Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 2, Issue 2, Article 16, 2001
SUBHARMONIC FUNCTIONS AND THEIR RIESZ MEASURE
RAPHAELE 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 , FRANCE
supper@math.u-strasbg.fr
Received 30 August, 2000; accepted 22 January, 2001
Communicated by A. Fiorenza
N
A BSTRACT. For
R subharmonic functions u in R , of Riesz measure µ, the growth of the function
s 7→ µ(s) = |ζ|≤s dµ(ζ) (s ≥ 0) is described and compared with the growth of u. It is
R
also shown
that,
if
u+ (x) [−ϕ0 (|x|2 )] dx < +∞ for some decreasing C 1 function ϕ ≥ 0,
RN
R
1
2
then RN |ζ|2 ϕ(|ζ| + 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).
Key words and phrases: subharmonic functions, order of growth, Riesz measure.
2000 Mathematics Subject Classification. 31A05, 31B05, 26D15, 28A75.
1. I NTRODUCTION
N
Let µ be the Riesz measure of some subharmonic
R function u in R (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 +∞
2−N −c
1−N −c
inf c :
s
dµ(s) = inf c :
s
µ(s) ds
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
∀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.
ISSN (electronic): 1443-5756
c 2001 Victoria University. All rights reserved.
030-00
2
R APHAELE S UPPER
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 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→ µ(s). For instance, when
Aγ
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 1.1. When N ≥ 3, throughout the paper we set C(γ, N ) =
and
γ
N −2
γ+N −2
D(B, γ, N ) = γ+Nγ −2 NBγ
, sometimes written merely D for brevity.
−2
Note that
γ
γ
C(γ, N ) =
N −2
N −2
γ+N −2
γ
−2
γ+N
N −2
γ+N −2
=
N −2
N −2
1+
γ
Nγ−2
≤e
γ+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:
Bγ
µ(s) ≤
C(γ, N ) sγ+N −2
N −2
1+
−2
! γ+N
N −2
A
γ
D · [µ(s)] γ+N −2
(see Theorem 3.4 and Corollary 3.5).
µ(s)
Bγ
It points out that lim sups→+∞ sγ+N
−2 is not greater than Beγ (when N = 2) or N −2 C(γ, N )
µ(s)
Bγ
(when N ≥ 3). Moreover, lim inf s→+∞ sγ+N
−2 does not exceed Bγ (if N = 2) or N −2 (if
N ≥ 3). This will follow from Theorems 4.2 and 4.5 which assert that the sets:
o
n
Aγ
γ µ(s)
s : µ(s) < Bγ s e
and


Bγ γ+N −2
s : µ(s) <
s

N −2
1+

−2
! γ+N
N −2 
A
γ
D · [µ(s)] γ+N −2

are unbounded in the cases when N = 2 and N ≥ 3 respectively.
The last section studies subharmonic functions u in RN (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 example in Theorem 6.1:
R
+
u (x) [−ϕ0 (|x|2 )] dx < +∞ (see Section 6.1 for more details on the decreasing function
RN
R
2 +1)
ϕ). The Riesz measure µ of u is then proved to verify: RN ϕ(|ζ|
dµ(ζ) < +∞. Propositions
|ζ|2
1
6.2 and 6.3 provide similar results under different L conditions.
J. Inequal. Pure and Appl. Math., 2(2) Art. 16, 2001
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S UBHARMONIC F UNCTIONS AND THEIR R IESZ M EASURE
3
2. S OME P RELIMINARIES
Lemma 2.1. If N = 2, then
Z
Z
r
log
dµ(ζ) ≤
|ζ|
|ζ|≤s
log
|ζ|≤r
r
dµ(ζ)
|ζ|
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) =
|ζ|≤s
Z
hr (ζ) dµ(ζ) +
|ζ|≤r
r<|ζ|≤s
|
hr (ζ) dµ(ζ) ≤
{z
}
hr (ζ) dµ(ζ).
|ζ|≤r
≤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|≤r
|ζ|≤s
s<|ζ|≤r
|ζ|≤s
|
{z
}
≥0
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
1
|ζ|N −2
Proof. As in the previous proof, with hr (ζ) =
−
1
r N −2
r
instead of log |ζ|
.
Lemma 2.3. If N = 2, then:
Z
0
r
µ(t)
dt =
t
Z
log
|ζ|≤r
r
dµ(ζ),
|ζ|
for any r > 0.
Proof. It follows from Fubini’s theorem that:
Z r Z r
Z r Z
Z
Z
µ(t)
1
dt
r
dt =
dµ(ζ) dt =
dµ(ζ) =
log
dµ(ζ).
t
|ζ|
0
0 t
|ζ|≤t
|ζ|≤r
|ζ| t
|ζ|≤r
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
J. Inequal. Pure and Appl. Math., 2(2) Art. 16, 2001
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4
R APHAELE S UPPER
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
3. E STIMATIONS OF THE R IESZ M EASURE
3.1. 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
1
µ(t)
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
2 π N/2
with SN the unit sphere in RN , dσ the area element on SN and σN = SN dσ = 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
so that
Z
1
µ(s)
dµ(ζ) ≤ A + B rγ − µ(s) log r = A +
log
|ζ|
γ
|ζ|≤s
Bγ γ
r − log rγ
µ(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
1
µ(s)
µ(s)
µ(s)
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.
J. Inequal. Pure and Appl. Math., 2(2) Art. 16, 2001
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S UBHARMONIC F UNCTIONS AND THEIR R IESZ M EASURE
5
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
The previous integral being ≤ log sγ , 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
R
Proof. Jensen’s inequality applies to (3.1) since |ζ|≤s dµ(ζ)
= 1, hence:
µ(s)
Z
µ(s)
Aγ
γ dµ(ζ)
≤ exp
· exp
log(|ζ| )
Beγ
µ(s)
µ(s)
|ζ|≤s
Z
Aγ
dµ(ζ)
≤ exp
|ζ|γ
.
µ(s)
µ(s)
|ζ|≤s
3.3. The case N ≥ 3.
Theorem 3.4. When N ≥ 3, the following estimation is valid for each s > 0:
Z
γ
1
γ+N −2 .
dµ(ζ)
≤
A
+
D
[µ(s)]
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
Z
|ζ|≤s
1
|ζ|N −2
dµ(ζ) ≤ A + B rγ +
µ(s)
rN −2
whose minimum (with s constant) is attained when Bγ rγ = (N − 2) rµ(s)
N −2 . In other words, this
N −2
γ+N
−2
Bγ 1
1
minimum is A + Nγ−2 + 1 rµ(s)
. Finally:
N −2 with r N −2 =
N −2 µ(s)
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:
−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
Proof. Let α =
1
sN −2
N −2
.
γ+N −2
According to Theorem 3.4, for any s > ε we have
1
1
A
D
D
A µ(s)α
≤
dµ(ζ) ≤
+
=
1+
.
µ(s) |ζ|≤s |ζ|N −2
µ(s)
µ(s)α
µ(s)α
D µ(s)
Z
Hence
α
[µ(s)] ≤ D s
J. Inequal. Pure and Appl. Math., 2(2) Art. 16, 2001
N −2
A
1
1+
D [µ(s)]1−α
.
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6
R APHAELE S UPPER
Now, it is obvious that 1 − α =
γ
γ+N −2
and D1/α =
Bγ
C(γ, N ).
N −2
N −2
Corollary 3.6. With N ≥ 3 and α = γ+N
, the following holds for each s > 0:
−2
Z
µ(s)α
A
α
µ(s) log
−
µ(s) ≤ (N − 2)
log |ζ| dµ(ζ)
D
D
|ζ|≤s
Proof. It follows from Jensen’s inequality that:
Z
Z
1
dµ(ζ)
1
dµ(ζ)
exp
log N −2
≤
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|
A
D
≤
+
,
µ(s)
µ(s)α
so that:
Z
dµ(ζ)
−(N − 2)
log |ζ|
≤ log
µ(s)
|ζ|≤s
A
D
+
µ(s)
µ(s)α
≤ log
D
µ(s)α
+
A µ(s)α
.
D µ(s)
4. G ROWTH
OF THE
R EPARTITION F UNCTION
4.1. 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 ν-integrablef )
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).
Aγ
Theorem 4.2. If N = 2 and A > γ2 , then the set of those s > 0 which satisfy µ(s) < Bγ sγ e µ(s)
is unbounded.
A proof is required only in the case where lims→+∞ µ(s) = +∞ (otherwise, Theorem 4.2
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.
4.3. Proof of Theorem 4.2 in the case of a continuous repartition function.
n
o
Aγ
γ µ(s)
is bounded and let s0 be one
Proof. Let us suppose that the set s > 0 : µ(s) < Bγ s e
of its majorants, chosen in such a way that s 7→ µ(s) is continuous on some neighbourhood of
[s0 , +∞[.
J. Inequal. Pure and Appl. Math., 2(2) Art. 16, 2001
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S UBHARMONIC F UNCTIONS AND THEIR R IESZ M EASURE
7
Aγ
A
− µ(s)
, such that:
Thus µ(s) ≥ Bγ sγ e µ(s) for all s ≥ s0 , that is: log s ≤ γ1 log µ(s)
Bγ
Z
Z
1
µ(|ζ|)
A
log |ζ| dµ(ζ) ≤
log
−
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) t
A
1
=
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 Theorem 3.1 that:
Beγ
Z
µ(s)
µ(s)
µ(s)
µ(s)
log
≤A+
log |ζ| dµ(ζ) +
log
− A log µ(s) + K(s0 ).
γ
Beγ
γ
Beγ
|ζ|<s0
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 4.2 in the general case, we will
introduce some notations which will also be useful in proving Theorem 4.5 (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 if the function s 7→ µ(s) is
n )−n
if this function is discontinuous at sn . Note that
continuous at point sn , and cn = µ(sµ(s
−
n )−µ (sn )
−
(sn )
1 − cn = µ(sn−µ
in case of discontinuity at sn .
−
n )−µ (sn )
For all 0 < t < s, let It and It,s be defined in RN by:
(
(
1 if t < |ζ| < s
It (ζ) = 1 if |ζ| = t
It,s (ζ) =
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
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
J. Inequal. Pure and Appl. Math., 2(2) Art. 16, 2001
if sk−1 < sk
if sk−1 = sk .
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8
R APHAELE S UPPER
Remark 4.3. 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
(1 − ck ) Isk +
In addition, notice that
Pn
k=1
k+l
X
1
Isj + ck+l Isk+l = Isk .
−
µ(s
j ) − µ (sj )
j=k+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
Z
h(ζ) dµ ≤
|ζ|≤sn
k=1
n+1 Z
X
h(ζ) dµk
if sn < sn+1
k=1
4.5. A reformulation of Theorem 4.2.
Aγ
Proposition 4.4. If N = 2 and A > γ2 , then n < Bγ(sn )γ e n for infinitely many n ∈ N∗ .
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
n+1 Z
X
log |ζ| dµk (ζ)
log |ζ| dµ(ζ) ≤
sm ≤|ζ|≤sn
k=m
≤
n+1
X
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
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
n+2
2
log(n + 2) ≤ A + log
− log(Beγ) + Km + (log sm )µ(sm )
A−
γ
γ
n
γ
|
{z
}
≤ γ2
The contradiction stems from the fact that there exists infinitely many n > m with sn < sn+1 .
Proof of Theorem 4.2 in the general case. Obviously, function s 7→ Bγ sγ is increasing. Thus,
Aγ
for any n such that n e− n < Bγ(sn )γ , there exists an open non–empty interval Jn (with upper
Aγ
Aγ
Aγ
bound sn ) such that n e− n < Bγ sγ < 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 4.2.
J. Inequal. Pure and Appl. Math., 2(2) Art. 16, 2001
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S UBHARMONIC F UNCTIONS AND THEIR R IESZ M EASURE
9
4.6. The case N ≥ 3.
Theorem 4.5. When N ≥ 3, the set of those s > 0 such that
µ(s) <
(4.1)
Bγ γ+N −2
s
N −2
1+
−2
! γ+N
N −2
A
γ
D · [µ(s)] γ+N −2
is unbounded.
Inequalities (4.1) and (4.2) are equivalent, with
γ
A
1
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
A
γD
α
N −2
µ(s) < s
1+
.
γ+N −2
D[µ(s)]1−α
α
γD
Now γ+N
= NBγ
so that formula (4.1) arises.
−2
−2
To prove Theorem 4.5, 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 4.5 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
1
A
D
≥ γ+N −2 µ(s) + µ(s)α for all s ≥ s0 . It follows that:
sN −2
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
D µ(s)
+∞.
γ
γ+N −2
, to:
Aγ
log µ(s0 ) + D µ(s0 )1−α .
γ+N −2
1
dµ(ζ) (Theorem
|ζ|≤s |ζ|N −2
Aγ log µ(s)
≤ A + K 0 (s0 ) for
γ+N −2
R
J. Inequal. Pure and Appl. Math., 2(2) Art. 16, 2001
3.4) leads, after cancellation of D µ(s)1−α =
any s ≥ s0 . A contradiction arises as s →
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10
R APHAELE S UPPER
4.8. A reformulation of Theorem 4.5.
N −2
Proposition 4.6. With N ≥ 3 and α = γ+N
, infinitely many n ∈ N∗ satisfy:
−2
γ
A
D
1
(4.3)
<
+ α .
−2
sN
γ+N −2 n
n
n
Proof. Suppose that there exists some m ∈ N such that sN1−2 ≥ γ+Nγ −2
n
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
A
n
+
D
nα
∀n > m. It
1
sN −2
k=m+1 k
n
X
A
D
γ
≥
+ α
γ + N − 2 k=m+1 k
k
Z n+1 A D
γ
+ 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
hence
γA log(n+1)
γ+N −2
0
≤ A + Km
. A contradiction arises as n → +∞.
Proof of Theorem 4.5 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
1
γ
A
D
< N −2 <
+
(∀s ∈ Jn ).
−2
sN
s
γ + N − 2 n nα
n
γ
D
A
D
1
A
D
A
Now, µ(s) < n for each s < sn , so that n + nα < µ(s) + µ(s)α . Hence sN −2 < γ+N −2 µ(s) + µ(s)α
∀s ∈ Jn and Theorem 4.5 follows.
5. S UM OF T WO R IESZ M EASURES
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 measure, then:
µε (s) = Bγ
sγ+N −2 + kε ∀s ≥ 1, where τN = max(1, N − 2) and kε is a constant depending
τN
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.
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S UBHARMONIC F UNCTIONS AND THEIR R IESZ M EASURE
11
Let θN = (N − 2)σN when N ≥ 3 and θ2 = 2π (see [2, p. 43]), since dµε = θ1N ∆uε dx =
1
∆uε rN −1 dr dσ, it is possible for all s ≥ 1 to compute
θN
Z s
s
σN
1
µε (s) = µε (1) +
Bγ(γ + N − 2)rγ+N −3 dr = µε (1) +
Bγ rγ+N −2 1
τN
1 θN
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 ).
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
τN
µ(s)
2Bγ
∀s ≥ 1. Note that lims→+∞ sγ+N
.
−2 = τ
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. In Theorems 4.2 and 4.5, one
µ(s)
B0γ
asserts that lim inf s→+∞ sγ+N
, which leads to 2B ≤ B 0 , hence a contradiction.
−2 ≤ τ
N
6. S UBHARMONIC F UNCTIONS S UBJECT TO C ONDITIONS OF L1 T YPE
6.1. 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, +∞[
N
such that lims→+∞ (log s)ϕ(s) = 0 (when N = 2) or 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 < +∞
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 6.1. 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→+∞
s→+∞
If a subharmonic function u in RN (harmonic in some neighbourhood of the origin, with u(0) =
R
R
−β|ζ|2
2
0) satisfies RN u+ (x) e−β|x| dx < +∞ then its Riesz measure µ verifies RN e |ζ|2 dµ(ζ) <
+∞. One thus encounters a result of [4, p. 88] for holomorphic functions in C.
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 0
|ζ|≤r
Since −ϕ0 (r2 ) ≥ 0, it follows that:
Z +∞ Z
0
r
dµ(ζ) [−ϕ0 (r2 )] r dr < +∞.
log
|ζ|
|ζ|≤r
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12
R APHAELE S UPPER
Fubini’s theorem transforms the above integral into:
Z Z +∞
r
0 2
log
[−ϕ (r )] r dr dµ(ζ).
|ζ|
R2
|ζ|
|
{z
}
:=I(ζ)≥0
Now,
1
I(ζ) =
4
Z
+∞
s
[−ϕ0 (s)] ds
2
|ζ|
2
|ζ|
R +∞
2
for any ζ ∈ R and an integration by parts leads to: 4 I(ζ) = |ζ|2
log
ϕ(s)
s
ds since
lims→+∞ (log s) ϕ(s) = 0 and lims→+∞ ϕ(s) = 0 as well. The positive function f : s 7→ ϕ(s)
s
R +∞
2 +1)
decreases for s > 0 so that b f (s) ds ≥ f (b+1) for all b > 0, hence: 4 I(ζ) ≥ ϕ(|ζ|
for
all
|ζ|2 +1
2
+1)
and 8 I(ζ) ≥ ϕ(|ζ|
≥ 0. Because of the harmonicity
|ζ|2
R
2
+1)
of u in a neighbourhood of the origin, |ζ|<1 ϕ(|ζ|
dµ(ζ) < +∞. The conclusion follows
|ζ|2
R
from |ζ|≥1 I(ζ) dµ(ζ) < +∞.
ζ ∈ R2 . If |ζ| ≥ 1, then
1
|ζ|2 +1
≥
1
2 |ζ|2
6.3. Proof of Theorem 6.1 in the case N ≥ 3.
Proof. Jensen–Privalov formula together with Lemma 2.4 lead to:
Z
Z
1
1
1
−
dµ(ζ) ≤
u(rx) dσx
|ζ|N −2 rN −2
σN SN
|ζ|≤r
∀r > 0.
Hence:
Z
0
+∞
Z
|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)
Now, for any ζ ∈ RN :
!
Z +∞ N −2
Z +∞
N
−1
2
r
1
s
0 ≤ J(ζ) =
− 1 [−ϕ0 (r2 )] r dr =
− 1 [−ϕ0 (s)] ds.
N −2
N −2
|ζ|
2
|ζ|
2
|ζ|
|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
Z
+∞
|ζ|2
N
s 2 −2
ϕ(s) ds.
|ζ|N −2
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.
J. Inequal. Pure and Appl. Math., 2(2) Art. 16, 2001
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S UBHARMONIC F UNCTIONS AND THEIR R IESZ M EASURE
13
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 < +∞
R2
then its Riesz measure µ satisfies:
Z
R
R2
ϕ(|ζ| + 1) dµ(ζ) < +∞ and
ϕ(|ζ|α + 1) log |ζ| dµ(ζ) < +∞
|ζ|≥1
holds for each α > 1.
Proof. As in Section 6.2:
R
R2
I(ζ) dµ(ζ) < +∞, here with
Z +∞
r
[−ϕ0 (r)] dr
I(ζ) =
r log
|ζ|
|ζ|
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). Since ϕ is non–
er
increasing and log |ζ|
≥ 1 for each r ≥ |ζ|, it follows that I(ζ) ≥ ϕ(|ζ| + 1) ∀ζ ∈ R2 .
Given α > 1, obviously |ζ|α ≥ |ζ| as soon as |ζ| ≥ 1, so that
Z +∞
Z +∞
er
I(ζ) ≥
ϕ(r) log
dr ≥ (α − 1)
ϕ(r) log |ζ| dr ≥ (α − 1)ϕ(|ζ|α + 1) log |ζ| ≥ 0.
|ζ|
α
α
|ζ|
|ζ|
R
The conclusion proceeds from |ζ|≥1 I(ζ) dµ(ζ) < +∞.
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) < +∞
RN
for any α ≥ 1.
R
Remark 6.4. 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
Z +∞ r
rN −2
0
J(ζ) =
− r [−ϕ (r)] dr =
(N − 1) N −2 − 1 ϕ(r) dr
|ζ|N −2
|ζ|
|ζ|
|ζ|
rN −2
|ζ|N −2
after an integration by parts. Obviously,
(N − 1)
rN −2
rN −2
−
1
≥
(N
−
2)
|ζ|N −2
|ζ|N −2
and
Z
+∞
J(ζ) ≥ (N − 2)
|ζ|
J. Inequal. Pure and Appl. Math., 2(2) Art. 16, 2001
≥ 1 for every r ≥ |ζ|, so that:
rN −2
ϕ(r) dr
|ζ|N −2
∀ζ ∈ RN .
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14
R APHAELE S UPPER
If |ζ| ≥ 1, then |ζ|α ≥ |ζ| since α ≥ 1, hence
Z
J(ζ) ≥ (N − 2)
+∞
rN −2
ϕ(r) dr
N −2
|ζ|α |ζ|
Z +∞
(α−1)(N −2)
≥ (N − 2) |ζ|
ϕ(r) dr
|ζ|α
(α−1)(N −2)
≥ (N − 2) |ζ|
ϕ(|ζ|α + 1).
R EFERENCES
[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.
J. Inequal. Pure and Appl. Math., 2(2) Art. 16, 2001
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