Journal of Inequalities in Pure and
Applied Mathematics
SUBHARMONIC FUNCTIONS AND THEIR RIESZ MEASURE
RAPHAELE 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
volume 2, issue 2, article 16,
2001.
Received 30 August, 2000;
accepted 22 January, 2001.
Communicated by: A. Fiorenza
Abstract
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c 2000 Victoria University
ISSN (electronic): 1443-5756
030-00
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Abstract
For subharmonic functions
u in RN , of Riesz measure µ, the growth of the
R
function s 7→ µ(s) = |ζ|≤s dµ(ζ) (s ≥ 0) is described and compared with the
R
growth of u. It is also shown that, ifR RN u+ (x) [−ϕ0 (|x|2 )] dx < +∞ for some
decreasing C 1 function ϕ ≥ 0, then RN |ζ|12 ϕ(|ζ|2 + 1) dµ(ζ) < +∞. Given two
subharmonic functions u1 and u2 , of Riesz measures µ1 and µ2 , with a growth
like ui (x) ≤ A + B|x|γ ∀x ∈ RN (i = 1, 2), it is proved that µ1 + µ2 is not
necessarily the Riesz measure of a subharmonic function u with such a growth
as u(x) ≤ A0 + B 0 |x|γ ∀x ∈ RN (here A > 0, A0 > 0 and 0 < B 0 < 2B).
Subharmonic Functions and
their Riesz Measure
Raphaele Supper
2000 Mathematics Subject Classification: 31A05, 31B05, 26D15, 28A75.
Key words: Subharmonic functions, order of growth, Riesz measure.
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Contents
Contents
1
2
3
4
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Some Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Estimations of the Riesz Measure . . . . . . . . . . . . . . . . . . . . . . .
3.1
Jensen–Privalov formula. . . . . . . . . . . . . . . . . . . . . . . .
3.2
The case N = 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3
The case N ≥ 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Growth of the Repartition Function . . . . . . . . . . . . . . . . . . . . . .
4.1
A measure on [0, +∞[, image of µ . . . . . . . . . . . . . . .
4.2
The case N = 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3
Proof of Theorem 4.1 in the case of a continuous
repartition function . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
7
10
10
10
12
15
15
15
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16
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Splitting measure µ . . . . . . . . . . . . . . . . . . . . . . . . . . . .
A reformulation of Theorem 4.1 . . . . . . . . . . . . . . . . .
The case N ≥ 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Proof of Theorem 4.3 in the case of a continuous
repartition function . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.8
A reformulation of Theorem 4.3 . . . . . . . . . . . . . . . . .
5
Sum of Two Riesz Measures . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
Subharmonic Functions Subject to Conditions of L1 Type . .
6.1
A weighted integral condition for subharmonic functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2
Proof of Theorem 6.1 in the case N = 2 . . . . . . . . . . .
6.3
Proof of Theorem 6.1 in the case N ≥ 3 . . . . . . . . . . .
References
4.4
4.5
4.6
4.7
17
19
20
21
23
25
27
27
28
29
Subharmonic Functions and
their Riesz Measure
Raphaele Supper
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1.
Introduction
Let µ be the Riesz measure of some subharmonic function u in RRN (N ∈ N,
N ≥ 2 and u non identically −∞, see [1, p. 104]) and µ(s) = |ζ|≤s dµ(ζ)
for any s ≥ 0 (where | · | denotes the Euclidean norm in RN ). The function
s 7→ µ(s) is non–decreasing since µ is a positive measure. The order of the
function s 7→ s2−N µ(s) is known to coincide with the convergence exponent of
µ:
Z
Z
+∞
+∞
s2−N −c dµ(s)
inf c :
s1−N −c µ(s) ds
= inf c :
1
1
(see [2, p. 66]) and does not exceed γ if u has a growth of the kind:
(1.1)
u(x) ≤ A + B|x|γ
Subharmonic Functions and
their Riesz Measure
Raphaele Supper
∀x ∈ RN
(with constants A ∈ R, B > 0 and γ > 0). This estimation of the growth of
µ(s) will be examined below, in Sections 3 and 4.
Definition 1.1. Given γ > 0 and B > 0, let SH(γ, B) stand for the set of all
subharmonic functions u in RN which are harmonic in some neighbourhood
of the origin with u(0) = 0 and which satisfy estimate (1.1) for some constant
A ∈ R.
In Proposition 5.2 (see Section 5), a counterexample is produced to show
that, given u1 and u2 two functions in this set SH(γ, B) and B 0 ∈ ]0, 2B[, the
sum of their respective Riesz measures µ1 and µ2 is not necessarily the Riesz
measure of a function of SH(γ, B 0 ).
Of course µ1 +µ2 is the Riesz measure associated with u1 +u2 ∈ SH(γ, 2B),
but µ1 + µ2 is also the Riesz measure of u1 + u2 − h for any harmonic function h
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in RN . This proposition means that there does not necessarily exist a harmonic
function h such that u1 + u2 − h ∈ SH(γ, B 0 ).
Let µ denote the Riesz measure of some function of SH(γ, B) with growth
(1.1). Sections 3 and 4 are devoted to the growth of the repartition function s 7→
Aγ
µ(s). For instance, when N = 2, we obtain the inequality: µ(s) ≤ Beγ sγ e µ(s)
(see Theorem 3.1 and Corollary 3.2).
−2
γ+N
N −2
γ+N −2
Notation . When N ≥ 3, throughout the paper we set C(γ, N ) =
γ
N −2
γ+N −2
Bγ γ+N −2
, sometimes written merely D for brevity.
and D(B, γ, N ) = γ
N −2
Subharmonic Functions and
their Riesz Measure
Raphaele Supper
Note that
γ+N −2
γ + N − 2 N −2
γ
γ
γ+N −2
N − 2 N −2
=
1+
N −2
γ
γ+N −2
.
≤ e
N −2
γ
γ
C(γ, N ) =
N −2
N −2
For N ≥ 3, we also obtain inequalities describing the growth of s 7→ µ(s)
and the constants involved in these estimations are given explicitely in terms of
A, B and γ. For example:
−2
! γ+N
N −2
Bγ
A
µ(s) ≤
C(γ, N ) sγ+N −2 1+
γ
N −2
D · [µ(s)] γ+N −2
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(see Theorem 3.4 and Corollary 3.5).
µ(s)
It points out that lim sups→+∞ sγ+N
−2 is not greater than Beγ (when N = 2)
µ(s)
Bγ
or N −2 C(γ, N ) (when N ≥ 3). Moreover, lim inf s→+∞ sγ+N
−2 does not exceed
Bγ
Bγ (if N = 2) or N −2 (if N ≥ 3). This will follow from Theorems 4.1 and 4.3
which assert that the sets:
n
o
Aγ
s : µ(s) < Bγ sγ e µ(s)
and


Bγ γ+N −2
s : µ(s) <
s

N −2
1+

−2
! γ+N
N −2 
A
Subharmonic Functions and
their Riesz Measure
γ
D · [µ(s)] γ+N −2

Raphaele Supper
are unbounded in the cases when N = 2 and N ≥ 3 respectively.
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N
The last section studies subharmonic functions u in R (harmonic in some
neighbourhood of the origin with u(0) = 0) such that the subharmonic function
u+ (defined by u+ (x) = max(u(x), 0) ∀x ∈ RN ) satisfies a L1 condition, for
in Theorem 6.1:
Rexample
+
u (x) [−ϕ0 (|x|2 )] dx < +∞ (see Section 6.1 for more details on the deRN
creasing function ϕ). The Riesz measure µ of u is then proved to verify:
R ϕ(|ζ|2 +1)
dµ(ζ) < +∞. Propositions 6.2 and 6.3 provide similar results un|ζ|2
RN
der different L1 conditions.
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2.
Some Preliminaries
Lemma 2.1. If N = 2, then
Z
Z
r
r
log
dµ(ζ) ≤
log
dµ(ζ)
|ζ|
|ζ|
|ζ|≤r
|ζ|≤s
for each r > 0 and each s > 0.
r
Proof. If r ≤ s, then hr (ζ) := log |ζ|
≤ 0 for r < |ζ| ≤ s, so that
Z
Z
Z
hr (ζ) dµ(ζ) =
hr (ζ) dµ(ζ)+
hr (ζ) dµ(ζ) ≤
hr (ζ) dµ(ζ).
|ζ|≤s
|ζ|≤r
r<|ζ|≤s
|ζ|≤r
|
{z
}
Subharmonic Functions and
their Riesz Measure
Z
≤0
If s < r, then hr (ζ) ≥ 0 for |ζ| ≤ r, hence
Z
Z
Z
Z
hr (ζ) dµ(ζ).
hr (ζ) dµ(ζ) =
hr (ζ) dµ(ζ)+
hr (ζ) dµ(ζ) ≥
|ζ|≤s
|ζ|≤r
|ζ|≤s
s<|ζ|≤r
|
{z
}
≥0
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Lemma 2.2. When N ≥ 3, the following majoration is valid for all r > 0 and
s > 0:
Z
Z
1
1
1
1
−
dµ(ζ) ≤
−
dµ(ζ).
|ζ|N −2 rN −2
|ζ|N −2 rN −2
|ζ|≤s
|ζ|≤r
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Proof. As in the previous proof, with hr (ζ) =
1
|ζ|N −2
−
1
r N −2
r
instead of log |ζ|
.
Lemma 2.3. If N = 2, then:
Z r
Z
µ(t)
r
dt =
log
dµ(ζ),
t
|ζ|
0
|ζ|≤r
for any r > 0.
Proof. It follows from Fubini’s theorem that:
Z r
Z r Z
µ(t)
1
dµ(ζ) dt
dt =
t
0 t
|ζ|≤t
0
Z r Z
dt
dµ(ζ)
=
|ζ|≤r
|ζ| t
Z
r
=
log
dµ(ζ).
|ζ|
|ζ|≤r
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their Riesz Measure
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Lemma 2.4. When N ≥ 3, then
Z r
Z
µ(t)
1
1
(N − 2)
dt =
−
dµ(ζ),
N −1
|ζ|N −2 rN −2
0 t
|ζ|≤r
for any r > 0
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Proof. As in the previous proof:
Z
Z r
Z r
µ(t)
N −2
(N − 2)
dt =
dµ(ζ) dt
N −1
N −1
0 t
0 t
|ζ|≤t
Z r
Z
N −2
=
dt dµ(ζ)
N −1
|ζ|≤r
|ζ| t
r
Z
−1
dµ(ζ)
=
N −2
|ζ|≤r t
|ζ|
Z
1
1
−
dµ(ζ).
=
|ζ|N −2 rN −2
|ζ|≤r
Subharmonic Functions and
their Riesz Measure
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3.
3.1.
Estimations of the Riesz Measure
Jensen–Privalov formula.
For any function u, subharmonic in RN , harmonic in some neighbourhood of
the origin, the Jensen–Privalov formula (see [2, p. 44]) holds for every r > 0:
Z 2π
Z r
µ(t)
1
iθ
u(r e ) dθ =
dt + u(0)
if N = 2
2π 0
t
0
Z
Z r
1
µ(t)
u(rx) dσx = (N − 2)
dt + u(0)
if N ≥ 3
N −1
σN SN
0 t
R
with SN the unit sphere in RN , dσ the area element on SN and σN = SN dσ =
2 π N/2
Γ(N/2)
(see [1, p. 29]). In all statements of both Sections 3 and 4, it will be
assumed that u ∈ SH(γ, B) and that its growth is indicated by (1.1).
3.2.
The case N = 2
Theorem 3.1. When N = 2, the following inequality holds for each s > 0:
Z
µ(s)
µ(s)
log
≤A+
log |ζ| dµ(ζ).
γ
Beγ
|ζ|≤s
Proof. For each r > 0 and each s > 0, it follows from Lemmas 2.1 and 2.3 that
Z
Z 2π
r
1
log
dµ(ζ) ≤
u(r eiθ ) dθ ≤ A + B rγ ,
|ζ|
2π
|ζ|≤s
0
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their Riesz Measure
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so that
Z
1
µ(s) Bγ γ
γ
γ
log
dµ(ζ) ≤ A + B r − µ(s) log r = A +
r − log r
|ζ|
γ
µ(s)
|ζ|≤s
:= ϕ(r).
Consider s constant, the minimum of ϕ is attained when Bγ rγ = µ(s), since
ϕ0 (r) = 1r (Bγ rγ − µ(s)). Finally, for each s > 0:
Z
µ(s)
µ(s)
µ(s)
1
µ(s)
log
dµ(ζ) ≤ A +
1 − log
=A−
log
|ζ|
γ
Bγ
γ
Beγ
|ζ|≤s
In Corollaries 3.2, 3.3 and 3.5, we set ε > 0 such that µ(s) > 0 ∀s > ε.
Aγ
Corollary 3.2. If N = 2, then µ(s) ≤ Beγ sγ e µ(s) for any s > ε.
Proof. Theorem 3.1 may be rewritten as:
Z
µ(s)
Aγ
dµ(ζ)
(3.1)
log
≤
+
log(|ζ|γ )
.
Beγ
µ(s)
µ(s)
|ζ|≤s
Subharmonic Functions and
their Riesz Measure
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γ
The previous integral being ≤ log s , Corollary 3.2 results.
Corollary 3.3. When N = 2, we have for every s > ε:
Z
Aγ
2
[µ(s)] ≤ Beγ exp
|ζ|γ dµ(ζ).
µ(s)
|ζ|≤s
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Proof. Jensen’s inequality applies to (3.1) since
R
dµ(ζ)
|ζ|≤s µ(s)
= 1, hence:
Z
Aγ
γ dµ(ζ)
· exp
log(|ζ| )
µ(s)
µ(s)
|ζ|≤s
Z
Aγ
dµ(ζ)
≤ exp
|ζ|γ
.
µ(s)
µ(s)
|ζ|≤s
µ(s)
≤ exp
Beγ
Subharmonic Functions and
their Riesz Measure
3.3.
The case N ≥ 3
Raphaele Supper
Theorem 3.4. When N ≥ 3, the following estimation is valid for each s > 0:
Z
γ
1
dµ(ζ) ≤ A + D [µ(s)] γ+N −2 .
N
−2
|ζ|≤s |ζ|
Proof. For all r > 0 and s > 0, Lemmas 2.2 and 2.4 lead to:
Z
Z
1
1
1
− N −2 dµ(ζ) ≤
u(rx) dσx ≤ A + B rγ ,
N −2
|ζ|
r
σ
N
|ζ|≤s
SN
that is
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Z
|ζ|≤s
1
|ζ|N −2
dµ(ζ) ≤ A + B rγ +
µ(s)
rN −2
γ
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µ(s)
.
rN −2
whose minimum (with s constant) is attained when Bγ r = (N − 2)
In
N −2
γ+N
−2
Bγ 1
1
other words, this minimum is A+ Nγ−2 + 1 rµ(s)
.
N −2 with r N −2 =
N −2 µ(s)
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Finally:
Z
1
|ζ|≤s
|ζ|N −2
dµ(ζ) ≤ A +
N −2
γ+N
−2
γ
N −2
Bγ
+1
(µ(s)) γ+N −2 .
γ
N −2
Corollary 3.5. When N ≥ 3, the following estimation holds for every s > ε:
Bγ
µ(s) ≤
C(γ, N ) sγ+N −2
N −2
N −2
.
γ+N −2
Proof. Let α =
1
sN −2
≤
1
µ(s)
Z
A
Subharmonic Functions and
their Riesz Measure
.
γ
D · [µ(s)] γ+N −2
Raphaele Supper
According to Theorem 3.4, for any s > ε we have
1
|ζ|≤s
1+
−2
! γ+N
N −2
|ζ|N −2
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dµ(ζ) ≤
Hence
[µ(s)]α ≤ D sN −2
Now, it is obvious that 1 − α =
A
D
D
+
=
α
µ(s) µ(s)
µ(s)α
γ
γ+N −2
1+
1
A
D [µ(s)]1−α
and D1/α =
α
1+
A µ(s)
D µ(s)
.
.
Bγ
C(γ, N ).
N −2
N −2
Corollary 3.6. With N ≥ 3 and α = γ+N
, the following holds for each
−2
s > 0:
Z
A
µ(s)α
α
µ(s) log
−
µ(s) ≤ (N − 2)
log |ζ| dµ(ζ)
D
D
|ζ|≤s
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Proof. It follows from Jensen’s inequality that:
Z
Z
1
dµ(ζ)
dµ(ζ)
1
log N −2
exp
≤
exp log N −2
|ζ|
µ(s)
|ζ|
µ(s)
|ζ|≤s
|ζ|≤s
Z
1 dµ(ζ)
=
N
−2 µ(s)
|ζ|≤s |ζ|
D
A
≤
+
,
µ(s)
µ(s)α
Subharmonic Functions and
their Riesz Measure
so that:
Z
A
D
+
µ(s) µ(s)α
D
A µ(s)α
≤ log
+
.
µ(s)α
D µ(s)
dµ(ζ)
−(N − 2)
log |ζ|
≤ log
µ(s)
|ζ|≤s
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4.
4.1.
Growth of the Repartition Function
A measure on [0, +∞[, image of µ
Let Φ : RN → [0, +∞[ be the measurable map defined by Φ(ζ) = µ(|ζ|) (the
function s 7→ µ(s) is increasing hence measurable on [0, +∞[). Let ν = Φ∗µ =
µ ◦ Φ−1 denote the measure image of µ under Φ (see [3, p. 80]):
Z +∞
Z
f (t) dν(t) =
f (Φ(ζ)) dµ(ζ)
0
RN
holds for any nonnegative measurable function f on [0, +∞[ (and for any νintegrablef )
Remark 4.1. If s 7→ µ(s) is continuous on some interval [a, +∞[ with a ≥ 0,
then ν(I) = c − b for any interval I with bounds b and c (c > b > µ(a)).
4.2.
The case N = 2
Up to the end of Section 4, µ stands for the Riesz measure associated with a
function of SH(γ, B) with growth (1.1).
Theorem 4.1. If N = 2 and A > γ2 , then the set of those s > 0 which satisfy
Aγ
µ(s) < Bγ sγ e µ(s) is unbounded.
A proof is required only in the case where lims→+∞ µ(s) = +∞ (otherwise,
Theorem 4.1 is obvious). When the function s 7→ µ(s) is continuous, at least on
some interval [a, +∞[ with a > 0, there is a direct proof which is quoted below
in Subsection 4.3. In this case, the assumption A > γ2 is no longer required.
The proof in the general case is the subject of Subsection 4.5.
Subharmonic Functions and
their Riesz Measure
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4.3.
Proof of Theorem 4.1 in the case of a continuous repartition function
n
o
Aγ
Proof. Let us suppose that the set s > 0 : µ(s) < Bγ sγ e µ(s) is bounded and
let s0 be one of its majorants, chosen in such a way that s 7→ µ(s) is continuous
on some neighbourhood of [s0 , +∞[.
Aγ
A
− µ(s)
,
Thus µ(s) ≥ Bγ sγ e µ(s) for all s ≥ s0 , that is: log s ≤ γ1 log µ(s)
Bγ
such that:
Z
Z
1
µ(|ζ|)
A
log
−
dµ(ζ)
log |ζ| dµ(ζ) ≤
γ
Bγ
µ(|ζ|)
s0 ≤|ζ|≤s
s0 ≤|ζ|≤s
Z µ(s) 1
t
A
=
log
−
dν(t)
γ
Bγ
t
µ(s0 )
Z µ(s) 1
t
A
=
log
−
dt
γ
Bγ
t
µ(s0 )
h
x iµ(s)/Bγ
µ(s)
= B x log
− A [log t]µ(s0 )
e µ(s0 )/Bγ
µ(s)
µ(s)
=
log
− A log µ(s) + K(s0 ),
γ
Beγ
0)
where K(s0 ) stands for A log µ(s0 ) − µ(sγ 0 ) log µ(s
. It follows from TheoBeγ
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their Riesz Measure
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rem 3.1 that:
Z
µ(s)
µ(s)
log
≤A+
log |ζ| dµ(ζ)
γ
Beγ
|ζ|<s0
µ(s)
µ(s)
+
log
− A log µ(s) + K(s0 ).
γ
Beγ
Finally: A log µ(s) ≤ A + K(s0 ) + µ(s0 ) log s0 for all s ≥ s0 . When s tends
to +∞, a contradiction arises.
4.4.
Splitting measure µ
Now, in order to prove Theorem 4.1 in the general case, we will introduce some
notations which will also be useful in proving Theorem 4.3 (where N ≥ 3).
That is why these notations are already given in RN for any N ∈ N, N ≥ 2.
It is still assumed that lims→+∞ µ(s) = +∞. Let (sn )n be the non–decreasing
sequence defined by: sn = inf{s > 0 : µ(s) ≥ n}. As the function s 7→ µ(s)
is right–continuous, we have µ(sn ) ≥ n for all n ∈ N. If this function is
continuous at some point sn , then µ(sn ) = n.
If sn < sn+1 , then µ(sn ) < n + 1. There are infinitely many integers n such
that sn < sn+1 because the measure dµ is finite on compact subsets of RN (see
[1, p. 81]).
R
For any s > 0, let µ− (s) = |ζ|<s dµ(ζ). The discontinuity points of s 7→
µ(s) are thus characterized by µ(s) > µ− (s). For every n ∈ N, let cn = 0
n )−n
if the function s 7→ µ(s) is continuous at point sn , and cn = µ(sµ(s
if
−
n )−µ (sn )
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−
(sn )
this function is discontinuous at sn . Note that 1 − cn = µ(sn−µ
in case of
−
n )−µ (sn )
discontinuity at sn .
For all 0 < t < s, let It and It,s be defined in RN by:
(
(
It (ζ) = 1 if |ζ| = t
It,s (ζ) = 1 if t < |ζ| < s
0 otherwise
0 otherwise
Let us write µ = µ1 + µ2 + · · · + µn + . . . , where measures µk are defined
such that
Z
Z
dµk (ζ) =
dµk (ζ) = 1
RN
sk−1 ≤|ζ|≤sk
Subharmonic Functions and
their Riesz Measure
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in the following way:
dµk = ( ck−1 Isk−1 + Isk−1 ,sk + (1 − ck ) Isk ) dµ
dµk =
1
Is dµ
µ(sk ) − µ− (sk ) k
if sk−1 < sk
if sk−1 = sk .
Remark 4.2. If sk−1 < sk = sk+1 = · · · = sk+l < sk+l+1 , then µ− (sk ) ≤ k <
k + l ≤ µ(sk ) and it is easy to check that
k+l
X
1
(1 − ck ) Isk +
Is + ck+l Isk+l = Isk .
µ(sj ) − µ− (sj ) j
j=k+1
P
In addition, notice that nk=1 µk (s) = min[n, µ(s)] and that, for any integrable
function h ≥ 0:
Z
n Z
X
h(ζ) dµ ≥
h(ζ) dµk
|ζ|≤sn
k=1
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Z
h(ζ) dµ ≤
|ζ|≤sn
4.5.
n+1 Z
X
h(ζ) dµk
if sn < sn+1
k=1
A reformulation of Theorem 4.1
Proposition 4.2. If N = 2 and A >
many n ∈ N∗ .
2
,
γ
then n < Bγ(sn )γ e
Aγ
n
for infinitely
Aγ
Proof. Suppose that there exists some integer m ∈ N∗ such that n ≥ Bγ(sn )γ e n
for each n ≥ m. It may be assumed that sm > sm−1 ≥ 1. For any n ≥ m satisfying sn < sn+1 , we have:
Z
log |ζ| dµ(ζ) ≤
sm ≤|ζ|≤sn
≤
n+1 Z
X
k=m
n+1
X
log |ζ| dµk (ζ)
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log sk
k=m
n+1 X
1
k
A
≤
log
−
γ
Bγ
k
k=m
Z n+2 t
A
1
≤
log
−
dt
γ
Bγ
t
m
n+2
n+2
=
log
− A log(n + 2) + Km
γ
Beγ
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with a constant Km independent from n. Since µ(sn ) ≥ n, Theorem 3.1 leads
to:
n
n
n+2
n+2
log
≤ A+(log sm )µ(sm )+
log
−A log(n+2)+Km
γ
Beγ
γ
Beγ
hence
2
n+2
n
2
− log(Beγ)+Km +(log sm )µ(sm )
log(n+2) ≤ A+ log
A−
γ
γ
γ
n
{z
}
|
≤ γ2
The contradiction stems from the fact that there exists infinitely many n > m
with sn < sn+1 .
Proof of Theorem 4.1 in the general case. Obviously, function s 7→ Bγ sγ is
Aγ
increasing. Thus, for any n such that n e− n < Bγ(sn )γ , there exists an open
Aγ
non–empty interval Jn (with upper bound sn ) such that n e− n < Bγ sγ <
Aγ
Aγ
Bγ(sn )γ ∀s ∈ Jn . Moreover µ(s) e− µ(s) < n e− n ∀s ∈ Jn (because µ(s) < n
for every s < sn ). Hence Theorem 4.1.
4.6.
The case N ≥ 3
Theorem 4.3. When N ≥ 3, the set of those s > 0 such that
−2
! γ+N
N −2
Bγ γ+N −2
A
(4.1)
µ(s) <
s
1+
γ
N −2
D · [µ(s)] γ+N −2
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their Riesz Measure
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is unbounded.
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Inequalities (4.1) and (4.2) are equivalent, with
1
A
γ
D
(4.2)
<
+
sN −2
γ + N − 2 µ(s) µ(s)α
and α =
N −2
γ+N −2
as in Section 3.3. Indeed, (4.2) may be rewritten
α
µ(s) < s
N −2
γD
γ+N −2
A
1+
D[µ(s)]1−α
.
α
γD
Now γ+N
= NBγ
so that formula (4.1) arises.
−2
−2
To prove Theorem 4.3, we can still assume lims→+∞ µ(s) = +∞. The case
where function s 7→ µ(s) is continuous (at least on some interval [a, +∞[ with
a > 0) is proved in Subsection 4.7 and the general case is proved in Subsection
4.8.
4.7.
Proof of Theorem 4.3 in the case of a continuous repartition function
Proof. Let us assume that there exists some s0 > 0 such that s 7→ µ(s) is
continuous on some neighbourhood of [s0 , +∞[ and that
γ
A
D
1
≥
+
sN −2
γ + N − 2 µ(s)
µ(s)α
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for all s ≥ s0 . It follows that:
Z
Z
dµ(ζ)
dµ(ζ)
≥
N
−2
N −2
|ζ|≤s |ζ|
s0 ≤|ζ|≤s |ζ|
Z
γ
A
D
≥
dµ(ζ)
+
γ + N − 2 s0 ≤|ζ|≤s µ(|ζ|) µ(|ζ|)α
Z µ(s) γ
A D
=
+ α dν(t)
γ + N − 2 µ(s0 ) t
t
Z µ(s) γ
A D
=
+ α dt
γ + N − 2 µ(s0 ) t
t
µ(s)
γ
D 1−α
=
A log t +
t
γ+N −2
1−α
µ(s0 )
=
Aγ log µ(s)
+ D µ(s)1−α − K 0 (s0 ),
γ+N −2
with
K 0 (s0 ) =
The majoration of
R
Aγ
log µ(s0 ) + D µ(s0 )1−α .
γ+N −2
1
|ζ|≤s |ζ|N −2
γ
γ+N −2
dµ(ζ) (Theorem 3.4) leads, after cancellation of
log µ(s)
D µ(s)1−α = D µ(s)
, to: Aγγ+N
≤ A + K 0 (s0 ) for any s ≥ s0 . A
−2
contradiction arises as s → +∞.
Subharmonic Functions and
their Riesz Measure
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4.8.
A reformulation of Theorem 4.3
(4.3)
1
−2
sN
n
N −2
,
γ+N −2
infinitely many n ∈ N∗ satisfy:
γ
A
D
<
+ α .
γ+N −2 n
n
Proposition 4.4. With N ≥ 3 and α =
Proof. Suppose that there exists some m ∈ N such that sN1−2 ≥
n
∀n > m. It then follows for all n > m:
Z
Z
n
X
1
1
dµ(ζ) ≥
dµk (ζ)
N
−2
N
−2
|ζ|
sm ≤|ζ|≤sn |ζ|
k=m+1
≥
n
X
γ
γ+N −2
A
n
+
D
nα
Subharmonic Functions and
their Riesz Measure
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1
sN −2
k=m+1 k
n
X
A
D
γ
≥
+ α
γ + N − 2 k=m+1 k
k
Z n+1
γ
A D
≥
+ α dt
γ + N − 2 m+1 t
t
γA log(n + 1)
0
=
+ D(n + 1)1−α − Km
γ+N −2
0
where the constant Km
does not depend on n. For those n > m such that
sn < sn+1 we have µ(sn ) < n + 1 and Theorem 3.4 provides us with:
Z
Z
1
1
dµ(ζ) ≤
dµ(ζ) ≤ A + D(n + 1)1−α
N
−2
N
−2
|ζ|
|ζ|
sm ≤|ζ|≤sn
|ζ|≤sn
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hence
γA log(n+1)
γ+N −2
0
≤ A + Km
. A contradiction arises as n → +∞.
Proof of Theorem 4.3 in the general case. Since the function s 7→ sN1−2 is decreasing, for each n ∈ N∗ satisfying (4.3) there exists an open interval Jn 6= ∅
(with right bound sn ) where
1
γ
A
D
1
< N −2 <
+
(∀s ∈ Jn ).
−2
sN
s
γ + N − 2 n nα
n
A
D
Now, µ(s) < n for each s < sn , so that An + nDα < µ(s)
+ µ(s)
α . Hence
1
A
D
< γ+Nγ −2 µ(s)
+ µ(s)
∀s ∈ Jn and Theorem 4.3 follows.
α
sN −2
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their Riesz Measure
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5.
Sum of Two Riesz Measures
Lemma 5.1. Given γ > 0, B > 0 and ε ∈ ]0, 1[, let uε be defined in RN by :
uε (x) = max{0, ϕε (|x|)}
∀x ∈ RN
with ϕε (r) = B rγ −B εγ ∀r ≥ 0. Then uε ∈ SH(γ, B). Let µε denote its Riesz
sγ+N −2 + kε ∀s ≥ 1, where τN = max(1, N − 2)
measure, then: µε (s) = Bγ
τN
and kε is a constant depending only on B, γ, N and ε.
Proof. Subharmonicity of uε = max(u1 , u2 ) will follow (see [1, p. 41]) from
the subharmonicity of both functions u1 and u2 defined in RN by u1 (x) =
ϕε (|x|) and u2 (x) ≡ 0: it is easy to verify that ∆u1 (x) = ϕ00ε (r) + N r−1 ϕ0ε (r) =
Bγrγ−2 (γ + N − 2) ≥ 0 (see [1, p. 26]). Obviously, uε has a growth of the kind
(1.1), uε (0) = 0 and uε is harmonic in the neighbourhood {x ∈ RN : |x| < ε}
of the origin.
Let θN = (N − 2)σN when N ≥ 3 and θ2 = 2π (see [2, p. 43]), since
dµε = θ1N ∆uε dx = θ1N ∆uε rN −1 dr dσ, it is possible for all s ≥ 1 to compute
Z
µε (s) = µε (1) +
1
s
s
σN
1
Bγ(γ + N − 2)rγ+N −3 dr = µε (1) + Bγ rγ+N −2 1
θN
τN
Proposition 5.2. Given γ > 0, B > 0 and 0 < B 0 < 2B, let µ1 and µ2
be the Riesz measures of two functions, respectively u1 and u2 , belonging to
SH(γ, B). Then µ1 + µ2 is not necessarily the Riesz measure associated with
a function of SH(γ, B 0 ).
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their Riesz Measure
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Proof. Given ε1 and ε2 ∈]0, 1[, let uε1 and uε2 ∈ SH(γ, B) be defined as in the
previous lemma and µ = µε1 + µε2 be the sum of their Riesz measures. Thus
µ(s)
2Bγ
sγ+N −2 + kε1 + kε2 ∀s ≥ 1. Note that lims→+∞ sγ+N
.
µ(s) = 2Bγ
−2 = τ
τN
N
Suppose that µ is the Riesz measure of some function u ∈ SH(γ, B 0 ) with
an estimate such as: u(x) ≤ A + B 0 |x|γ (∀x ∈ RN ) for some constant A ∈ R.
µ(s)
B0γ
In Theorems 4.1 and 4.3, one asserts that lim inf s→+∞ sγ+N
, which
−2 ≤
τN
0
leads to 2B ≤ B , hence a contradiction.
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6.
6.1.
Subharmonic Functions Subject to Conditions
of L1 Type
A weighted integral condition for subharmonic functions.
Theorem 6.1. Given N ∈ N (N ≥ 2) and a positive non–increasing C 1
function ϕ on [0, +∞[ such that lims→+∞ (log s)ϕ(s) = 0 (when N = 2) or
N
lims→+∞ s 2 −1 ϕ(s) = 0 (when N ≥ 3), let u be a subharmonic function in RN ,
harmonic in some neighbourhood of the origin with u(0) = 0, such that:
Z
u+ (x) [−ϕ0 (|x|2 )] dx < +∞
Subharmonic Functions and
their Riesz Measure
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RN
where the subharmonic function u+ is defined by u+ (x) = max(u(x), 0) ∀x ∈
RN . Then the Riesz measure µ of u verifies:
Z
ϕ(|ζ|2 + 1)
dµ(ζ) < +∞.
|ζ|2
RN
Example 1. With N ≥ 2, β > 0 and ϕ defined by ϕ(s) = e−βs ∀s > 0,
obviously
N
lim (log s)ϕ(s) = lim s 2 −1 ϕ(s) = 0.
s→+∞
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s→+∞
If a subharmonic function u inR RN (harmonic in some neighbourhood of the
2
origin, with u(0) = 0) satisfies RN u+ (x) e−β|x| dx < +∞ then its Riesz meaR
−β|ζ|2
sure µ verifies RN e |ζ|2 dµ(ζ) < +∞. One thus encounters a result of [4, p.
88] for holomorphic functions in C.
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6.2.
Proof of Theorem 6.1 in the case N = 2
Proof. Abiding by Jensen’s formula (Subsection 3.1) and by Lemma 2.3:
Z
Z 2π
r
1
log
dµ(ζ) ≤
u+ (r eiθ ) dθ
∀r > 0.
|ζ|
2π
|ζ|≤r
0
Since −ϕ0 (r2 ) ≥ 0, it follows that:
Z +∞ Z
r
log
dµ(ζ) [−ϕ0 (r2 )] r dr < +∞.
|ζ|
0
|ζ|≤r
Fubini’s theorem transforms the above integral into:
Z Z +∞
r
0 2
[−ϕ (r )] r dr dµ(ζ).
log
|ζ|
R2
|ζ|
|
{z
}
:=I(ζ)≥0
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Now,
I(ζ) =
1
4
Z
+∞
log
|ζ|2
s
[−ϕ0 (s)] ds
|ζ|2
R +∞
for any ζ ∈ R2 and an integration by parts leads to: 4 I(ζ) = |ζ|2 ϕ(s)
ds since
s
lims→+∞ (log s) ϕ(s) = 0 and lims→+∞ ϕ(s) = 0 as well. The positive funcR +∞
tion f : s 7→ ϕ(s)
decreases for s > 0 so that b f (s) ds ≥ f (b + 1) for all
s
2 +1)
1
b > 0, hence: 4 I(ζ) ≥ ϕ(|ζ|
for all ζ ∈ R2 . If |ζ| ≥ 1, then |ζ|21+1 ≥ 2 |ζ|
2
|ζ|2 +1
and 8 I(ζ) ≥
Subharmonic Functions and
their Riesz Measure
ϕ(|ζ|2 +1)
|ζ|2
R
≥ 0. Because of the harmonicity of u in a neighbour-
ϕ(|ζ|2 +1)
|ζ|2
|ζ|<1
hood of the origin,
R
I(ζ) dµ(ζ) < +∞.
|ζ|≥1
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dµ(ζ) < +∞. The conclusion follows from
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Proof of Theorem 6.1 in the case N ≥ 3
6.3.
Proof. Jensen–Privalov formula together with Lemma 2.4 lead to:
Z
Z
1
1
1
−
dµ(ζ) ≤
u(rx) dσx
∀r > 0.
|ζ|N −2 rN −2
σN SN
|ζ|≤r
Hence:
Z +∞ Z
0
|ζ|≤r
1
|ζ|N −2
−
1
dµ(ζ) [−ϕ0 (r2 )] rN −1 dr < +∞.
rN −2
Taking Fubini’s theorem into account, this integral becomes:
Z +∞ Z
1
1
0 2
N −1
−
[−ϕ (r )] r
dr dµ(ζ).
|ζ|N −2 rN −2
IRN
|ζ|
|
{z
}
:=J(ζ)
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Now, for any ζ ∈ RN :
0 ≤ J(ζ)
Z +∞ N −2
r
=
− 1 [−ϕ0 (r2 )] r dr
N −2
|ζ|
|ζ|
!
Z
N
1 +∞ s 2 −1
=
− 1 [−ϕ0 (s)] ds.
N
−2
2 |ζ|2
|ζ|
N
Since lims→+∞ s 2 −1 − |ζ|N −2 ϕ(s) = 0, an integration by parts leads to:
N −2
2 J(ζ) =
2
Subharmonic Functions and
their Riesz Measure
Z
+∞
|ζ|2
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N
s 2 −2
ϕ(s) ds.
|ζ|N −2
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N
Obviously, s 2 −2 ≥ |ζ|N −4 for all s ≥ |ζ|2 , so that:
Z +∞
4
1
ϕ(|ζ|2 + 1)
J(ζ) ≥ 2
ϕ(s) ds ≥
≥ 0.
N −2
|ζ| |ζ|2
|ζ|2
Propositions 6.2 and 6.3 will be proved by using the same method.
Proposition 6.2. Let ϕ be a positive C 1 non–increasing function on [0, +∞[
such that
limr→+∞ r ϕ(r) log r = 0. If a subharmonic function u in R2 (harmonic in
some neighbourhood of the origin with u(0) = 0) verifies:
Z
u+ (x) [−ϕ0 (|x|)] dx < +∞
Subharmonic Functions and
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R2
R
then its Riesz measure µ satisfies: R2 ϕ(|ζ| + 1) dµ(ζ) < +∞ and
Z
ϕ(|ζ|α + 1) log |ζ| dµ(ζ) < +∞
|ζ|≥1
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holds for each α > 1.
Proof. As in Section 6.2:
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R
R2
I(ζ) dµ(ζ) < +∞, here with
Z
+∞
I(ζ) =
r log
|ζ|
r
[−ϕ0 (r)] dr
|ζ|
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R +∞
er
which turns into I(ζ) = |ζ| ϕ(r) log |ζ|
dr after an integration by parts which
uses
limr→+∞ r ϕ(r) log r = 0 (this garantees that limr→+∞ r ϕ(r) = 0 as well).
er
Since ϕ is non–increasing and log |ζ|
≥ 1 for each r ≥ |ζ|, it follows that
2
I(ζ) ≥ ϕ(|ζ| + 1) ∀ζ ∈ R .
Given α > 1, obviously |ζ|α ≥ |ζ| as soon as |ζ| ≥ 1, so that
Z +∞
er
I(ζ) ≥
ϕ(r) log
dr
|ζ|
|ζ|α
Z +∞
ϕ(r) log |ζ| dr ≥ (α − 1)ϕ(|ζ|α + 1) log |ζ|
≥ (α − 1)
Subharmonic Functions and
their Riesz Measure
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|ζ|α
≥ 0.
The conclusion proceeds from
R
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I(ζ) dµ(ζ) < +∞.
|ζ|≥1
Proposition 6.3. Given N ∈ N, N ≥ 3, let ϕ be a positive non–increasing
C 1 function in [0, +∞[ such that limr→+∞ rN −1 ϕ(r) = 0. If a subharmonic
function u in RN (harmonic in some neighbourhood of the origin with u(0) = 0)
verifies:
Z
u+ (x) [−ϕ0 (|x|)] dx < +∞
RN
then its Riesz measure µ satisfies
Z
ϕ(|ζ|α + 1) |ζ|(α−1)(N −2) dµ(ζ) < +∞
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RN
J. Ineq. Pure and Appl. Math. 2(2) Art. 16, 2001
for any α ≥ 1.
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R
Remark 6.1. When α = 1, we encounter RN ϕ(|ζ| + 1) dµ(ζ) < +∞ again.
R
Proof. As in Section 6.3: RN J(ζ) dµ(ζ) < +∞, here with
Z +∞ N −1
r
J(ζ) =
− r [−ϕ0 (r)] dr
N −2
|ζ|
|ζ|
Z +∞ rN −2
=
(N − 1) N −2 − 1 ϕ(r) dr
|ζ|
|ζ|
after an integration by parts. Obviously,
rN −2
|ζ|N −2
≥ 1 for every r ≥ |ζ|, so that:
rN −2
rN −2
(N − 1) N −2 − 1 ≥ (N − 2) N −2
|ζ|
|ζ|
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and
Z
+∞
J(ζ) ≥ (N − 2)
|ζ|
rN −2
ϕ(r) dr
|ζ|N −2
∀ζ ∈ RN .
α
If |ζ| ≥ 1, then |ζ| ≥ |ζ| since α ≥ 1, hence
Z +∞ N −2
r
J(ζ) ≥ (N − 2)
ϕ(r) dr
N −2
|ζ|α |ζ|
Z +∞
(α−1)(N −2)
≥ (N − 2) |ζ|
ϕ(r) dr
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|ζ|α
≥ (N − 2) |ζ|(α−1)(N −2) ϕ(|ζ|α + 1).
Page 32 of 33
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References
[1] W.K. HAYMAN AND P.B. KENNEDY, Subharmonic functions. Vol. I, London Mathematical Society Monographs, No. 9. Academic Press, London–
New York, 1976.
[2] L.I. RONKIN, Functions of completely regular growth, Mathematics and
its Applications (Soviet Series), 81. Kluwer Academic Publishers’ Group,
Dordrecht, 1992.
[3] D.W. STROOCK, A Concise Introduction to the Theory of Integration,
Third edition, Birkhäuser Boston, Inc., Boston, MA, 1999.
[4] K.H. ZHU, Zeros of functions in Fock spaces, Complex Variables, Theory
Appl., 21(1–2) (1993), 87–98.
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J. Ineq. Pure and Appl. Math. 2(2) Art. 16, 2001
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