Subharmonic Almost Periodic Functions of Slow Growth
We obtain a complete description of the Riesz measures of almost periodic subharmonic functions with at most of linear growth on C. As a consequence we get a complete description of zero sets for the class of entire functions of exponential type with almost periodic modula.
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irk-123456789-1064402016-09-29T03:02:09Z Subharmonic Almost Periodic Functions of Slow Growth Favorov, S.Yu. Rakhnin, A.V. We obtain a complete description of the Riesz measures of almost periodic subharmonic functions with at most of linear growth on C. As a consequence we get a complete description of zero sets for the class of entire functions of exponential type with almost periodic modula. 2007 Article Subharmonic Almost Periodic Functions of Slow Growth / S.Yu. Favorov, A.V. Rakhnin // Журнал математической физики, анализа, геометрии. — 2007. — Т. 3, № 1. — С. 109-127. — Бібліогр.: 13 назв. — англ. 1812-9471 http://dspace.nbuv.gov.ua/handle/123456789/106440 en Журнал математической физики, анализа, геометрии Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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We obtain a complete description of the Riesz measures of almost periodic subharmonic functions with at most of linear growth on C. As a consequence we get a complete description of zero sets for the class of entire functions of exponential type with almost periodic modula. |
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Article |
| author |
Favorov, S.Yu. Rakhnin, A.V. |
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Favorov, S.Yu. Rakhnin, A.V. Subharmonic Almost Periodic Functions of Slow Growth Журнал математической физики, анализа, геометрии |
| author_facet |
Favorov, S.Yu. Rakhnin, A.V. |
| author_sort |
Favorov, S.Yu. |
| title |
Subharmonic Almost Periodic Functions of Slow Growth |
| title_short |
Subharmonic Almost Periodic Functions of Slow Growth |
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Subharmonic Almost Periodic Functions of Slow Growth |
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Subharmonic Almost Periodic Functions of Slow Growth |
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Subharmonic Almost Periodic Functions of Slow Growth |
| title_sort |
subharmonic almost periodic functions of slow growth |
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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2007 |
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http://dspace.nbuv.gov.ua/handle/123456789/106440 |
| citation_txt |
Subharmonic Almost Periodic Functions of Slow Growth / S.Yu. Favorov, A.V. Rakhnin // Журнал математической физики, анализа, геометрии. — 2007. — Т. 3, № 1. — С. 109-127. — Бібліогр.: 13 назв. — англ. |
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Журнал математической физики, анализа, геометрии |
| work_keys_str_mv |
AT favorovsyu subharmonicalmostperiodicfunctionsofslowgrowth AT rakhninav subharmonicalmostperiodicfunctionsofslowgrowth |
| first_indexed |
2025-07-07T18:29:39Z |
| last_indexed |
2025-07-07T18:29:39Z |
| _version_ |
1837013905252024320 |
| fulltext |
Journal of Mathematical Physics, Analysis, Geometry
2007, vol. 3, No. 1, pp. 109�127
Subharmonic Almost Periodic Functions of Slow Growth
S.Yu. Favorov and A.V. Rakhnin
Department of Mechanics and Mathematics, V.N. Karazin Kharkiv National University
4 Svobody Sq., Kharkiv, 61077, Ukraine
E-mail:Sergey.Ju.Favorov@univer.kharkov.ua
rakhnin@gmail.com
Received August 7, 2006
We obtain a complete description of the Riesz measures of almost periodic
subharmonic functions with at most of linear growth on C . As a consequence
we get a complete description of zero sets for the class of entire functions of
exponential type with almost periodic modula.
Key words: subharmonic function, almost periodic function, Riesz mea-
sure, zero set, entire function of exponential type.
Mathematics Subject Classi�cation 2000: 31A05 (primary); 42A75, 30D15
(secondary).
Bohr's Theorem (see [2], or [10, Ch. 6, � 1]) implies that any almost periodic
function? on real axis R with the bounded spectrum is just the restriction to
R of an entire almost periodic function f of exponential type. Moreover, f has
no zeros outside some strip jImzj � H if and only if supremum and in�mum of
the spectrum f also belong to the spectrum. In [9] (see also [10, Appendix VI])
M.G. Krein and B.Ya. Levin obtained a complete description of zeros of functions
from the last class. Namely, a set fakgk2Z in a horizontal strip of the �nite width
is just a zero set of an entire almost periodic function f of exponential type if and
only if the set is almost periodic and has the representation
ak = dk + (k); k 2 Z; (1)
where d is a constant, the function (k) is bounded, and the values
Sn = lim
r!1
X
jkj<r
[ (k + n)� (k)]
k
k2 + 1
(2)
are bounded uniformly in n 2 Z.
?Explicit de�nitions of almost periodicity for the functions, measures, and discrete sets see
in [10, Ch. 6 and Appendix VI], [11], or Sect. 3 of the present paper.
c
S.Yu. Favorov and A.V. Rakhnin, 2007
S.Yu. Favorov and A.V. Rakhnin
It can be proved that almost periodicity of fakg yields representation (1) and
a �nite limit in (2) for every �xed n 2 Z. Also, one can obtain a complete
description of zero sets for the class of entire functions of exponential type with
almost periodic modulus and zeros in a horizontal strip of �nite width: we should
only replace the Sn by ReSn:
Observe that every entire function of exponential type bounded on R has the
form
f(z) = Cei�z lim
r!1
Y
jakj<r
�
1� z
ak
�
; � 2 R (3)
([10, Ch. 5]; for simplicity we suppose 0 62 fakg); so we have an explicit represen-
tation for the functions from the classes mentioned above.
Note that in [5] one of the authors of the present paper obtained a complete
description of zero sets for holomorphic almost periodic functions on the strip
and on the plane without any growth conditions. An implicit representation for
a special case of almost periodic holomorphic functions was obtained earlier in
[3]. Besides, it was proved in [3] that zero sets of holomorphic functions with the
almost periodic modulus on the strip (or on the plane) are just almost periodic
discrete sets. This result is the consequence of a more general one: every almost
periodic measure on a strip is just the Riesz measure of some subharmonic almost
periodic function on the strip.
In Sect. 3 of our paper we obtain a complete description of the Riesz measures
for almost periodic subharmonic functions of normal type with respect to order
1 (note that it is the smallest growth for the bounded on R subharmonic func-
tion). In particular, we consider the case of periodic subharmonic functions. As
a consequence, we get a complete description of zero sets for the class of entire
functions of exponential type with the almost periodic modulus without any ad-
ditional requirements on distributions of zeros. Note that representation (1) with
a bounded function (k) is incorrect here, therefore the methods used in paper
[9] do not work in our case. The integral representation from [3] creates an almost
periodic subharmonic function with a given almost periodic Riesz measure, but
does not allow to control the growth of the function, therefore it is not �t for our
problem as well.
We make use of a subharmonic analogue of the representation
log jf(z)j =
1Z
0
n(0; t)� n(z; t)
t
dt� �y + log jCj; (4)
for functions of the form (3) (see [6] or review [7, p. 45]); here n(c; t) is a number
of zeros in the disc fz : jz � cj � tg. We obtain this analogue in Sect. 2 of our
paper. Also, we get a complete description of the Riesz measures for the bounded
on R subharmonic functions with at most of linear growth on C .
110 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 1
Subharmonic Almost Periodic Functions of Slow Growth
Here we base on a subharmonic analogue of (3) as well. Of course, this ana-
logue can be obtained by repeating all the steps of the proof (3) for entire functions
in [10], nevertheless we prefer to give a short proof in Sect. 1, using Azarin's the-
ory of limit sets for subharmonic functions [1]. The idea of the proof belongs to
Prof. A.F. Grishin, and the Authors are very grateful to him.
1. In this section we prove the following theorem:
Theorem 1. Let v(z) be a subharmonic function on C such that
v+(z) = O(jzj) as z !1 (5)
and
sup
x2R
v(x) <1: (6)
Then
v(z) = lim
R!1
Z
jwj<R
(log jz � wj � log+ jwj)d�(w) +A1y +A2: (7)
Here z = x+ iy, � = 1
2�
4v is the Riesz measure of v, A1; A2 2 R, and the limit
exists uniformly on compact subsets in C .
Note that the condition (5) means that v is at most of normal type with
respect to the order 1.
Our proof of Th. 1 is based on Azarin's theory of limit sets [1]. Thus, if
a subharmonic function v satis�es (5), then:
a) The family vt(z) = t�1v(tz), t > 1 is a relatively compact set in the space
of distributions D0(C ); in other words, for every sequence of functions from this
family there is a subsequence converging to a subharmonic function.
Note that the convergence in D0(C ) is weak on all functions from the class
of in�nitely smooth compactly supported functions D(C ); moreover, the class of
subharmonic functions is closed with respect to this convergence.
b) If
v1 = lim
t!1
vt(z); (8)
then the Riesz measure �1 of the function v1 satis�es the equality
�1 = lim
t!1
�t; (9)
where �t(E) = t�1�(tE) for the Borel subsets of C ; limits (9) exist in the sense of
weak convergence on the continuous compactly supported functions on C . More-
over, in this case there exists
lim
R!1
Z
1�jwj<R
d�(w)
w
6=1: (10)
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 1 111
S.Yu. Favorov and A.V. Rakhnin
If a subharmonic function satis�es (5) and (8), then it is called a completely
regular growth (with respect to the order 1).
In what follows we need a simple criterion of the compactness for the family
of subharmonic functions (see[1]).
Lemma A. A family fu�g of subharmonic functions on C is a relatively
compact set in the space of distributions D0(C ) if and only if
a) sup� supz2K u�(z) <1 for all compacta K � C ,
b) inf� supz2K0
u�(z) > �1 for some compact set K0 � C .
Also, we need the following variant of the Fragmen�Lindelof theorem.
Theorem FL. If a function v is subharmonic in the neighborhood of closure
of the upper half-plain C
+ = fz = x + iy : y > 0g and satis�es conditions (5),
(6), then for all z 2 C
+
v(z) � sup
x2R
v(x) + �+y;
with �+ = lim sup
y!+1
y�1v(iy).
The proof of this statement is the same as for holomorphic on C + and contin-
uous on C + functions (see, for example, [8, p. 28]).
First, let us prove a subharmonic analogue of the Cartwright theorem (a holo-
morphic case see, for example, in [10, Ch. V]).
Theorem 2. Assume that a subharmonic function v on C satis�es (5) and
(6). By de�nition, put
�� = lim sup
y!�1
v(iy)
jyj : (11)
Then v(z) is a completely regular growth; moreover, the function v1 from (8) has
the form
v1(z) =
�
�+y; y � 0;
��jyj; y < 0;
(12)
R e m a r k. From Theorem FL it follows that if a subharmonic function v on
C satis�es the conditions of Th. 2 with �+ � 0 and �� � 0, then v is a constant.
The proof is based on the following lemma
Lemma 1. Let u < 0 be a subharmonic function on C
+ . Then for every
R <1 and r 2 (0; R=2)
u(ir)
r
� C
R3
Z
jz�iRj<R
u(z)dm2(z); (13)
where C is an absolute constant and m2 is a plain Lebesque measure.
112 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 1
Subharmonic Almost Periodic Functions of Slow Growth
P r o o f. Since Poisson formula for the disc B(iR=2; R=2) = fz : jz�iR=2j <
R=2g, we have
u(ir) � 1
2�
2�Z
0
u(iR=2+ei�R=2)
(R=2)2 � (R=2� r)2
(R=2)2 + (R=2� r)2 �R(R=2� r) cos(�=2 + �)
d�:
Using the inequality u < 0, replace the interval of integration by [�=4; 3�=4]. We
obtain
u(ir) � r
(R� r)
1
4
sup
�2[�=4;3�=4]
u
�
iR
2
+
R
2
ei�
�
: (14)
Since for all � 2 [�=4; 3�=4]
B(iR;R=2) � B(iR=2 + ei�R=2; (1 +
p
2
2
)R=2) � C
+ ;
we have
u(
iR
2
+
R
2
ei�) � 4
�R2(1 +
p
2
2
)2
Z
jz�iR=2�ei�R=2j�(1+
p
2
2
)R=2
u(z)dm2(z)
� 8
�R2(3 + 2
p
2)
Z
jz�iRj<R=2
u(z)dm2(z):
Then replace the average over the disc B(iR;R=2) by the average over the disc
B(iR;R). So, the assertion of the lemma follows from (14).
P r o o f o f T h e o r e m 2. Without loss of generality, it can be assumed
that sup
x2R
v(x) = 0: Put u(z) = v(z) � �+y. From Theorem FL it follows that
u(z) < 0 on C + , then
lim sup
y!+1
u(iy)
y
= 0: (15)
Fix z0 2 C
+ . Let ', 0 � ' � 1, be an in�nite di�erentiable and compactly
supported function on C
+ , depending only on jz � z0j. Apply Lemma 1 for the
function ut(z) = u(tz)t�1. If supp' � B(iR;R), then we get
u(itr)
tr
� C
R3
Z
jz�iRj<R
ut(z)dm2(z) � C
R3
Z
C+
ut(z)'(z)dm2(z):
Since (15), we see that for any " > 0 and t > t(") there is r 2 (0; R=2) such that
u(itr) � �"tr. We obtainZ
C+
ut(z)'(z)dm2(z) � �"R
3
C
:
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 1 113
S.Yu. Favorov and A.V. Rakhnin
Hence, Z
C+
ut(z)'(z)dm2(z)! 0 as t!1:
Therefore, vt(z) = ut(z) + �+y ! �+y in the space D0(C +). Similarly, vt(z) !
���y in the space D0(C �), where C � = fz = x + iy : y < 0g. Every limit
function for vt(z) is always subharmonic, therefore we get (12). So limit (8)
exists. Theorem 2 is proved.
Consequence. The Riesz measure of the limit function v1(z) equals
�+ + ��
2�
m1(x);
where m1 is the Lebesgue measure on R.
P r o o f o f T h e o r e m 1. From the Jensen�Privalov formula for the
subharmonic function, we get the estimate for r � 1
�(B(0; r)) � C1r; (16)
where constant C1 depends only on v. Using the Brelot�Hadamard theorem
for the subharmonic function (see, for example, [12]), we obtain that there is
a harmonic polynomial H(z) of degree 1 such that
v(z) =
Z
jwj<1
log jz � wjd�(w) +
Z
jwj�1
log
����1� z
w
���+Re
z
w
�
d�(w) +H(z): (17)
Denote by v0(z) the �rst integral in (17). Since (10), we get
v(z) = v0(z) + lim
R!1
Z
1�jwj<R
log
���1� z
w
��� d�(w) +A0x+A1y +A2: (18)
The application of Th. 2 yields that the function v is a completely regular growth,
hence the measures �t converge weakly to the measure
�1 =
�+ + ��
2�
m1: (19)
Let �0 be a restriction of the measure � to C n B(0; 1). Obviously, the measures
�0t converge weakly to the measure �1 as well. Therefore,
vt(z) = v0t (z) + lim
R!1
Z
jwj<R
log
���1� z
w
��� d�0t(w) +A0x+A1y +
A2
t
: (20)
114 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 1
Subharmonic Almost Periodic Functions of Slow Growth
Pass to a limit in (20) as t!1 in the space D0(C ). First, by Th. 2, the func-
tions vt(z) converge to the function v1(z) from (12). Since v0(z) = O(log jzj) as
jzj ! 1, we see that v0t (z) ! 0. By (16), we obtain �0t(B(0; R)) � C1R for all
t � 1 and R > 0. Therefore, we get uniformly in t � 1,
Z
jwj�R
d�0t(w)
jwj2 = 2
1Z
R
�t(B(0; s))
s3
ds� �t
0(R)
R2
! 0; (21)
as R!1. Also, by (10), uniformly in t � 1, R0 � R
Z
R�jwj�R0
d�0t(w)
w
=
Z
Rt�jwj�R0t
d�(w)
w
! 0; (22)
as R!1. For jzj < C and a su�ciently large jwj
���log
���1� z
w
���+Re
z
w
��� � jzj2
jwj2 :
Therefore, taking into account (21) and (22), we obtain for all ' 2 D(C ) uniformly
in t � 1
Z
C
0
B@ lim
R!1
Z
jwj�R
log
���1� z
w
��� d�t0(w)
1
CA'(z)dm2(z)
= lim
R!1
Z
jwj�R
0
@Z
C
log
���1� z
w
���'(z)dm2(z)
1
A d�t
0(w): (23)
Note that the measure �1 does not charge any circle jwj = R, therefore the
restrictions of the measures �t
0(w) to any disc B(0; R) converge weakly to the
restriction of the measure �1(w). The function
R
log jz � wj'(z)dm2(z) is con-
tinuous in the variable w, so we have
lim
t!1
Z
jwj�R
Z
log jz�wj'(z)dm2(z)d�t(w) =
Z
jwj�R
Z
log jz�wj'(z)dm2(z)d�1(w):
(24)
By the same reason for each Æ > 0
lim
t!1
Z
�jwj�R
log jwj�t(w) =
Z
�jwj�R
log jwj�1(w): (25)
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 1 115
S.Yu. Favorov and A.V. Rakhnin
Furthermore,
Z
jwj�Æ
log jwj�0t(w) = log Æ
�0(B(0; Æt))
t
�
ÆZ
0
�0(B(0; st))
st
ds: (26)
Since �0(B(0; r)) � C1r for all r > 0, we see that (26) tends to zero as Æ ! 0
uniformly in t � 1. Combining (23)�(25), and (26), we get the equality
v1(z) = lim
R!1
Z
jwj�R
log
���1� z
w
��� d�1(w) +A0x+A1y:
Take y = 0. Since v1(x) = 0 and
lim
R!1
Z
juj�R
log
���1� x
u
��� du = 0
for all x 2 R, we obtain A0 = 0. Now the assertion of Th. 1 follows from (18).
2. Here we get a complete description of the Riesz measures for subharmonic
functions with at most of linear growth on C (i.e., not exceeding C(jzj+ 1) with
C <1) and with some additional conditions (bounded on R or with the compact
family of translations along R). Holomorphic analogues of the corresponding
theorems were obtained earlier by one of the Authors in [6].
First, prove some lemmas.
Lemma 2. If a measure � on C satis�es the condition
�(B(0; R + 1))� �(B(0; R)) = o(R) as R!1; (27)
and the limit
lim
R!1
Z
jwj<R
(log+ jz � wj � log+ jwj)d�(w)
exists at some point z 2 C , then the limit equals
1Z
1
�(B(0; t))� �(B(z; t))
t
dt:
P r o o f. For all z 2 C and R 2 (jzj+ 1;1) we have
Z
jwj<R
log+ jz � wjd�(w) �
Z
jwj<R
log+ jwjd�(w) = (logR)�(B(z;R))
116 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 1
Subharmonic Almost Periodic Functions of Slow Growth
�
RZ
1
�(B(z; t))
t
dt� (logR)�(B(0; R)) +
RZ
1
�(B(0; t))
t
dt
+
Z
jwj<R; jw�zj�R
log+ jz � wjd�(w) �
Z
jwj�R; jw�zj<R
log+ jz � wjd�(w)
=
RZ
1
�(B(0; t)) � �(B(z; t))
t
dt
+
Z
jwj<R; jw�zj�R
log
����z � w
R
���� d�(w)�
Z
jwj�R; jw�zj<R
log
����z � w
R
���� d�(w): (28)
If jwj < R and jz � wj � R or jwj � R and jz � wj < R, then we have
1� jzj
R
�
����z � w
R
���� � 1 +
jzj
R
:
Therefore the integrand functions of the last two integrals in (28) are O(1=R) as
R!1. The domains of integrations are the subsets of the ring R� jzj � jwj �
R+ jzj, hence, by (27), these integrals tend to 0 as R!1. Lemma 2 is proved.
Lemma 3. Assume that a measure � satis�es (10), (16), and (27), and
V (z) = lim
R!1
Z
jwj<R
(log jz � wj � log+ jwj)d�(w): (29)
Then V is a subharmonic function with the Riesz measure � and
V (z) =
1Z
1
�(B(0; t)) � �(B(z; t))
t
dt+
Z
jw�zj<1
log jz � wjd�(w) (30)
for all z 2 C . Furthermore, the function
~V (z) =
1
2�
2�Z
0
V (z + ei�)d� (31)
satis�es the equality
~V (z) =
1Z
1
�(B(0; t)) � �(B(z; t))
t
dt: (32)
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 1 117
S.Yu. Favorov and A.V. Rakhnin
P r o o f. It follows from (16) that the integral in (17) is a subharmonic
function on C with the Riesz measure �; besides, it satis�es (5) (see, for example,
[12, Ch. 1]). If, in addition, � satis�es (10), then the limit in (29) exists uniformly
on bounded sets, and the function V coincides with the integral in (17) up to
a linear term; so it has the same properties as well.
Using the equality
R 2�
0
log ja+ ei�jd� = 2� log+ jaj, we get
~V (z) = lim
R!1
Z
jwj<R
(log+ jz � wj � log+ jwj)d�(w):
and V (z) = ~V (z) +
R
jw�zj<1
log jz � wjd�(w). Then Lemma 2 implies (32) and
(30). Lemma 3 is proved.
Lemma 4. Assume that a subharmonic on C function v satis�es (5). Then
the family of translations fv(z + h)gh2R is a relatively compact subset in D0(C ) if
and only if the function
~v(z) =
1
2�
Z 2�
0
v(z + ei�)d� (33)
is bounded on R; this function is bounded simultaneously with the function
v̂(z) =
1
�
Z
jwj<1
v(z + w)dm2(w): (34)
P r o o f. If the family fv(z+h)gh2R is a relatively compact subset, then it is
uniformly bounded from above on compacta in C and the function v is uniformly
bounded from above on every strip jyj < H. Then by Lemma A there is a compact
subset K0 of C such that
sup
K0
v(z + h) � C2; 8h 2 R:
Take d > supK0
jzj. Then for any h 2 R there is a point z(h), jz(h)j < d, such
that
1
(d+ 1)2�
Z
jw�z(h)j�d+1
v(w + h)dm2(w) � v(z(h) + h) > C2 � 1:
Further, we have
Z
jw�z(h)j�d+1
v(w + h)dm2(w) �
Z
jwj�1
v(w + h)dm2(w) + (d2 + 2d)� sup
jyj<d+1
v(z):
118 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 1
Subharmonic Almost Periodic Functions of Slow Growth
Therefore, infh2R v̂(h) > �1. Since ~v(h) � v̂(h), we see that the functions v̂ and
~v are bounded uniformly from below on R. It is clear that these functions are
bounded uniformly from above on R as well.
On the other hand, if ~v(z) is bounded from below on R, then
infh2R supjwj=1 v(h+w) > �1; if ~v(h) is bounded from above on R, then v(h) is
bounded from above on R as well. Since Theorem FL, we see that v(z) is bounded
from above on every strip jyj < H. It follows from Lemma A that fv(z + h)gh2R
is a relatively compact set. Hence v̂(x) is bounded on R.
Now we can prove the theorems mentioned above.
Theorem 3. For a measure � on C to be the Riesz measure for some sub-
harmonic function satisfying conditions (5) and (6), it is necessary and su�cient
that the conditions (10), (16), (27), and
sup
x2R
1Z
1
�(B(0; t))� �(B(x; t))
t
dt <1: (35)
were ful�lled.
P r o o f. If a subharmonic function v satis�es (5) and (6), then, by Th. 2,
its Riesz measure � satis�es (10) and the measures �t converge to the measure
�1 = (�+ + ��)(2�)�1m1. The last measure does not charge any circle jwj = R,
therefore (9) implies that �(B(0; R)) = CR+ o(R) as R!1. Hence we get (16)
and (27). By Theorem 1, v(z) = V (z) + A1y + A2. So the function V is also
bounded from above on R. Since Theorem FL, we see that the same is true for
the function ~V from (31). Now Lemma 3 implies (35).
Conversely, if a measure � satis�es (10), (16), and (27), then, by Lemma 3,
the subharmonic function V has the Riesz measure � and satis�es (5) and (6).
Theorem 3 is proved.
Theorem 4. For a measure � on C to be the Riesz measure for some sub-
harmonic function v with the property (5) such that the family of translations
fv(z + h)gh2R is a relatively compact subset in D0(C ), it is necessary and su�-
cient that the conditions (10), (16), (27), and
sup
x2R
������
1Z
1
�(B(0; t)) � �(B(x; t))
t
dt
������ <1 (36)
were ful�lled.
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 1 119
S.Yu. Favorov and A.V. Rakhnin
P r o o f. If a function v satis�es (5), and the function ~v from (33) is bounded
on R, then v is bounded from above on R. Theorem 3 implies conditions (10),
(16), and (27) for the Riesz measure � of the function v. By Theorem 1, ~v equals
~V from (31) on R up to a constant term, therefore Lemmas 3 and 4 imply (36).
Conversely, if a measure � satis�es (10), (16), (27), and (36), then V from
(29) satis�es (5), and the function ~V is bounded on R.
Hence the assertion of Th. 4 follows from Lemma 4.
3. A continuous function F (z) on a closed strip fz = x+iy : x 2 R; jyj � Hg
with H � 0 is almost periodic if the family of translations fF (z + h)gh2R is
a relatively compact set with respect to the topology of uniform convergence on
the strip; a function is almost periodic on an open strip (in particular, on C ), if
it is almost periodic on every closed substrip of a �nite width.
A measure (maybe complex) � on C is called almost periodic if for any test-
function ' 2 D(C ) the convolution
R
'(w+t)d�(w) is an almost periodic function
in t 2 R ([11]).
The following statement is valid:
Theorem R (Theorem 1.8 [11]). For a measure � to be almost periodic
it is necessary and su�cient to ful�l the following condition: for any sequence
fhng � R there should exist a subsequence fh0ng such that the convolutions
R
'(w+
x + h0n)d�(w) would converge uniformly with respect to x 2 R and ' 2 L, where
L is a compact subset in D(C ). Moreover, for a measure to be almost periodic it
is su�cient to check this condition only for all single-point sets L � D(C ).
If we take L = f'(z + iy)gjyj�H for some ' 2 D(C ), we obtain that the
convolutions
R
'(w+z+h0n)d�(w) actually converge uniformly on any strip jyj �
H, hence the function
R
'(w + z)d�(w) is almost periodic on C .
Further, a subharmonic function v on C is called almost periodic, if the measure
v(z)dm2(z) is almost periodic (see [3]; an equivalent de�nition see in [4]).
It follows from the de�nition that the Riesz measure of an almost periodic
subharmonic function is also almost periodic. Conversely, every almost periodic
measure is the Riesz measure of some almost periodic subharmonic function (see
[3], where the case of the strip is studied as well). Note that the family of trans-
lations fv(z + h)gh2R is a relatively compact subset of D0(C ) for every almost
periodic subharmonic function v on C . Note also that any almost periodic sub-
harmonic function is bounded from above on every horizontal strip of a �nite
width (see [3]).
Here we obtain the following result:
Theorem 5. A necessary and su�cient condition for a measure � on C to be
the Riesz measure of some almost periodic subharmonic function at most of linear
growth is that the measure should be almost periodic and satisfy (16), (27), (10),
and (36).
120 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 1
Subharmonic Almost Periodic Functions of Slow Growth
The proof of the theorem is based on the following lemmas.
Lemma 5. Suppose subharmonic functions v1(z) and v2(z) on C with the
common Riesz measure satisfy (5) and (6); then v1(z) = v2(z)+p1+p2y. Further,
if
sup
R
v1(x) = sup
R
v2(x) and �+(v1) = �+(v2);
where �+ is de�ned in (11), then v1 � v2.
P r o o f. The �rst part follows from Th. 1, the second one is evident.
Lemma 6. Assume that a subharmonic on C function v satis�es (5) and
a family fv(z+h)gh2R is a relatively compact subset of D0(C ); if v(z+hn)! v�(z)
in the space D0(C ), then the family fv�(z + h)gh2R is a relatively compact subset
as well and
sup
x2R
v�(x) � sup
x2R
v(x); (37)
inf
t2R
Z
jz�tj<1
v�(z)dm2(z) � inf
t2R
Z
jz�tj<1
v(z)dm2(z); (38)
�+(v
�) � �+(v); ��(v
�) � ��(v): (39)
P r o o f. Put M = sup
x2R
v(x). By Theorem FL, we get for any " > 0
v(z) �M + 2"maxf�+(v); ��(v)g; jyj < 2":
Let ' be a function from D(C ) such that ' depends only on jzj, ' � 0, '(z) = 0
for jzj � ",
R
'(z)dm2(z) = 1. Then
(v � ')(z) �M + 2"maxf�+(v); ��(v)g; jyj < ":
Therefore,
(v� � ')(z) �M + 2"maxf�+(v); ��(v)g; jyj < ":
Note that v� is a subharmonic function, hence (v� � ')(z) � v�(z). Since " is
arbitrary, we obtain (37). By the same argument, for all y 2 R
sup
x2R
v�(x+ iy) � sup
x2R
v(x+ iy):
Therefore we obtain (5) and (39).
Further, the functions v(z + hn) are integrable on every disc and uniformly
bounded from above, therefore we can replace the convergence of measures v(z+
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 1 121
S.Yu. Favorov and A.V. Rakhnin
hn)dm2(z) in the sense of distributions by the weak convergence of measures.
Since the limit measure v�(z)dm2(z) does not charge any circle, we have
lim
n!1
Z
jz�tj<1
v(z + hn)dm2(z) =
Z
jz�tj<1
v�(z)dm2(z)
for each t 2 C . Hence we get (38). Taking into account Lemma 4, we see that the
family fv�(z + h)gh2R is a compact subset of D0(C ). The lemma is proved.
Lemma 7. Under the conditions of the previous lemma, suppose that the Riesz
measure � of the function v(z) is almost periodic. Then inequalities (37)�(39)
turn into equalities, the Riesz measure �� of the function v�(z) becomes almost
periodic, and there is a subsequence fhn0g such that for every ' 2 D(C )
lim
n0!1
sup
t2R
����
Z
'(w � t� hn0)d�(w) �
Z
'(w � t)d��(w)
���� = 0: (40)
P r o o f. For all ' 2 D(C ) we have
lim
n!1
Z
'(z�hn)v(z)dm2(z) = lim
n!1
Z
'(z)v(z+hn)dm2(z) =
Z
'(z)v�(z)dm2(z):
Since � = (2�)�14v, we obtain
lim
n!1
Z
'(z � hn)d�(z) =
Z
'(z)d��(z): (41)
From Theorem R it follows that there is a subsequence fhn0g such that for
any ' 2 D(C ) the almost periodic functions
R
'(z � t� hn0)d�(z) converge to an
almost periodic function uniformly in t 2 R. If we replace z by z� t in (41), then
we get (40). Consequently, the function
R
'(w � t)d��(w) is almost periodic in
t 2 R, and �� is an almost periodic measure.
Passing to a subsequence again if necessary, we may assume that the functions
v�(z�hn) converge in the space D0(C ) to some subharmonic function v��(z) with
the Riesz measure ���. Therefore,
lim
n0!1
Z
'(z + hn0)d�
�(z) =
Z
'(z)d���(z):
On the other hand, it follows from (40) that
lim
n!1
����
Z
'(w)d�(w) �
Z
'(w + hn)d�
�(w)
���� = 0:
122 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 1
Subharmonic Almost Periodic Functions of Slow Growth
Here ' is an arbitrary function from D(C ), hence, ��� = �. By Lemma 5, we get
v��(z) = v(z) +D1 +D2y.
Since (37) is valid for the pairs v; v� and v�; v��, we get D1 � 0. Then (38) for
the pairs v; v� and v�; v�� implies D1 � 0, and we obtain D1 = 0 and the equality
in (37) and (38). By the same way, we obtain the equalities in (39). Lemma 7 is
proved.
P r o o f o f T h e o r e m 5. The necessity follows immediately from Th. 4.
Let us prove a su�ciency. Suppose � satis�es the conditions of Th. 5. Let V be
the function from (29), and let fhng � R be an arbitrary sequence. It follows
from Th. 4 that the family fV (z + hn)g is a relatively compact subset of D0(C ).
Therefore we can assume without loss of generality that V (z + hn) ! V �(z) in
D0(C ). To prove the Th. 5, we need to check that
Z
'(z � t� hn)V (z)dm2(z) !
Z
'(z � t)V �(z)dm2(z)
uniformly in t 2 R for any ' 2 D(C ).
Assume the contrary. Then there is '0 2 D(C ), "0 > 0, and ftng � R such
that����
Z
'0(w)V (w + hn + tn)dm2(w)�
Z
'0(w)V
�(w + tn)dm2(w)
���� � "0: (42)
(If necessary we can replace the sequence fhng by a subsequence.)
We may assume also that V (z + hn + tn)! V ��(z), V �(z + tn)! V ���(z) in
D0(C ) as n ! 1. By ��, ���, ���� denote the Riesz measures of the functions
V �, V ��, V ���, respectively. Then we have
lim
n!1
Z
'(w � hn � tn)d�(w) =
Z
'(w)d���(w); (43)
lim
n!1
Z
'(w � tn)d�
�(w) =
Z
'(w)d����(w) (44)
for any ' 2 D(C ).
On the other hand, the measure � satis�es (40). Hence the integrals in the
left-hand sides of (43) and (44) have the same limit, and ��� = ����.
By Lemma 7 we obtain
sup
R
V ��(x) = sup
R
V (x) = sup
R
V �(x) = sup
R
V ���(x)
and
�+(V
��) = �+(V ) = �+(V
�) = �+(V
���):
Using Lemma 5, we get V �� � V ���. This contradicts (42). Theorem 5 is proved.
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 1 123
S.Yu. Favorov and A.V. Rakhnin
Now let d be a divisor in C , i.e., a sequence fakg � C without �nite limit
points such that any value may appear with a �nite multiplicity. A divisor is
called almost periodic if the discrete measure supported at the points ak with the
mass at every point being equal the multiplicity of the point in the sequence is
almost periodic (see [12], [3]; in [3] there is an equivalent geometric de�nition).
Moreover, almost periodic divisors are just the divisors of entire functions? with
almost periodic modulus in every substrip fz = x+ iy : jyj < Hg. ([3]).
Theorem 6. For a divisor fakg to be the divisor of an entire function of
exponential type with the almost periodic modulus, it is necessary and su�cient
to ful�l the following conditions:
a) the divisor should be almost periodic,
b) there should be a �nite limit
lim
R!1
X
jakj<R
1
ak
;
c) n(0; t) = O(t),
d) n(0; t+ 1)� n(0; t) = o(t),
e)
sup
x2R
������
1Z
1
n(0; t)� n(x; t)
t
dt
������ <1;
here n(c; t) = cardfk : jak � cj � tg.
P r o o f. By Theorem 6, conditions a)�e) mean just the existence of an
almost periodic subharmonic function v at most of linear growth with the Riesz
measure supported at the points fakg with the mass being equal the multiplicity
of the point in the sequence. Then v(z) = log jf(z)j for an entire function f of
exponential type such that the divisor of f is fakg. Since v is almost periodic, we
get that jf j is almost periodic (see [3]). The theorem is proved.
Now we consider a periodic case.
Theorem 7. A necessary and su�cient condition for a measure � on C to be
the Riesz measure of some periodic subharmonic function with period 1 at most
of linear growth is that the measure should be stable with respect to the translation
on 1 and
�fz = x+ iy : 0 � x < 1; y 2 Rg <1: (45)
?A divisor fakg is the divisor of a holomorphic function f when zeros of f coincide with the
values fakg and the multiplicity of every zero equals the multiplicity of corresponding ak.
124 Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 1
Subharmonic Almost Periodic Functions of Slow Growth
P r o o f. Let v be a subharmonic function such that v(z + 1) = v(z). It is
clear that its Riesz measure is stable with respect to the translation on 1. By
Theorem 1, it follows that � satis�es (16). Using the equality
�fz = x+ iy : 0 � x < 1; jyj < ng =
1
2n
�fz = x+ iy : �n � x < n; jyj < ng;
(46)
we get (45).
Conversely, let � be stable with respect to the translation on 1 and satisfy
(45). Using (46), we obtain (16). Then for any r > 0
�fz : r � jzj < r + 1g �
X
n2Z; jnj�r+1
�fz : x 2 [0; 1); r � jz + nj < r + 1g: (47)
Fix Æ 2 (0; 1). For jnj < r(1� Æ)� 1 we have
fz : x 2 [0; 1); r � jz + nj < r + 1g � fz : x 2 [0; 1); jyj > rÆg
Hence (47) is majorized by
2r(1� Æ)�fz : x 2 [0; 1); jyj > rÆg+ (2Ær + 5)�fz : x 2 [0; 1); y 2 Rg: (48)
It follows from (45) that for any " > 0 there exist Æ > 0 and r0 < 1 such that
for r � r0 (48) is less than r". This yields (27).
Further, take R > r > 1. We have�������
Z
r<jzj<R
d�(z)
z
�
Z
r<j[x]+iyj<R
d�(z)
z
�������
� 1
r � 1
�fz : r � 1 < jzj < r + 1g+ 1
R� 1
�fz : R� 1 < jzj < R+ 1g; (49)
where [x] is the integral part of real x. By (27), the right-hand side of (49) tends
to zero as r !1. Then we obtainZ
r<j[x]+iyj<R
d�(z)
z
=
X
n2Z: jnj�R
Z
x2[0;1); r<jn+iyj<R
d�(z)
z + n
=
Z
x2[0;1); y2R
X
n2Z: r<jn+iyj<R
z + n
jz + nj2d�(z): (50)
Now for any x 2 [0; 1), y 2 R we have
X
n2Z: jn+iyj>r
jzj
jz + nj2 �
X
n2NSf0g: n2>r2�y2
1 + jyj
n2 + y2
+
X
n2N: n2>r2�y2
1 + jyj
(n� 1)2 + y2
Journal of Mathematical Physics, Analysis, Geometry, 2007, vol. 3, No. 1 125
S.Yu. Favorov and A.V. Rakhnin
� 2
1Z
p
maxf0;r2�y2g
1 + jyj
t2 + y2
dt+
2(1 + jyj)
y2
�r(y); (51)
here �r(y) is a characteristic function of the interval (
p
r2 � 1;1). Besides,
�
X
n2Z: r<jn+iyj<R
n
jz + nj2 =
X
n2N: r<jn+iyj<R
n
�
1
jz � nj2 �
1
jz + nj2
�
=
X
n2N: r2<n2+y2<R2
4n2x
((n� x)2 + y2)((n+ x)2 + y2)
: (52)
It is easy to see that the right-hand side of (52) is also majorized by (51). Both
terms monotonically decrease to 0 as r ! 1, hence (50) tends to 0 as r ! 1
uniformly in R. It follows from (49) and (50) that (10) is valid. Finally, by (31)
and (32), the integral
1Z
1
�(B(0; t)) � �(R(z; t))
t
dt
is bounded for z = x 2 [0; 1]. Since � is stable with respect to the translation on
1, we get (36). Now the assertion of Th. 7 follows from Th. 5.
Consequence. The necessary and su�cient conditions for a divisor fakg to
be the divisor of an entire periodic (with period 1) function of exponential type
bounded on real axis are that the divisor should be periodic with period 1 and its
restriction to the strip fz : 0 � x < 1; y 2 Rg should be �nite.
In this case the corresponding function is a �nite product of the elementary
functions
p
1� cos 2�(z �
k) with Re
k 2 [0; 1); it is unique up to a multiplier
Cei�z, � 2 R.
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