|²)]^A+P² ^−τ[ξ(1− |b|²)]^τ sinceτ ≥0and ^A+P₂ −τ ≥0. Finally:

I_A,P,T(a, b)≤ L·Γ ^A+P₂

(1− |a|²)^A+P² ^−τ(1− |b|²)^τ Z 1

0

ξ^P^−τ−1(1−ξ)^A+τ−^A+P² ⁻¹dξ.

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This integral converges sinceP −τ >0and A+τ− A+P

2 = A−P

2 +τ >0.

Now the result follows with K =L·Γ

A+P 2

B

P −τ,A−P 2 +τ

=LΓ(P −τ) Γ

A−P 2 +τ

.

Proof of Theorem4.1in the caseN = 3. Here

I_A,P,T(a, b) = Z π

0

Z 1 0

(1−r²)^T r²(sinθ1)JP(s, t, b1, b2)

(1− |a|s)^A dr dθ₁, where

J_P(s, t, b₁, b₂) = Z 2π

0

dθ₂

(1−s b₁−t b₂ cosθ₂)^P = 2I_P(s, t, b₁, b₂)

with I_P(s, t, b₁, b₂) as in Proposition 3.6, with N = 3. Hence I_A,P,T(a, b) has the same expression as in Formula (4.1), withN = 3, sinceσ₁ = 2. Thus the proof ends in the same manner as that above in the caseN >3.

Proof of Theorem 4.1 in the caseN = 2. Now x1 = s = r cosθ and x2 = t = r sinθ:

I_A,P,T(a, b) = Z 2π

0

Z 1 0

(1−r²)^T r dr dθ (1− |a|s)^A(1−s b₁−t b₂)^P

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= Z

B2

X

`∈N

Γ(`+A)

`! Γ(A) (|a|s)^`X

n∈N

(t b₂)ⁿ n! Γ(P)

Γ(n+P)

(1−s b₁)^n+P (1−s²−t²)^T ds dt

= X

(`,n,j)∈N³

Γ(`+A)|a|^`(b2)ⁿΓ(j+n+P) (b1)^j

`! Γ(A)n! Γ(P)j!

Z

B2

s^`+jtⁿ(1−s²−t²)^Tds dt.

The last integral vanishes when n is odd or ` +j odd. Otherwise (n = 2k and

`+j = 2h), it equals

2 Z

H

s^`+jtⁿ(1−s²−t²)^T ds dt= Γ h+¹₂

Γ k+ ¹₂

Γ(T + 1) Γ(k+h+T + 2)

by Lemma3.4and turns into

n! (2h)!πΓ(T + 1) 2^2h+2kh!k! Γ(k+h+T + 2)

according to [3, p. 40]. Thus I_A,P,T(a, b)is again recognized as Formula (4.2) now withN = 2and the proof ends as for the caseN >3.

We now present an example of a family of functions {f_a}_a which is uniformly bounded above inY_α,β,γ:

Corollary 4.2. Givenβ >−^N+1₂ (N ≥2) letα= ^N+1₂ +βandγ >max(α,−1−β).

For anya∈B_N letf_adenote the function defined by:f_a(x) = _(1−hx,ai)¹ 2α,∀x∈B_N. Thenf_a ∈ Y_α,β,γ,∀a ∈B_N. Moreover, there existsK > 0such thatMY_α,β,γ(f_a) ≤ K,∀a ∈B_N.

Remark 4. This constantKis the same as that in the previous theorem, withA= 2α, P = 2γ andT =β+γ.

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Proof. With the above choices for parametersA, P,T, we actually have: P > A >

0,T >−1and

A+P = 2α+ 2γ =N + 1 + 2β+ 2γ =N + 1 + 2T >1.

The conditions0 ≤ τ ≤ α+γ together withγ−α < τ <2γ reduce to: γ −α <

τ ≤α+γ. Let

(4.3) J_b(f_a) = (1− |b|²)^α Z

BN

(1− |x|²)^βf_a(x) (1− |Φ_b(x)|²)^γdx.

Now

J_b(f_a) = (1− |b|²)^α+γ Z

BN

(1− |x|²)^β+γ

(1− hx, ai)^N+1+2β(1− hx, bi)^2γ dx

≤K ∀a ∈B_N,∀b∈B_N

according to Theorem4.1applied withτ =α+γ = ^A+P₂ . 4.1. Proof of Theorem2.2in the case (i)

GivenR ∈]0,1[, the subharmonicity ofgprovides for anya∈B_N the majoration:

g(a)≤ 1 V_a

Z

B(a,Ra)

g(x)dx

withVathe volume ofB(a, Ra). From Lemma3.3, it is clear that:

1≤

21 +R 1−R

1− |x|² 1− hx, ai

A

∀x∈B(a, R_a)

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withA = 2α > 0. Nowg(x)≥0,∀x ∈B_N. Withf_aas in Corollary4.2, this leads to:

V_ag(a)≤

21 +R 1−R

AZ

B(a,Ra)

(1− |x|²)^Af_a(x)g(x)dx.

Now

A=α+β+N + 1

2 =α+β+N − N −1 2 , thus

V_ag(a)≤

21 +R 1−R

AZ

B(a,Ra)

(1− |x|²)^λ+α+β+Nfa(x)g(x) (1− |x|²)^λ+^N−1² dx

≤C⁰K

21 +R 1−R

AZ

B(a,Ra)

dx (1− |x|²)^λ+^N−1² from Corollary4.2. Lemmas 1 and 5 of [5] provide

1− |x|² 1− |a|²

λ+^N−1₂

≥C_λ+^N−1

2 ∀x∈B(a, R_a), with C_λ+^N−1

2 defined in the same pattern as C_β in the proof of Proposition 3.1.

Finally:

V_ag(a)≤ C⁰K C_λ+^N−1

2

21 +R 1−R

A

V_a

(1− |a|²)^λ+^N−1² , thus

MX

λ+N−1 2

(g)≤ C⁰K C_λ+^N−1

2

21 +R 1−R

2α

∀R ∈]0,1[.

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The majorant is an increasing function with respect toR. LettingRtend toward0⁺, we get:

MX

λ+N−1 2

(g)≤ C⁰K C_λ+^N−1

2

2^2α.

4.2. Proof of Theorem2.2in the case (ii) Here we work withf_adefined by:

f_a(x) = 1

(1− hx, ai)^A where A=α+β+N.

Theorem4.1applies once again, withA = N + 1 + 2β > ^N−1₂ > 0,P = 2γ > 0 andT = β+γ >−1(becauseγ >−1−β). ConditionA+P = N + 1 + 2T is fulfilled too. Moreoverτ :=α+γ =β+γ+ 1satisfies both0≤τ ≤β+γ+^N₂⁺¹ (obviously0< β+γ+ 1and1< ^N₂⁺¹) andγ−β− ^N+1₂ < τ <2γ:

τ −γ+β+N + 1

2 = 2β+ N + 3

2 >0 and 2γ−τ =γ−1−β >0.

With such a choice forτ we have A+P

2 −τ = N + 1

2 −1 = N −1 2 , thus

(4.4) I_A,P,T(a, b)≤ K

(1− |a|²)^N+1² ⁻¹(1− |b|²)^α+γ ∀a∈B_N,∀b∈B_N. Hence,J_b(f_a)defined in Formula (4.3) now satisfies

(4.5) J_b(f_a)≤ K

(1− |a|²)^N−1² ∀a∈B_N,∀b ∈B_N.

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In other words,

(4.6) MY_α,β,γ(f_a)≤ K

(1− |a|²)^N−1² ∀a ∈B_N. This implies:

(4.7) MX_λ+α+β+N(g f_a)≤ C⁰K

(1− |a|²)^N−1² ∀a∈B_N.

WithRandV_aas in the previous proof, we obtain here:

V_ag(a)≤

21 +R 1−R

AZ

B(a,Ra)

(1− |x|²)^λ+α+β+Nf_a(x)g(x) (1− |x|²)^λ dx

≤ C⁰K (1− |a|²)^N−1²

21 +R 1−R

AZ

B(a,Ra)

dx (1− |x|²)^λ and the last integral is majorized by _C ^V^a

λ(1−|a|²)^λ withCλ defined similarly toCβ in the proof of Proposition3.1. Finally:

MX

λ+N−1 2

(g)≤ C⁰K

C_λ 2^N+1+2β.

4.3. Proof of Theorem2.2in the case (iii) Heref_ais defined by:

f_a(x) = 1

(1− hx, ai)^A(1− |x|²)^β+γ ∀x∈B_N,

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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 chooseτ satisfying both 0≤τ ≤ N + 1

2 and 2γ− N + 1

2 < τ <2γ.

Now

τ := N + 1

2 = A+P

2 =α+γ fulfills the last condition since:

2γ−τ = 2

γ−N + 1 4

>0 and τ −2γ+ N + 1 2 = 2

N + 1 2 −γ

>0.

Formula (4.3) impliesJ_b(f_a)≤Kfor alla∈B_N and allb ∈B_N. ThusMY_α,β,γ(f_a)≤ K,∀a ∈B_N. As before,

V_ag(a)≤

21 +R 1−R

AZ

B(a,Ra)

(1− |x|²)^A+β+γg(x) (1− hx, ai)^A(1− |x|²)^β+γdx.

Now

A+β+γ =N + 1−γ+β

=N + 1 +α− N + 1 2 +β

=α+β+N − N −1 2 ,

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whence

V_ag(a)≤

21 +R 1−R

AZ

B(a,Ra)

(1− |x|²)^α+β+Nf_a(x)g(x) (1− |x|²)^N−1² dx

≤C⁰K

21 +R 1−R

AZ

B(a,Ra)

dx (1− |x|²)^λ+^N−1² and the proof ends as in the case (i). Here

MX

λ+N−1 2

(g)≤ C⁰K C_λ+^N−1

2

2^N+1−2γ.

4.4. Proof of Theorem2.2in the case (iv) Heref_ais defined by:

f_a(x) = 1

(1− hx, ai)^A(1− |x|²)^β ∀x∈B_N,

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 ≤ ^N₂⁺¹ + γ and γ− ^N+1₂ < τ <2γ). Hence Inequalities (4.4), (4.5), (4.6) and (4.7) follow. Now (4.8) V_ag(a)≤

21 +R 1−R

AZ

B(a,Ra)

(1− |x|²)^A+βf_a(x)g(x)dx.

SinceA+β =α+β+N, this turns into:

V_ag(a)≤

21 +R 1−R

AZ

B(a,Ra)

M_X_λ+α+β+N(g f_a) (1− |x|²)^λ dx

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and the proof ends as in the case (ii), here with:

MX

λ+N−1 2

(g)≤ C⁰K C_λ 2^N+1.

4.5. Proof of Theorem2.2in the case (v) Heref_ais defined by:

f_a(x) = 1

(1− hx, ai)^A(1− |x|²)^γ ∀x∈B_N, where

A=N+ 1 + 2(β−γ)> N+ 1−N + 3

2 = N −1 2 >0.

WithP = 2γ >0andT =β, the conditionA+P =N + 1 + 2T of Theorem4.1 is fulfilled. Moreoverτ :=α+γ = 1 +βsatisfies

0≤τ ≤ N + 1

2 +β and 2γ−N + 1

2 −β < τ <2γ since:

2γ−τ = 2γ−(1 +β)>0 and τ−2γ+N + 1

2 +β =−2γ+N + 3

2 + 2β >0.

Again

A+P

2 −τ = N + 1

2 −1 = N −1 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:

MX

λ+N−1 2

(g)≤ C⁰K

C_λ 2^N^{+1+2(β−γ)}.

4.6. Proof of Theorem2.2in the case (vi) Heref_ais defined by:

f_a(x) = 1

(1− hx, ai)^A(1− |x|²)^α ∀x∈B_N

with A = N + β > ^N₂⁻¹ > 0, P = 2γ > 0, T = ^β−1₂ +γ > −1 (actually T + 1 = ^β+1₂ +γ >0). The use of Theorem4.1is allowed since

A+P =N + 1 +β−1 + 2γ =N + 1 + 2T.

Now τ := α+γ = ^β+1₂ +γ satisfies0 ≤ τ ≤ ^N^+β₂ +γ (because of γ > −^β+1₂ ).

Moreoverγ −^N^+β₂ < τ <2γis fulfilled too since β+ 1

2 < γ and β+ 1 + (N +β) = 1 +N + 2β >0.

In addition,

A+P

2 −τ = N +β

2 − β+ 1

2 = N −1 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

MX

λ+N−1 2

(g)≤ C⁰K C_λ 2^N+β.