A Property of Azarin's Limit Set of Subharmonic Functions
Let v(z) be a subharmonic function of order ρ > 0, and Fr(v) be the limit set in the sense of Azarin. Let z be fixed and I(z) = {u(z) : u is in Fr(v)}. We prove that I(z) is either a closed interval or a semiclosed interval which does not contain its infimum.
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України
2008
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| Цитувати: | A Property of Azarin's Limit Set of Subharmonic Functions / A. Chouigui, A.F. Grishin // Журнал математической физики, анализа, геометрии. — 2008. — Т. 4, № 3. — С. 346-357. — Бібліогр.: 8 назв. — англ. |
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irk-123456789-1065112016-09-30T03:02:57Z A Property of Azarin's Limit Set of Subharmonic Functions Chouigui, A. Grishin, A.F. Let v(z) be a subharmonic function of order ρ > 0, and Fr(v) be the limit set in the sense of Azarin. Let z be fixed and I(z) = {u(z) : u is in Fr(v)}. We prove that I(z) is either a closed interval or a semiclosed interval which does not contain its infimum. 2008 Article A Property of Azarin's Limit Set of Subharmonic Functions / A. Chouigui, A.F. Grishin // Журнал математической физики, анализа, геометрии. — 2008. — Т. 4, № 3. — С. 346-357. — Бібліогр.: 8 назв. — англ. 1812-9471 http://dspace.nbuv.gov.ua/handle/123456789/106511 en Журнал математической физики, анализа, геометрии Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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Digital Library of Periodicals of National Academy of Sciences of Ukraine |
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DSpace DC |
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English |
| description |
Let v(z) be a subharmonic function of order ρ > 0, and Fr(v) be the limit set in the sense of Azarin. Let z be fixed and I(z) = {u(z) : u is in Fr(v)}. We prove that I(z) is either a closed interval or a semiclosed interval which does not contain its infimum. |
| format |
Article |
| author |
Chouigui, A. Grishin, A.F. |
| spellingShingle |
Chouigui, A. Grishin, A.F. A Property of Azarin's Limit Set of Subharmonic Functions Журнал математической физики, анализа, геометрии |
| author_facet |
Chouigui, A. Grishin, A.F. |
| author_sort |
Chouigui, A. |
| title |
A Property of Azarin's Limit Set of Subharmonic Functions |
| title_short |
A Property of Azarin's Limit Set of Subharmonic Functions |
| title_full |
A Property of Azarin's Limit Set of Subharmonic Functions |
| title_fullStr |
A Property of Azarin's Limit Set of Subharmonic Functions |
| title_full_unstemmed |
A Property of Azarin's Limit Set of Subharmonic Functions |
| title_sort |
property of azarin's limit set of subharmonic functions |
| publisher |
Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
| publishDate |
2008 |
| url |
http://dspace.nbuv.gov.ua/handle/123456789/106511 |
| citation_txt |
A Property of Azarin's Limit Set of Subharmonic Functions / A. Chouigui, A.F. Grishin // Журнал математической физики, анализа, геометрии. — 2008. — Т. 4, № 3. — С. 346-357. — Бібліогр.: 8 назв. — англ. |
| series |
Журнал математической физики, анализа, геометрии |
| work_keys_str_mv |
AT chouiguia apropertyofazarinslimitsetofsubharmonicfunctions AT grishinaf apropertyofazarinslimitsetofsubharmonicfunctions AT chouiguia propertyofazarinslimitsetofsubharmonicfunctions AT grishinaf propertyofazarinslimitsetofsubharmonicfunctions |
| first_indexed |
2025-07-07T18:35:19Z |
| last_indexed |
2025-07-07T18:35:19Z |
| _version_ |
1837014262345629696 |
| fulltext |
Journal of Mathematical Physics, Analysis, Geometry
2008, vol. 4, No. 3, pp. 346�357
A Property of Azarin's Limit Set of Subharmonic
Functions
A. Chouigui and A.F. Grishin
Department of Mechanics and Mathematics V.N. Karazin Kharkiv National University
4 Svobody Sq., Kharkiv, 61077, Ukraine
E-mail:grishin@univer.kharkov.ua
a_chouigui@yahoo.fr
Received November 22, 2007
Let v(z) be a subharmonic function of order � > 0, and Fr(v) be the
limit set in the sense of Azarin. Let z be �xed and I(z) = fu(z) : u 2 Fr(v)g.
We prove that I(z) is either a closed interval or a semiclosed interval which
does not contain its in�mum.
Key words: subharmonic function, limit set of Azarin, indicator of growth.
Mathematics Subject Classi�cation 2000: 31A05.
The following de�nitions are needed to state our main result. The de�nition
and properties of proximate order �(r) in the sense of Valiron can be found in [1].
We denote V (r) = r
�(r).
A subharmonic function v is of proximate order �(r) if
lim
z!1
v(z)
V (jzj)
<1:
Let v(z) be a subharmonic function of proximate order �(r), � = lim�(r) 2
(0;1) (r !1), and let
vt(z) =
v(tz)
V (t)
be a trajectory of Azarin of subharmonic function v. The limit set of Azarin Fr(v)
is de�ned as a set of functions given by
u(z) = lim
n!1
vtn(z)
for some sequence (tn), tn ! +1:
The limit is taken in the sense of distributions. This means that
lim
n!1
ZZ
vtn(z)'(z)dm2(z) =
ZZ
u(z)'(z)dm2(z)
c
A. Chouigui and A.F. Grishin, 2008
A Property of Azarin's Limit Set of Subharmonic Functions
for any test function ', where m2 is a two-dimensional Lebesgue measure.
Let
h(�) = lim
r!1
v(rei�)
V (r)
; h(�) = lim
r!1
� v(rei�)
V (r)
:= sup
E
lim
r !1
r2E
inf
v(rei�)
V (r)
;
where E � (0;1) runs over all sets of zero linear density which is de�ned by
densE = lim
r!1
mes(E \ [0; r])
r
:
The function h is called an indicator of function v and the function h is called
a lower indicator of function v. In 1979 V.S. Azarin [2] proved that
H(z) := sup fu(z) : u 2 Fr(v)g = h(�)r�; z = re
i�
;
H(z) := inf fu(z) : u 2 Fr(v)g = h(�)r�:
See [3] for other properties of the limit set.
Denote I(z) = fu(z) : u 2 Fr(v)g. We prove the following re�ned version of
Azarin's theorem.
Theorem 1. For each z, (h(�)r�; h(�)r�] is a subset of I(z), and I(z) is
a subset of [h(�)r�; h(�)r�]:
We give an example of a subharmonic function v such that h(�)r�2I(z) for
some z. The case h(�)r� 2 I(z) is possible as well.
P r o o f of Theorem 1. Denote
(Ftu)(z) =
u(tz)
t�
:
V.S. Azarin [2] proved that Ft(Fr(v))(z) � Fr(v). The map Ft : Fr(v) ! Fr(v)
is one-to-one. We denote H(z) := sup fu(z) : u 2 Fr(v)g, H(z) := inf fu(z) :
u 2 Fr(v)g. We have
H(tz) = sup fu(tz) : u 2 Fr(v)g = t
� sup
�
u(tz)
t�
: u 2 Fr(v)
�
= t
�
H(z):
Thus H(rei�) = r
�
H(ei�): Analogously, H(rei�) = r
�
H(ei�): For every " > 0
there exists (see, for example, [2]) a number R such that v(rei�) < (h(�)+ ")V (r)
is valid for r 2 [R;1) and for any �.
Consequently,
vt(z) =
v(trei�)
V (t)
< (h(�) + ")
V (tr)
V (t)
; tr > R:
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3 347
A. Chouigui and A.F. Grishin
It is known [1] that
V (tr)
V (t)
� r
�
; 0 < a � r � b <1;
where the double arrow means a uniform convergence on the given set. Hence
there exist numbers R1 > 0 and �0 2 (0; 1) such that
vt(z) � h(�) + 2"; z 2 C(ei�; �0); t � R1: (1)
Here C(ei�; �0) is an open disk centered at ei� with the radius �0. Let u be
an arbitrary function from Fr(v). It follows from the de�nition of �ne topology
[4] that the set E = fz : u(z) > u(ei�) � "g is a �ne neighborhood of ei�. Then
there exists a compact set K such that K � E \ C(ei�; �0) and capK > 0.
Therefore there exists a positive measure � such that �(K) > 0; supp(�) � K,
and the potential
b(z) =
ZZ
ln jz � �jd�(�)
is continuous ([5], corollary to Th. 3.7).
Further we need the following results. Let vn(z) be a sequence of subharmonic
functions converging in the sense of distributions to a distribution w. Then w is
a regular distribution and may be represented by a subharmonic function w(z).
We recall that the distribution T
T' =
ZZ
f(z)'(z)dm2(z);
where f is a locally integrable function, is called a regular distribution. Let �n and
� be the Riesz masses of vn and w, respectively. We have �n = 1
2��vn; � = 1
2��w,
where � is the Laplace operator. Di�erentiation is continuous in the space of
distributions. It follows that �n ! � in the sense of distributions. Theorem 0.4
[5] states that �n converges weakly to � as a sequence of Radon measures. This
means that (�n; ') ! (�; ') for any continuous compactly supported function '.
In addition, if a compact set K is Jordan measurable with respect to the measure
� (this means that �(@K) = 0), then �n converges weakly to � as a sequence of
elements of the Banach space C�(K). This means that (�n; ') ! (�; ') for any
function ' which is continuous on K. If K = B(z0; R) and �(@B(z0; R)) = 0,
then
lim
n!1
ZZ
B(z0;R)
ln jz � �jd�n(�) =
ZZ
B(z0;R)
ln jz � �jd�(�)
in the sense of distributions. We have the Riesz representation
vn(z) =
ZZ
B(z0;R)
ln jz � �jd�n(�) + un(z);
348 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3
A Property of Azarin's Limit Set of Subharmonic Functions
where un is a harmonic function in disk C(z0; R) = fz : jz � z0j < Rg : It is clear
that (un) is a convergent sequence in the sense of distributions. Then the sequence
(un) is uniformly convergent on every compact setK � C(z0; R) by Th. 4.4.2 ([6]).
Thus, modulo the uniformly convergent sequence (un) of harmonic functions, the
(vn) is a sequence of potentials, and so many classical results from potential theory
may be extended to (vn). In particular,
lim
n!1
ZZ
vtn(z)d�(z) =
ZZ
w(z)d�(z): (2)
The proof of an analogous proposition for potentials appeared in [5, Th. 3.8].
Now we have�
u(ei�)� "
�
�(K) �
ZZ
u(z)d�(z) �
�
h(�) + 2"
�
�(K):
The left-hand side follows from the inequality u(z) > u(ei�) � " for z 2 K;
and the right-hand side follows from (1). This gives u(ei�) � h(�) for any u 2
Fr(v); H(ei�) � h(�):
Further we prove that there exists a function u0 2 Fr(v) such that u0(e
i�) =
h(�): Since
h(�) = lim
r!1
v(rei�)
V (r)
;
then there exists a sequence (tn), tn ! 1 as n ! 1; such that the sequence of
real numbers vtn(e
i�) converges to h(�) as n!1:
The set fvt(z) : t 2 0;1g is compact in the sense of distributions, see [7,
Th. 2.7.1.1]. Hence we can �nd a subsequence tn
k
such that vtn
k
(z) ! u0(z) in
the sense of distributions.
According to the principle of ascent (for potentials it is Th. 1.3 [5]),
h(�) = lim
k!1
vtn
k
(ei�) � u0(e
i�) � H(ei�):
This yields
h(�) = H(ei�); u0(e
i�) = h(�):
Our next step is to prove the inequality u(ei�) � h(�) for u 2 Fr(v). Let
u 2 Fr(v) and " > 0. It is evident that we may assume h(�) > �1. Since u is
upper semicontinuous, then there exists � 2 (0; 1) such that u(z) < u(ei�) + "
for z 2 B(ei�; �). Let tn ! 1 as n ! 1 be such a sequence that vtn ! u as
a sequence of distributions. Then, for � 2 (0; 1)
lim
n!1
1+�Z
1��
���u(tei�)� vtn(te
i�)
��� dt! 0: (3)
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3 349
A. Chouigui and A.F. Grishin
Using the similar arguments as those used to prove (2), one can reduce the
proposition to the following. Let a sequence �n of Borel measures in disk B(z0; R)
converge weakly to a measure �. Then the real sequence
an =
1+�Z
1��
�������
ZZ
B(z0;R)
ln jtei� � �jd(�n � �)(�)
������� dt
converges to zero.
We have
an =
ZZ
B(z0;R)
0
@ 1+�Z
1��
hn(t) ln jte
i� � �jdt
1
Ad(�n � �)(�);
where
hn(t) = sign
ZZ
B(z0;R)
ln jtei� � �jd(�n � �)(�):
The function hn(t) is measurable and jhn(t)j � 1. Consider a family of functions
Hn(�) =
1+�Z
1��
hn(t) ln jte
i� � �jdt; n = 1; 2; : : : :
The inequality
jHn(�)j �
1+�Z
1��
���ln jtei� � �j
��� dt
shows that the family Hn(�) is uniformly bounded in B(z0; R).
Further,
jHn(�2)�Hn(�1)j �
1+�Z
1��
����ln j tei� � �2
tei� � �1
j
���� dt
=
1+�Z
1��
max
�
ln j
te
i� � �2
tei� � �1
j; ln j
te
i� � �1
tei� � �2
j
�
dt
�
1+�Z
1��
ln
�
1 +
j�2 � �1j
jtei� � �1j
�
dt+
1+�Z
1��
ln
�
1 +
j�2 � �1j
jtei� � �2j
�
dt = J1 + J2:
350 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3
A Property of Azarin's Limit Set of Subharmonic Functions
In the integral J1 we introduce a new variable r by the formula r = jtei�� �1j.
We obtain
J1 =
1Z
0
ln
�
1 +
j�2 � �1j
r
�
d�(r);
where �(r) = mes
�
[(1� �)ei� ; (1 + �)ei�] \B(�1; r)
�
.
The function �(r) is constant in [R;1), where R = max
�
j(1� �)ei� � �1j ;
j(1 + �)ei� � �1j
�
. The inequality �(r) � 2r is obvious. From the properties given
above it follows that
J1 =
RZ
0
ln
�
1 +
j�2 � �1j
r
�
d�(r) � 2
RZ
0
ln
�
1 +
j�2 � �1j
r
�
dr:
It can be shown by integrating by parts. The integral J2 is estimated in a similar
way. Be speci�c about what estimates show equicontinuity Hn(�). Arzela�Ascoli's
theorem gives a compactness of the family Hn(�). Consequently, the sequence
an =
ZZ
B(z0;R)
Hn(�)d(�n � �)(�)
converges to zero. Formula (3) is proved.
If An =
�
t 2 [1� �; 1 + �] :
��u(tei�)� vtn(te
i�)
�� � "
; then mes(An) ! 0 as
n ! 1. If B(") =
�
r 2 (0;1) : v(rei�) < h(�)� "
, then the formula for h(�)
gives that the linear density of B(") is zero.
For � 2 (0; 1) we denote
Bn =
�
t 2 [1� �; 1 + �] : (h(�)� ")
V (tnt)
V (tn)
� vtn(te
i�)
�
:
If t 2 Bn, then
vtn(te
i�) =
v(tnte
i�)
V (tn)
� (h(�)� ")
V (tnt)
V (tn)
;
v(tnte
i�) � (h(�)� ") V (tnt):
It follows that tnt 2 B("), tnBn � B("), mes(tnBn) � mes (B(") \ [(1� �)tn;
(1 + �)tn]) ; and
mes(Bn) �
mes (B(") \ [0; (1 + �)tn])
tn
:
Now the property that density of B(") is zero implies mes(Bn) ! 0 as n ! 1:
We have�
h(�)� "
�
V (tnt)
V (tn)
< vtn(te
i�) � u(tei�) +
���u(tei�)� vtn(te
i�)
��� < u(ei�) + 2"
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3 351
A. Chouigui and A.F. Grishin
for t 2 [1� �; 1 + �] n (An [Bn). The convergence
V (tnt)
V (tn)
� t
�
; t 2 [1� �; 1 + �]
leads to the inequality
(h(�)� ")
V (tnt)
V (tn)
> h(�)� 2"
for su�ciently large n and small �. Thus we obtain the claimed inequality u(ei�)
� h(�):
Now is the �nal step of the proof. With � �xed, we denote
A(r; �) =
1
rV (r)
(1+�)rZ
r
v(tei�)dt:
The function A(r; �) is continuous and bounded in the variable r 2 [1;1). Then
the limit set, i.e., the set of all subsequential limits, of A(r; �) as r ! +1 is
a closed interval J(�) = [A(�); B(�)]. We claim that
J(�) =
8<
:
1+�Z
1
u(tei�)dt : u 2 Fr(v)
9=
; : (4)
In fact, let
a(�) = lim
n!1
A(rn; �) = lim
n!1
1+�Z
1
vrn(te
i�)dt:
In addition, we may assume that vrn(z) ! u(z) in the sense of distributions.
Then the equality
lim
n!1
1+�Z
1
vrn(te
i�)dt =
1+�Z
1
u(tei�)dt;
which is a special case of (2), gives a(�) =
1+�R
1
u(tei�)dt. Clearly, for any u 2 Fr(v)
the value of integral
1+�R
1
u(tei�)dt belongs to the interval J(�). Relation (4) is
proved. Note that it also follows from the results obtained by V.S. Azarin [2].
According to Theorem 2 [8],
lim
�!+0
A(�)
�
= h(�); lim
�!+0
B(�)
�
= h(�):
352 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3
A Property of Azarin's Limit Set of Subharmonic Functions
If h(�) < h < h(�); then the inequalities
A(�)
�
< h
(1 + �)�+1 � 1
(�+ 1)�
<
B(�)
�
are valid for all su�ciently small �. Therefore there exists a strictly positive
number � and a function u 2 Fr(v) such that
h
(1 + �)�+1 � 1
�+ 1
=
1+�Z
1
u(tei�)dt;
1+�Z
1
�
u(tei�)� ht
�
�
dt = 0:
We claim that there exists t0 2 [1; 1 + �] with u(t0e
i�) = ht
�
0. Consider the
function w(z) = u(z)� hjzj�: Further we will assume that � is �xed and consider
w(tei�) as a function in variable t. We have either w(tei�) = 0 almost everywhere
on the interval [1; 1 + �] or the function w(tei�) has strictly positive and strictly
negative values on this interval. In the �rst case t0 is obtained. We consider
the second case. The function w(tei�) is upper semicontinuous. Hence the set
E =
�
t > 0 : w(tei�) < 0
is open. Under the assumption E is nonempty, the set
E meets the interval [1; 1 + �]. We have
E =
1[
k=1
(ak; bk);
where (ak; bk) is a disjoint system of intervals. There exists k such that (ak; bk)\
[1; 1 + �] 6= ?: The point ak or the point bk necessarily belongs to the interval
(1; 1+�), assume that bk 2 (1; 1+�). Because bk2E, the inequality w(bke
i�) � 0
is valid. The function w(z) is continuous in �ne topology. Hence there exists
a �ne neighborhood G of bke
i� such that
w(bke
i�) = lim
z ! bke
i�
z 2 G
w(z):
According to the theorem of Lebesgue and Beurling ([4, Prop. IX.6]), the point
bke
i� is a limit point for the set G \ (ake
i�
; bke
i�). This gives w(bke
i�) � 0 and
then w(bke
i�) = 0: Thus bk is the required point t0. For the function
u
(1)(z) =
u(t0z)
t
�
0
2 Fr(v)
we have u(1)(ei�) = h.
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3 353
A. Chouigui and A.F. Grishin
Now the assertions of the theorem follow from the above. The theorem is
proved.
We produce a subharmonic function v such that h(0) = H(1)2I(1) and con-
struct the limit set of Azarin of this function in the form
Fr(v) = fut(z) : t 2 (0;1)g: (5)
Consider the function
a(z) =
1X
n=1
1
n3
ln
����1� z
1� e�n
���� :
We have
a(z) =
1X
n=1
1
n3
!
ln jzj+
1X
n=1
1
n3
ln
1
1� e�n
+O
�
1
jzj
�
; z !1:
On every interval
�
�1; 1� e
�1
�
;
�
1� e
�k
; 1� e
�k�1
�
; k = 1; 2; : : : ; (1;1);
the function a(x) is strictly concave since
a
00(x) = �
1X
n=1
1
n3
1
(x� 1 + e�n)2
:
Let xn 2
�
1� e
�n
; 1� e
�n�1
�
be such a point that
a(xn) = max
�
a(x) : x 2
�
1� e
�n
; 1� e
�n�1
�
:
Then the function a(x) increases on the interval (1� e
�n
; xn) and decreases
on the interval
�
xn; 1� e
�n�1
�
. First we prove the relation
a(xn)! a(1) (6)
as n!1: Let �k = 1� 1
2
�
1 + 1
e
�
e
�k
: We have
a(1)� a(�k) = �
1X
n=1
1
n3
ln
����1� 1
2
�
1 +
1
e
�
e
n�k
���� :
From the inequalities�������
X
1�n� k
2
1
n3
ln
�
1�
1
2
�
1 +
1
e
�
e
n�k
�������� � (1 +
1
e
)e�k=2
1X
n=1
1
n3
;
354 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3
A Property of Azarin's Limit Set of Subharmonic Functions
�������
X
k
2
<n�k
1
n3
ln
�
1�
1
2
�
1 +
1
e
�
e
n�k
�������� �
X
k
2
<n�k
1
n3
ln
2
1� 1
e
;
�����
1X
n=k+1
1
n3
ln
����1� 1
2
�
1 +
1
e
�
e
n�k
����
����� �
1X
n=k+1
1
n2
it follows that a(�k) ! a(1); as k ! 1: Since �k 2
�
1� e
�k
; 1� e
�k�1
�
, then
a(xk) � a(�k);
lim
k!1
a(xk) � a(1):
The upper semicontinuity of the function a(z) yields
lim
k!1
a(xk) � lim
z!1
a(z) � a(1);
and (6) follows.
Introduce � and �1 with 0 < � < �1 < 1: It follows from (6) that
�� = inf
a(xn)
x
�1
n
> �1:
Let �0 be the number in the interval
�
0; 1� 1
e
�
such that a(�0) = ����10 .
The existence of �0 follows from the inequality a1(t) = a(t) + �t
�1 > 0 in the
right neighborhood of zero and the inequality a1(t) < 0 in the left neighborhood
of 1 � 1
e
. The function a1(t) is strictly concave on the interval
�
0; (1 � 1
e
)
�
, and
a1(0) = 0. Any strictly concave function has at most two zeros. This proves that
�0 is unique. We choose c > 2� and denote
A1 = (�0; 1�
1
e
); Ak = (xk�1; 1 � e
�k); k = 2; 3; : : : :
Let sk be the unique point t 2 Ak such that a(t) = �ct�1 . We denote a2(t) =
a(t) + ct
�1 and consider the case k � 2:
We have a2(xk�1) > 0; a2(t) < 0 in the left neighborhood of 1 � e
�k.
This shows that sk exists. Analogously, there exists s1k 2
�
1� e
�k+1
; xk�1
�
such that a2(s1k) = 0. The function a2(t) is strictly concave in each interval�
1� e
�k+1
; 1� e
�k
�
. This proves that sk is unique. The existence and unique-
ness of s1 is proved in the same way as for �0.
We have a(z) = Re�(z);
�(z) =
1X
n=1
1
n3
ln
�
1�
z
1� e�n
�
;
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3 355
A. Chouigui and A.F. Grishin
�
0(z) =
1X
n=1
1
n3
1
z � 1 + e�n
:
A level-line of a(z) is a real-analytic curve if it does not meet zeros and poles of
�
0(z). We wish to �nd zeros of �0(z). The following identity holds
Im�0(z) = �y
1X
n=1
1
n3jz � 1 + e�nj2
: (7)
It gives that all zeros of �0(z) are real. Now it is easy to verify that the set of
zeros of �0(z) is fxn; n = 1; 2; : : :g :
Let �k be the unique point in the interval
�
1� e
�k
; xk
�
such that a(�k) =
a(sk). The inequality a(x) < a(sk) is realized on the interval (sk; �k). From the
identity @a
@y
(z) = �Im�0(z) and (7) it follows that the function a(x; y) strictly
increases on (0;1) and strictly decreases on (�1; 0) in the variable y. Therefore
there exist functions y1(x) > 0 and y2(x) < 0 on the interval (sk; �k) such that
a(x; y1(x)) = a(sk); a(x; y2(x)) = a(sk):
The collection of curves z = x + iy1(x); z = x + iy2(x); x 2 (sk; �k), and
points z = sk; z = �k is a closed Jordan curve Lk that is a level curve of a(z).
It is a real analytic curve. Let Gk be a bounded domain with boundary Lk.
Let u(z) be a function such that u(z) = a(z) if z2
1S
k=1
Gk, and u(z) = a(sk)
if z 2 Gk. The function u(z) is subharmonic. It is important for us that the
inequalities
u(x) � �cx�1 ; u(x) > �cx� (8)
are realized on the semi-axis (0;1).
Consider the Azarin trajectory of function u,
ut(z) =
u(tz)
t�
; t 2 (0;1):
One can prove that ut(z) ! 0 in the sense of distributions when t ! 0 or
t!1.
Theorem 9 [3] asserts that there exists a subharmonic function v of order �
such that
Fr(v) = fut(z) : t 2 (0;1)g [ f0g:
It follows from (8) that
ut(1) =
u(t)
t�
> �c:
356 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3
A Property of Azarin's Limit Set of Subharmonic Functions
In addition,
us
k
(1) =
u(sk)
s
�
k
= �cs�1��
k
! �c
as k !1:
This gives H(1) = �c; H(1)2I(1): The function v is a required example.
Acknowledgements. The authors thank Prof. D. Drasin, Prof. S. Merenkov
and the reviewer for the help in preparing this paper.
References
[1] B.Ja. Levin, Distribution of Zeros of Entire Functions. AMS, Providence, RI, 1980.
[2] V.S. Azarin, On Asymptotic Behavior of Subharmonic Functions of Finite Order.
� Mat. Sb. 108 (1979), No. 2, 147�169. (Russian)
[3] V.S. Azarin, Limits Sets of Entire and Subharmonic Functions. In: A.A Gold-
berg, B.Ya. Levin, and I.V Ostrovskii, Entire and Meromorphic Functions. Springer,
Berlin, 1997, 48�66.
[4] M. Brelot, On Topologies and Boundaries in Potential Theory. Springer�Verlag,
Berlin, Heidelberg, New York, 1971.
[5] N.S. Landkof, Foundation of Modern Potential Theory. Springer, Berlin, Heidel-
berg, New York, 1972.
[6] L. H�ormander, The Analysis of Partial Linear Di�erential Operators. Springer,
Berlin, Heidelberg, New York, Tokyo, 1, 1983.
[7] V.S. Azarin, The Theory of Growth of Subharmonic Functions. (Lectures). Kharkov
State Univ., Kharkov, 1978. (Russian)
[8] A.F. Grishin and T.I. Malyutina, New Formulas for Indicators of Subharmonic
Functions. � Mat. �z., analiz, geom. 12 (2005), 25�72. (Russian)
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3 357
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