On Subharmonic Functions of the First Order with Restrictions on the Real Axis
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України
2008
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On Subharmonic Functions of the First Order with Restrictions on the Real Axis |
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On Subharmonic Functions of the First Order with Restrictions on the Real Axis Poedintseva, I.V. |
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On Subharmonic Functions of the First Order with Restrictions on the Real Axis |
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On Subharmonic Functions of the First Order with Restrictions on the Real Axis |
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On Subharmonic Functions of the First Order with Restrictions on the Real Axis |
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On Subharmonic Functions of the First Order with Restrictions on the Real Axis |
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on subharmonic functions of the first order with restrictions on the real axis |
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Poedintseva, I.V. |
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Poedintseva, I.V. |
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2008 |
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English |
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Журнал математической физики, анализа, геометрии |
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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Article |
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1812-9471 |
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https://nasplib.isofts.kiev.ua/handle/123456789/106514 |
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On Subharmonic Functions of the First Order with Restrictions on the Real Axis / I.V. Poedintseva // Журнал математической физики, анализа, геометрии. — 2008. — Т. 4, № 3. — С. 380-394. — Бібліогр.: 11 назв. — англ. |
| work_keys_str_mv |
AT poedintsevaiv onsubharmonicfunctionsofthefirstorderwithrestrictionsontherealaxis |
| first_indexed |
2025-11-26T23:36:10Z |
| last_indexed |
2025-11-26T23:36:10Z |
| _version_ |
1850781524358594560 |
| fulltext |
Journal of Mathematical Physics, Analysis, Geometry
2008, vol. 4, No. 3, pp. 380�394
On Subharmonic Functions of the First Order
with Restrictions on the Real Axis
I.V. Poedintseva
Department of Mechanics and Mathematics V.N. Karazin Kharkiv National University
4 Svobody Sq., Kharkiv, 61077, Ukraine
E-mail:Irina.V.Poedintseva@univer.kharkov.ua
Received December 17, 2007
Subharmonic functions v of the �rst proximate order �(r) with the in-
tegral
RR
0
t
1��(t)(v(t)+v(�t))
1+t2
dt bounded with respect to R are studied. This is
an extension of a result by N.I. Akhiezer, who studied the case �(r) � 1,
v(z) = ln jf(z)j, where f(z) is an entire function.
Key words: proximate order, functions of the class A, functions of com-
pletely regular growth.
Mathematics Subject Classi�cation 2000: 31A05.
According to the de�nition given by N.N. Meiman [1], a function f is of the
class A if
1X
k=1
����= 1
ak
���� <1; (1)
where ak are zeros of f . Functions of the class A play an important role in the
theory of entire functions. In applications they are most often of exponential type,
that is, of the �rst order and normal type.
In [2] N.I. Akhiezer proved that an entire function of exponential type belongs
to the class A if the integral
RZ
0
ln jF (x)F (�x)j
1 + x2
dx (2)
is bounded with respect to R.
The Cartwright class of entire functions of exponential type with the bounded
integral
1Z
�1
ln+ jF (t)j
1 + t2
dt (3)
c
I.V. Poedintseva, 2008
On Subharmonic Functions of the First Order with Restrictions on the Real Axis
is also well known. From the results of [3] it follows that the functions of the
Cartwright class are of the class A and of completely regular growth.
A chapter of book [4] and a section of review [5] are devoted to the functions of
the classA. The integrals of form (3) were also studied in the book by P. Koosis [6].
In the present paper the concepts of proximate and formal orders are widely used.
A di�erentiable function �(r) on the half-axis (0;1) is said to be of proximate
order in the sense of Valiron if it satis�es the following two conditions:
1) lim
r!+1
�(r) = � 2 (�1;1),
2) lim
r!+1
r ln r�0(r) = 0.
Properties of the proximate order are presented, e.g., in [4]. Some more results
can be found in [7].
The function r�(r) is denoted by V (r).
The proximate order �(r) is said to be a formal order of a function v subhar-
monic in the plane C (in the upper half-plane C +) if there exists a constant M1
such that
v(rei�) �M1V (r); r � 1
for � 2 [0; 2�] (� 2 (0; �)).
A class of functions of formal order �(r) is denoted by SF (�(r)).
We say that a measure � has a formal order �(r) if there exists a constant M2
such that for all su�ciently large r
�(B(0; r)) �M2V (r); B(0; r) = fz : jzj � rg:
A class of measures of formal order �(r) is denoted by M(�(r)).
We should note that the Riesz measure � of subharmonic function v(z) of the
formal order �(r) is also of the formal order �(r).
The de�nitions above are in [8].
For the subharmonic functions v of formal order �(r), �(r) ! 1 as r ! 1,
the following two integrals
RZ
0
W (t)
1 + t2
(v(t) + v(�t))dt (4)
and ZZ
B(0; R)
1Z
0
W (t)
1 + t2
ln
����1� t2
�2
���� dtd�(�); (5)
are considered, where W (t) = t1��(t), � is the Riesz measure of the function v(z).
The �rst integral is analogical to (2), and the second one to the partial sum of
the series from (1).
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3 381
I.V. Poedintseva
The proximate order �(r) was introduced to obtain the estimates of functions
in a neighborhood of in�nity. Therefore, the behavior of �(r), as r ! 0, is
not signi�cant. In addition, the formulations and proves of many theorems are
simpli�ed if we consider the integrals of the form
aZ
0
V (t)K(x; t)dt:
For example, integral (4) arises in the presented paper. Without any restric-
tions on the behavior of �(r) in a neighborhood of zero, the integral (4), generally
speaking, is divergent. And in the concrete formulations we ought to replace the
low limit of integration by 1.
However, sometimes the appropriate restrictions on the behavior of �(r) in
a neighborhood of zero are more preferable. In our case the condition W (r)W
�
1
r
�
= 1 turns out to be convenient.
In Theorem 1 it is proved that the integrals (4) and (5) are bounded, or are
not bounded simultaneously. This is the main result of the paper, the remaining
statements are its corollaries. In particular, under some additional restrictions on
the function v(z) we obtain that (5) is bounded if and only if the following two
integrals are convergent:
ZZ
CnB(0; r0)
W (�)
�
j sin �j d�(�) (6)
and
ZZ
CnB(0; r0)
W 0(�) d�(�); (7)
where � = �ei�, and r0 is an arbitrary �xed strictly positive number.
If �(r) � 1, then the second integral vanishes. In this particular case Th. 1
transforms into the extension of the Akhiezer theorem for the class of subharmonic
functions. Thus, one can say that Th. 1 generalizes the Akhiezer theorem to the
case of subharmonic functions of a proximate order �(r), �(r)! 1 as r ! +1.
We should point out that the convergence conditions for (6) and (7) are
independent. Hence, the case of a general proximate order �(r) and the case
�(r) � 1 are signi�cantly di�erent.
Note that many statements on entire or subharmonic functions of order � can
be easily extended to the functions of order �(r). However, Akhiezer's proof can
not be extended to the case of general proximate order.
382 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3
On Subharmonic Functions of the First Order with Restrictions on the Real Axis
Now we give some notation for the sets used in the paper:
B(z0; R) = fz : jz � z0j � Rg;
K(R1; R2) = fz : R1 � jzj < R2g:
Further we will formulate the de�nition of the limit set of measure according
to V.S. Azarin [10].
Consider the transformation (�)t of measure � 2M(�(r)), which is de�ned by
the equality
�t(E) =
�(tE)
V (t)
for any Borel set E.
The Azarin limit set Fr[�] of a measure � 2M(�(r)) is de�ned as the set of
measures given by the condition
� = lim
n!1
�tn for some sequence (tn); tn ! +1:
Here the limit is taken in the sense of distributions. This means that for a test
function '(z) (that is, compactly supported and in�nitely di�erentiable) the fol-
lowing equality holds:
lim
n!1
ZZ
'(z)d�tn (z) =
ZZ
'(z)d�(z):
A de�nition of the normal set is also needed.
A set E is said to be normal for a measure � if �(@E) = 0.
Let v(z) be a subharmonic function and � be its Riesz measure,
v1(z) = v(z) �
ZZ
B(0;1)
ln jz � �jd�(�); v2(z) = v1(z)� v1(0):
Note that the function v1(z) is harmonic in the disk jzj < 1. Also, let �1 be
a restriction of measure � on the exterior of an open unit disk. The measure �1
is the Riesz measure of subharmonic function v2.
Integral (4) for the function v2 is bounded together with the same integral for
the function v. Moreover, the Azarin limit sets of measures � and �1 coincide.
Hence, further we may assume that the function v(z) is harmonic in the disk
jzj < 1 and v(0) = 0, unless the contrary is stated.
Lemma 1.The limit set of the integral
I1(R) =
ZZ
K(R;1)
RZ
0
W (t)
t2
ln
����1� t2
�2
���� dtd�(�)
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3 383
I.V. Poedintseva
in the direction R! +1 is of the form8><
>:
ZZ
K(1;1)
1Z
0
1
u2
ln
����1� u2
�1
2
���� dud�(�1); � 2 E1
�
9>=
>; : (8)
Here � is the Riesz measure of subharmonic function of the class SF (�(r)), E1
� is
the set of measures � which coincides with the limit set in the direction t!1 of
the family of measures �1t , where �
1
t is the restriction of the measure �t to the set
fz : jzj � 1g.
P r o o f. First of all, let us give some helpful inequalities.
From the Poisson�Jensen formula for a subharmonic function v under the
assumption v(0) = 0 it follows that there exists a constant M3 such that
RZ
0
�(t)
t
dt �M3V (R): (9)
From (9) in a standard way (see, for example, [4, Ch. 1, �5]) we get
�(R) �M4V (R): (10)
Let us formulate another useful for us statement ([7, Th. 4]).
If �(r) is an arbitrary zero proximate order for which V (r)V
�
1
r
�
= 1, then
there exists a zero proximate order �1(r) such that
V (xt) � V (x)V1(t); x > 0; t > 0; (11)
where V1(t) = t�1(t).
As it was said above, we assume that
W (r)W
�
1
r
�
= 1:
It allows us to apply (11) to the function W (r).
Then consider the integral I1(R). Replacing � by R�1 in the exterior integral
and t by uR in the inner integral, we obtain
I1(R) =
ZZ
K(1;1)
1Z
0
W (uR)
W (R)
1
u2
ln
����1� u2
�1
2
���� dud�R(�1)
=
ZZ
C
1Z
0
W (uR)
W (R)
1
u2
ln
����1� u2
�1
2
���� dud�1R(�1):
384 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3
On Subharmonic Functions of the First Order with Restrictions on the Real Axis
Choose a sequence (Rn), Rn ! +1, such that its limit (I1(Rn)) exists (it may
be improper limit).
Since the sequence
�
�1Rn
�
is bounded, then we may additionally assume ([10,
Th. 1.2.1]) that it converges as the sequence of distributions to a measure �. Note
that � 2 E1
�.
Represent the integral I1(Rn) in the form of the following sum:
I1(Rn) =
ZZ
B(0;N)
1Z
0
W (uRn)
W (Rn)
1
u2
ln
����1� u2
�1
2
���� dud�1Rn(�1)
+
ZZ
K(N;1)
1Z
0
W (uRn)
W (Rn)
1
u2
ln
����1� u2
�1
2
���� dud�1Rn(�1) = I1
1
(Rn) + I2
1
(Rn): (12)
Here N is chosen so that the support of measure � does not contain the circle
jzj = N .
First we prove that
lim
n!1
I1
1
(Rn) =
ZZ
B(0;N)
1Z
0
1
u2
ln
����1� u2
�1
2
���� dud�(�1): (13)
To end this we take an arbitrary Æ 2 (0; 1) and represent the integral I1
1
(Rn) in
the form of the following sum:
I1
1
(Rn) =
ZZ
B(0;N)
ÆZ
0
W (uRn)
W (Rn)
1
u2
ln
����1� u2
�1
2
���� dud�1Rn(�1)
+
ZZ
B(0;N)
1Z
Æ
W (uRn)
W (Rn)
1
u2
ln
����1� u2
�1
2
���� dud�1Rn(�1) = I
1;1
1
(n; Æ) + I
1;2
1
(n; Æ):
Let us consider the second summand in the right-hand part of the equality.
Since the disk B(0; N) is normal with respect to measure �, then Th. 0.50 [9]
yields
lim
n!1
I
1;2
1
(n; Æ) =
ZZ
B(0;N)
1Z
Æ
1
u2
ln
����1� u2
�1
2
���� dud�(�1):
To prove this equality we have to apply the following property of proximate
order ([4, Ch. 1, x 12]):
W (uR)
W (R)
� 1; R! +1; u 2 [a; b]; 0 < a < b < +1: (14)
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3 385
I.V. Poedintseva
Using (11) for I
1;1
1
(n; Æ), we get the estimates
���I1;1
1
(n; Æ)
��� �
ZZ
B(0;N)
ÆZ
0
V1(u)
u2
����ln
����1� u2
�1
2
����
���� dud�1Rn(�1)
�M5
ÆZ
0
V1(u)du
ZZ
B(0;N)
d�1Rn(�1)
j�1j2
�M6
ÆZ
0
V1(u)du:
From this it follows that I
1;1
1
(n; Æ) tends to zero uniformly with respect to n
as Æ ! +0. Together with the equality
lim
Æ!0
ZZ
B(0;N)
ÆZ
0
1
u2
ln
����1� u2
�1
2
���� dud�1(�1) = 0
it gives the demanded relation (13).
Now we estimate the summand I2
1
(Rn).
��I2
1
(Rn)
�� �
ZZ
K(N;1)
1Z
0
W (uRn)
W (Rn)
2
�12
dud�1Rn(�1)
� 2
ZZ
K(N;1)
1Z
0
V1(u)du
d�Rn
�2
1
�M7
1Z
N
d�1Rn(�1)
�12
; �1 = j�1j:
Integrating by parts and using (10) and (11) for � 2 (0; 1) and N > N(�),
we obtain
1Z
N
d�R(�1)
�12
=
�R(�1)
�12
����
1
N
+ 2
1Z
N
�R(�1)
�13
d�1 � 2
1Z
N
�R(�1)
�13
d�1
= 2
1Z
N
1
V (R)
�(R�1)
�13
d�1 � 2M4
1Z
N
V (R�1)
V (R)
d�1
�13
� 2M4
1Z
N
d�1
�12��
=
M8
N1��
:
Thus, for I2
1
(Rn) we have
��I2
1
(Rn)
�� =
�������
ZZ
K(N;1)
1Z
0
W (uRn)
W (Rn)
1
u2
ln
����1� u2
�1
2
���� dud�Rn(�1)
�������
�
M9
N1��
:
386 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3
On Subharmonic Functions of the First Order with Restrictions on the Real Axis
The following estimate also holds:�������
ZZ
K(N;1)
1Z
0
ln
����1� u2
�2
1
���� d�(�)
�������
�
M10
N1��
:
Taking into account (12), we get
������I1(Rn)�
ZZ
C
1Z
0
1
u2
ln
����1� u2
�1
2
���� dud�(�1)
������
�
�������
I1(Rn)�
ZZ
B(0;N)
1Z
0
1
u2
ln
����1� u2
�1
2
���� dud�(�1)
�������
+
M11
N1��
:
Passing to the limit (it means the upper limit) �rst, when n!1, and then,
when N ! 1, we obtain that the limit set of integral I1(R) in the direction
R! +1 consists of the values given by formula (8).
Conversely, if we take the sequence (Rn) such that the sequence (�1Rn) con-
verges to the measure � in the sense of distributions, then we �nd that the right-
hand part of formula (8) is held in the limit set of integral I1(R) in the direction
R! +1. The lemma is proved.
R e m a r k 1. Since for a function �1 2 E
1
� the inequality �1(B(0; R)) �MR
holds for all R > 0 ([10, Th. 1.2.2]), then the limit set of integral I1(R) in the
direction R ! +1 is bounded, and hence, the value I1(R) is bounded on the
half-axis [1;1).
Lemma 2. The limit set of the integral
I2(R) =
ZZ
B(0; R)
1Z
R
W (t)
t2
ln
����1� t2
�2
���� dtd�(�)
in the direction R! +1 is of the form8><
>:
ZZ
B(0;1)
1Z
1
1
u2
ln
����1� u2
�1
2
���� dud�(�1); � 2 E2
�
9>=
>; :
Here � is the Riesz measure of subharmonic function of the class SF (�(r)), E2
� is
the set of measures �, which coincides with the limit set in the direction t! +1
of the family of measures �2t , where �
2
t is the restriction of measure �t to the set
f� : j�j � 1g.
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3 387
I.V. Poedintseva
The proof of Lem. 2 is not presented here, as it is completely analogous to
that of Lem. 1.
R e m a r k 2. The value I2(R) is bounded.
Theorem 1. Let v(z) 2 SF (�(r)), �(r) ! 1 as r ! +1, in the plane C .
Then
RZ
0
W (t)
1 + t2
(v(t) + v(�t))dt =
ZZ
B(0; R)
1Z
0
W (t)
1 + t2
ln
����1� t2
�2
���� dtd�(�) + '(R); (15)
where �(�) is the Riesz measure of function v, '(R) is the function bounded on
(0;1).
P r o o f. As it was pointed out above, in the proof we may assume that the
support of measure � does not contain the set B(0; 1) and v(0) = 0. In this case
in the formulation of the theorem the integrals (1) and (2) can be replaced by the
integrals
RZ
0
W (t)
t2
(v(t) + v(�t))dt (16)
and ZZ
B(0; R)
1Z
0
W (t)
t2
ln
����1� t2
�2
���� dtd�(�): (17)
An analogical statement was proved by Levin ([4, Ch. 5, x 1]). In our case his
proof is valid.
By the Brelot�Hadamar theorem ([11, Ch. 4, x 2]) for the function v (under
our additional assumptions) the formula
v(z) =
ZZ
C
<
�
ln
�
1�
z
�
�
+
z
�
�
d�(�) + c =z (18)
is true with some real constant c. From this it follows that
RZ
0
W (t)
t2
(v(t) + v(�t))dt =
RZ
0
Z
C
W (t)
t2
ln
����1� t2
�2
���� d�(�)dt:
Let us interchange the order of integration in the last integral. The function
RR
0
W (t)
t2
ln
���1� t2
�2
��� dt is continuous in the variable � on the set C n f0g and of the
form O
�
1
j�j2
�
as � !1. Hence, the integrand from the integral in the right-hand
388 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3
On Subharmonic Functions of the First Order with Restrictions on the Real Axis
part of the last equality falls into the L1-space by measure dt � d�. Due to the
Fubini theorem we have
RZ
0
W (t)
t2
(v(t) + v(�t))dt =
ZZ
C
RZ
0
W (t)
t2
ln
����1� t2
�2
���� dtd�(�): (19)
Now we represent the last integral in this equality as the following sum:
ZZ
C
RZ
0
W (t)
t2
ln
����1� t2
�2
���� dtd�(�) =
ZZ
B(0; R)
1Z
0
W (t)
t2
ln
����1� t2
�2
���� dtd�(�)
+
ZZ
K(R;1)
RZ
0
W (t)
t2
ln
����1� t2
�2
���� dtd�(�)�
ZZ
B(0; R)
1Z
R
W (t)
t2
ln
����1� t2
�2
���� dtd�(�):
Then from (19) we get
RZ
0
W (t)
t2
(v(t) + v(�t)) dt =
ZZ
B(0; R)
1Z
0
W (t)
t2
ln
����1� t2
�2
���� dtd�(�) + I1(R)� I2(R);
(20)
where I1(R), I2(R) are the integrals de�ned in Lems. 1 and 2, respectively. From
Remarks 1 and 2 to the corresponding lemmas one can see that the value '(R) =
I1(R) � I2(R) is bounded. Together with (20) it gives us the assertion of the
theorem.
Let us notice that if �(r) � 1, then we obtain the Akhiezer theorem ([2],
p.475).
Further there will be given a corollary of Th. 1.
Theorem 2. Let v(z) be a subharmonic function of completely regular growth
with respect to the proximate order �(r), �(r) ! 1 as r ! +1, in C . Then the
convergence of the integral
1Z
0
W (t)
1 + t2
(v(t) + v(�t))dt (21)
is equivalent to the convergence of the integral
ZZ
C
1Z
0
W (t)
1 + t2
ln
����1� t2
�2
���� dtd�(�): (22)
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3 389
I.V. Poedintseva
P r o o f. Since the function v is of completely regular growth, then the limit
set of its Riesz measure is single-element ([10, Th. 1.3.1]). Moreover, the support
of the limit measure does not contain any circle. According to Th. 0.50 [9] in this
case E1
� and E2
� are single-element sets, and hence, each of the two values I1(R)
and I2(R) has its limit as R ! +1. Now equality (20) implies the assertion of
the theorem.
It turns out that under some additional restrictions on the function �(r) the
boundness of (4) and (5) implies the existence of the limits of these integrals as
R! +1.
Theorem 3. Let v(z) 2 SF (�(r)), �(r) ! 1 as r ! +1, in C . Also let the
following two conditions be valid:
1) �(r) = �(r)� 1 + r�
0
(r) ln r � 0,
2) rW 0(r) =W1(r), where W1(r) = r�1(r), �1(r) is some zero proximate order.
Then the integral
RZ
0
W (t)
1 + t2
(v(t) + v(�t))dt (23)
is bounded from above if and only if the following two integrals,ZZ
C
W (�)
�
j sin �jd�(�) (24)
and ZZ
C
W1(�)
�
d�(�); (25)
are convergent. Here � is the restriction of the Riesz measure of the function v to
the ring K(1;1).
P r o o f. As it was said above, without loss of generality, we assume that
the function v is harmonic in the unit disk and v(0) = 0. In this case instead of
(23) we may consider the integral
RZ
0
W (t)
t2
(v(t) + v(�t))dt: (26)
Denote
w(�) =
1Z
0
W (t)
t2
ln
����1� t2
�2
���� dt: (27)
The function w(�) satis�es the following asymptotic formula:
w(�) = j sin �j
W (�)
�
+ (1 + o(1))H(�)
W1(�)
�
; (28)
390 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3
On Subharmonic Functions of the First Order with Restrictions on the Real Axis
where
H(�) =
1Z
0
�
1
u2
ln
���1� u2e2i�
���� cos �
u
ln
����u� ei�
u+ ei�
����
�
du:
Now let us evaluate H(�). Integrating by parts, for the �rst summand of the
integrand in the last integral we obtain
1Z
0
1
u2
ln
���1� u2e2i�
��� du = 2<
1Z
0
du
u2 � e�2i�
= <
1Z
�1
du
u2 � e�2i�
:
The evaluation by the residue theorem gives us
1Z
0
1
u2
ln
���1� u2e2i�
��� du = �jsin�j:
Integrating by parts again, for the second summand in the integrand we �nd
1Z
0
cos �
u
ln
����u� ei�
u+ ei�
���� du = �2 cos � <
1Z
0
ei� lnu
u2 � e2i�
du:
Applying again the residue theorem, we get
1Z
0
cos �
u
ln
����u� ei�
u+ ei�
���� du = ��
��
2
� �
�
cos �; � 2 (0; �):
From the evenness of the integral in the variable � it follows that for � 2 (��; 0)
1Z
0
cos �
u
ln
����u� ei�
u+ ei�
���� du = ��
��
2
+ �
�
cos �:
And, �nally, using the continuity of H(�) on the segment [��; �], we arrive
at the formula
H(�) =
8<
:
�
�
�
2
+ �
�
cos � � � sin �; � 2 [��; 0];
�
�
�
2
� �
�
cos � + � sin �; � 2 [0; �]:
It is easy to see that
� � H(�) �
�2
2
:
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3 391
I.V. Poedintseva
Taking into account these inequalities, as well as the relation
W1(r) = rW 0(r) = �W (r)�(r) � 0;
from formula (20) and representation (28) the assertion of the theorem follows
immediately.
It remains to prove formula (28). From (27), integrating by parts we get
w(�) = �
1Z
0
W (t)d (t; �) = lim
"!+0
0
@�W (t) (t; �)j1" +
1Z
"
W 0(t) (t; �)dt
1
A ;
where
(t; �) =
1Z
t
1
x2
ln
����1� x2
�2
���� dx:
It is easy to verify that for t > 0 and � 2 (0; �)
(t; �) = <
�
1
t
ln
�
1�
t2
�2
�
�
1
�
ln
�
t� �
t+ �
��
=
1
t
ln
����1� t2
�2
����� cos �
�
ln
���� t� �
t+ �
����+ sin �
�
arctg
2t� sin �
t2 � �2
+
� sin �
�
�[0;�)(t):
To get this one should use the following relations:
= ln
�
t� �
t+ �
�
= = ln(t� �)(t+ �) = = ln(t2 � �2 � 2i�t sin �)
= � arctg
2t� sin �
t2 � �2
+ k(t)�; lim
t!+1
= ln
�
t� �
t+ �
�
= 0
and the continuity of function in the variable t on the half-axis (0;1).
Thus, we obtain
w(�) = � sin �
W (�)
�
+
sin �
�
1Z
0
W1(t)
t
arctg
2t� sin �
t2 � �2
dt
+
1Z
0
W1(t)
�
1
t2
ln
����1� t2
�2
����� cos �
t�
ln
���� t� �
t+ �
����
�
dt: (29)
Consider the second summand in the right-hand part of (29). We denote this
integral by I5(�). Changing the variable t = u� in it, we have
I5(�) = sin �
W1(�)
�
1Z
0
W1(u�)
W1(�)
1
u
arctg
2u sin �
u2 � 1
du:
392 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3
On Subharmonic Functions of the First Order with Restrictions on the Real Axis
After having changed the variable u = 1
v
one can make sure that
1Z
0
1
u
arctg
2u sin �
u2 � 1
du = 0:
Using the last equality, as well as the mentioned above property of a proximate
order (14) and inequality (11), and reasoning as in the proof of Lem. 1, we arrive
at the following asymptotic formula:
I5(�) = o(1)
W1(�)
�
; � !1:
Then consider the last summand in the right-hand part of (29). Denote this
integral by I6(�) and change the variable t = u� in it. We obtain
I6(�) =
W1(�)
�
1Z
0
W1(u�)
W1(�)
�
1
u2
ln
���1� u2e2i�
���� cos �
u
ln
����u� ei�
u+ ei�
����
�
du:
From here we derive the following asymptotic formula:
I6(�) =
W1(�)
�
(H(�) + o(1)) ; � !1:
And now, due to the evenness of function w(�ei�) in the variable � from (29)
there follows the demanded representation. The theorem is proved.
Theorem 4. Let v(z) 2 SF (�(r)), �(r) ! 1 as r !1, in the plane C and
W1(r) = rW 0(r).
Then the convergence of integrals (24) and (25) implies the boundness
of integral (23).
P r o o f. The assertion of the theorem follows immediately from (28) and
(20).
If we additionally assume that v(z) is the function of completely regular
growth, then using Th. 2 one can get the following corollaries of Ths. 3 and 4.
Theorem 5. Let v(z) be a subharmonic function of completely regular growth
with respect to the proximate order �(r), �(r) ! 1 as r ! +1, in the plane C .
Also let the following two conditions be valid:
1) �(r) = �(r)� 1 + r�
0
(r) ln r � 0,
2) �(r) is such that rW 0(r) =W1(r), where W1(r) = r�1(r), �1(r) is some zero
proximate order.
Then the convergence of integral (23) is equivalent to the convergence of inte-
grals (24) and (25).
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3 393
I.V. Poedintseva
P r o o f. The assertion of the theorem follows immediately from (28) and
Th. 2.
Theorem 6. Let v(z) be a subharmonic function of completely regular growth
with respect to the proximate order �(r), �(r) ! 1 as r ! +1, in the plane C .
Also let W1(r) = rW 0(r).
Then the convergence of integral (23) follows from the convergence of (24)
and (25) .
P r o o f. The assertion of the theorem follows immediately from (28) and
Th. 2.
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394 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3
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