Approximation of Subharmonic Functions in the Unit Disk
I would like to thank Prof. O. Skaskiv, who read the paper and made valuable suggestions, as well as other participants of the Lviv seminar on the theory of analytic functions for their valuable comments that contributed to the improvement of the initial version of the paper. I wish also to thank th...
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irk-123456789-1065032016-09-30T03:02:52Z Approximation of Subharmonic Functions in the Unit Disk Chyzhykov, I.E. I would like to thank Prof. O. Skaskiv, who read the paper and made valuable suggestions, as well as other participants of the Lviv seminar on the theory of analytic functions for their valuable comments that contributed to the improvement of the initial version of the paper. I wish also to thank the anonymous referee for a number of useful comments and corrections. 2008 Article Approximation of Subharmonic Functions in the Unit Disk / I.E. Chyzhykov // Журнал математической физики, анализа, геометрии. — 2008. — Т. 4, № 2. — С. 211-236. — Бібліогр.: 11 назв. — англ. 1812-9471 http://dspace.nbuv.gov.ua/handle/123456789/106503 en Журнал математической физики, анализа, геометрии Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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I would like to thank Prof. O. Skaskiv, who read the paper and made valuable suggestions, as well as other participants of the Lviv seminar on the theory of analytic functions for their valuable comments that contributed to the improvement of the initial version of the paper. I wish also to thank the anonymous referee for a number of useful comments and corrections. |
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Chyzhykov, I.E. |
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Chyzhykov, I.E. Approximation of Subharmonic Functions in the Unit Disk Журнал математической физики, анализа, геометрии |
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Chyzhykov, I.E. |
| author_sort |
Chyzhykov, I.E. |
| title |
Approximation of Subharmonic Functions in the Unit Disk |
| title_short |
Approximation of Subharmonic Functions in the Unit Disk |
| title_full |
Approximation of Subharmonic Functions in the Unit Disk |
| title_fullStr |
Approximation of Subharmonic Functions in the Unit Disk |
| title_full_unstemmed |
Approximation of Subharmonic Functions in the Unit Disk |
| title_sort |
approximation of subharmonic functions in the unit disk |
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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2008 |
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http://dspace.nbuv.gov.ua/handle/123456789/106503 |
| citation_txt |
Approximation of Subharmonic Functions in the Unit Disk / I.E. Chyzhykov // Журнал математической физики, анализа, геометрии. — 2008. — Т. 4, № 2. — С. 211-236. — Бібліогр.: 11 назв. — англ. |
| series |
Журнал математической физики, анализа, геометрии |
| work_keys_str_mv |
AT chyzhykovie approximationofsubharmonicfunctionsintheunitdisk |
| first_indexed |
2025-07-07T18:34:36Z |
| last_indexed |
2025-07-07T18:34:36Z |
| _version_ |
1837014218223648768 |
| fulltext |
Journal of Mathematical Physics, Analysis, Geometry
2008, vol. 4, No. 2, pp. 211�236
Approximation of Subharmonic Functions
in the Unit Disk
I.E. Chyzhykov
Faculty of Mechanics and Mathematics, Ivan Franko Lviv National University
Universytets'ka, 1, Lviv, 79000, Ukraine
E-mail:ichyzh@lviv.farlep.net
Received October 9, 2006
We prove that if u is a subharmonic function in D = fjzj < 1g, then
there exists an absolute constant C and an analytic function f in D such
that
R
D
ju(z)� log jf(z)jj dm(z) < C, where m denotes the plane Lebesgue
measure. We also (following the arguments of Lyubarskii and Malinnikova)
answer Sodin's question, namely, we show that the logarithmic potential of
measure � supported in a square Q, with �(Q) being an integer N , admits
approximations by the subharmonic function log jP (z)j, where P is a poly-
nomial with
R
Q
jU�(z) � log jP (z)jjdxdy = O(1), independent of N and �.
We also consider uniform approximations.
Key words: subharmonic function, approximation, Riesz measure, ana-
lytic function.
Mathematics Subject Classi�cation 2000: 31A05, 30E10.
1. Introduction
We use the standard notions of subharmonic function theory [1]. Let U(E; t) =
f� 2 C : dist (�; E) < tg, E � C , t > 0, where dist (z;E)
def
= inf�2E jz � �j, and
U(z; t) � U(fzg; t) for z 2 C . A class of subharmonic functions in a domain
G � C is denoted by SH(G). For a subharmonic function u 2 SH(U(0; R)),
0 < R � +1, we write B(r; u) = maxfu(z) : jzj = rg, 0 < r < R and de-
�ne the order �[u] by �[u] = lim sup
r!+1
logB(r; u)= log r if R = 1 and by �[u] =
lim sup
r!R
logB(r; u)= log 1
R�r if R <1.
Let also �u denote the Riesz measure associated with the subharmonic func-
tion u, n(r; u) = �u(U(0; r)), let m be the planar Lebesgue measure and l be
the Lebesgue measure on the positive ray. For an analytic function f in D we
write Zf = fz 2 D : f(z) = 0g. The symbol C(�) with indices stands for some
c
I.E. Chyzhykov, 2008
I.E. Chyzhykov
positive constants depending only on the values in brackets. We write a � b
if C1a � b � C2a for some positive constants C1 and C2, and a(r) � b(r) if
limr!R a(r)=b(r) = 1.
An important result was proved by R.S. Yulmukhametov [2]. For any function
u 2 SH(C ) of order � 2 (0;+1), and � > �, there exists an entire function f and
a set E� � C such that
��u(z) � log jf(z)j�� � C(�) log jzj; z !1; z 62 E�; (1.1)
and E� can be covered by a family of disks U(zj ; tj), j 2 N, with
P
jzjj>R tj =
O(R���), (R! +1).
If u 2 SH(D ), a counterpart of (1.1) holds with log 1
1�jzj instead of log jzj and
an appropriate choice of E�.
From the recent result by Yu. Lyubarskii and Eu. Malinnikova [3] it follows
that for L1 approximation relative to planar measure, we may drop the assumption
that u has a �nite order of growth and obtain sharp estimates.
Theorem A ([3]). Let u 2 SH(C ). Then, for each q > 1=2, there exists
R0 > 0 and an entire function f such that
1
�R2
Z
jzj<R
��u(z) � log jf(z)j
��dm(z) < q logR; R > R0: (1.2)
An example constructed in [3] shows that we cannot take q < 1=2 in estimate
(1.2). The case q = 1=2 remains open.
The following theorem complements this result.
Let � be a class of slowly growing functions : [1;+1) ! (1;+1) (in par-
ticular, (2r) � (r) as r ! +1).
Theorem B ([4]). Let u 2 SH(C ), � = �u. If for some 2 � there exists a
constant R1 satisfying the condition
(8R > R1) : �(fz : R < jzj � R (R)g) > 1; (1.3)
then there exists an entire function f such that (R � R1)Z
jzj<R
��u(z)� log jf(z)j�� dm(z) = O(R2 log (R)): (1.4)
R e m a r k 1.1. In the case (r) � q > 1 we obtain Th. 1 [3].
The following example and Th. C show (see [4] for details) that estimate (1.4)
is sharp in the class of subharmonic functions satisfying (1.3).
212 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2
Approximation of Subharmonic Functions in the Unit Disk
For ' 2 �, let
u(z) = u'(z) =
1
2
+1X
k=1
log
���1� z
rk
���;
where r0 = 2, rk+1 = rk'(rk), k 2 N [ f0g. Thus, �u satis�es condition (1.3)
with (x) = '3(x).
Theorem C. Let 2 � be such that (r) ! +1 (r ! +1). There exists
no entire function f for whichZ
jzj<R
��u (z) � log jf(z)j�� dm(z) = o(R2 log (R)); R!1:
A further question arises naturally: Are there the counterparts of Ths. A and
B for subharmonic functions in the unit disk? We have the following theorem.
Theorem 1. Let u 2 SH(D ). There exists an absolute constant C and an
analytic function f in D such thatZ
D
��u(z)� log jf(z)j�� dm(z) < C: (1.5)
For a measurable set E � [0; 1) we de�ne the density
D1E = lim
R"1
l(E \ [R; 1))
1�R
:
Corollary 1. Let u 2 SH(D ), " > 0. There exists an analytic function f in D
and E � [0; 1), D1E < ", such that
2�Z
0
��u(rei�)� log jf(rei�)j��d� = O
� 1
1� r
�
; r " 1; r 62 E: (1.6)
The relationship (1.6) is equivalent to the condition
T (r; u) � T (r; log jf j) = O((1� r)�1); r " 1; r 62 E;
where T (r; v) is the Nevanlinna characteristic of a subharmonic function v. The
author does not know whether (1.6) is the best possible.
R e m a r k 1.2. No restrictions on the Riesz measure �u or the growth of u
are required in Th. 1.
R e m a r k 1.3. It is clear that (1.5) is sharp in the class SH(D ), but can be
improved under growth restrictions.
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2 213
I.E. Chyzhykov
Theorem D (M.O. Hirnyk [5]). Let u 2 SH(D ), �[u] < +1. Then there
exists an analytic function f in D such that
2�Z
0
��u(rei�)� log jf(rei�)j��d� = O
�
log2
1
1� r
�
; r " 1:
Theorem 1 does not allow to conclude that
u(z)� log jf(z)j = O(1); z 2 D nE (1.7)
for any �small� set E.
Su�cient conditions for (1.7) in the complex plane were obtained in [3] by
using the so-called notion of a locally regular measure admitting a partition of
slow variation.
We also prove a counterpart of Th. 30 of [3] using a similar concept. The cor-
responding Th. 3 will be formulated in Sect. 3. Here we formulate an application
of Th. 3.
Theorem 2. Let
j = (z = zj(t) : t 2 [0; 1]), 1 � j � m, be the smooth
Jordan curves in U(0; 1) such that arg zj(t) = �j(jzj(t)j) � �j(r), jzj(1)j = 1,
j�0j(r)j � K for r0 � r < 1 and some constants r0 2 (0; 1), K > 0, 1 � j � m.
Let u 2 SH(D ), supp�u �
Sm
j=1[
j ], �u([
j ] \ [
k]) = 0, j 6= k, and
�u
���
[
j ]
(U(0; r)) =
�j
(1� r)�(r)
;
where �j is a positive constant, �(r) = �
�
1
1�r
�
, �(R) is a proximate order [7],
�(R)! � > 0 as R! +1.
Then there exists an analytic function f such that for all " > 0
log jf(z)j � u(z) = O(1); (1.8)
z 62 E" = f� 2 D : dist (�; Zf ) � "(1 � j�j)1+�(r)g, where
log jf(z)j � u(z) � C; (1.9)
for some C > 0 and all z 2 D . Moreover,
Zf �
[
�2Sj [
j ]
U(�; 2(1� j�j)1+�(r));
and
T (r; u)� T (r; f) = O(1); r " 1: (1.10)
214 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2
Approximation of Subharmonic Functions in the Unit Disk
R e m a r k 1.4. Obviously, we cannot obtain a lower estimate for the
left-hand side of (1.9) for all z, because it equals �1 on Zf .
The theorems similar to Th. 2 are proved in [6, Ch.10, Ths. 10.16, 10.20]. The
di�erence is that in [6] only the weaker estimates are obtained for approximation
in a more general settings.
2. Proof of Theorem 1
2.1. Preliminaries
Let u 2 SH(D ). Then the Riesz measure �u is �nite on the compact subsets
of D . In order to apply a partition theorem (Th. E) we have to modify the
Riesz measure. By subtracting an integer-valued discrete measure ~� from �u
we may arrange that �(fpg) = (�u � ~�)(fpg) < 1 for any point p 2 D . The
measure ~� corresponds to the zeros of an entire function g. Thus we can consider
~u = u � log jgj, �~u = �. According to Lem. 1 [4], in any neighborhood of the
origin there exists a point z0 with the following properties:
a) on each line L� going through z0 there is at most one point �� such that
�(f��g) > 0, while �(L� n f��g) = 0;
b) on each circle K� with center z0 there exists at most one point �� such that
�(f��g) > 0, while �(K� n f��g) = 0.
As it follows from the proof of Lem. 1 [4], the set of points z0 not satisfying
conditions a) and b) has a planar measure zero. A similar assertion holds for
the polar set u(z0) = �1 [1, Ch.5.9, Th. 5.32]. Therefore, we can assume that
properties a), b) hold, and u(z0) 6= �1.
Then consider the subharmonic function u0(z) = u
�
z0�z
1�z�z0
�
� u(w(z)), u0(0) =
u(z0). Since jw0(z)j = 1�jz0j2
j1�z�z0j2 , we have jjw0(z)j � 1j � 3jz0j for jz0j � 1=2.
The Jacobian of the transformation w(z) is jw0(z)j2, consequently, this change
of variables does not change relation (1.5).
Let
u3(z) =
Z
U(0;1=2)
log jz � �j d�u(�): (2.1)
The subharmonic function u(z)� u3(z) is harmonic in U(0; 1=2).
Let q 2 (0; 1) be such that
12X
j=1
qj > 11: (2.2)
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2 215
I.E. Chyzhykov
We de�ne (n 2 f0; 1; : : : g)
Rn = 1� qn=2; An = f� : Rn � j�j < Rn+1g; Mn =Mn(q) =
h 2�
log Rn+1
Rn
i
;
An;m =
n
� 2 An :
2�m
Mn
� arg0 � <
2�(m+ 1)
Mn
o
; 0 � m �Mn � 1:
Represent �u
���
An;m
= �
(1)
n;m + �
(2)
n;m such that:
i) supp�
(j)
n;m � An;m, j 2 f1; 2g;
ii) �
(1)
n;m(An;m) 2 2Z+, 0 � �
(2)
n;m(An;m) < 2.
Let
�(j)n =
Mn�1X
m=0
�(j)n;m; ~�(j) =
X
n
�(j)n ; j 2 f1; 2g:
Property ii) implies
�(2)n (An) � 13
(1� q)(1�Rn)
; n! +1; (2.3)
as follows from the asymptotic equality
log
Rn+1
Rn
� (1� q)(1�Rn); n! +1; (2.4)
and the de�nition of Mn.
Let
u2(z) =
Z
D
log
���E�1� j�j2
1� ��z
; 1
����d~�(2)(�); (2.5)
where E(w; p) = (1 � w) expfw + w2=2 + � � � + wp=pg, p 2 N is the Weierstrass
primary factor.
Lemma 1. Let u2 2 SH(D ), and
T (r; u2) = O
�
log2
1
1� r
�
; r " 1;
Z
D
ju2(z)j dm(z) < C1(q):
216 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2
Approximation of Subharmonic Functions in the Unit Disk
P r o o f o f L e m m a 1. The following estimates for log jE(w; p)j are well
known (cf. [7, Ch.1, �4, Lem. 2], [1, Ch.4.1, Lem. 4.2]):
j logE(w; 1)j � jwj2
2(1� jwj) ; jwj < 1;
log jE(w; 1)j � 6ejwj2; w 2 C :
(2.6)
First we prove the convergence of integral in (2.5). For �xed Rn let jzj � Rn. We
choose p such that qp < 1=4. Then for j�j � Rn+p we have
1� j�j2
j1� ��zj �
2(1 � j�j)
1� jzj � 2(1�Rn+p)
1�Rn
<
1
2
:
Hence, using the �rst estimate in (2.6), (2.3) and the de�nition of Rn, we obtain
Z
j�j�Rn+p
���log
���E�1� j�j2
1� ��z
; 1
����
��� d~�(2)(�) �
Z
j�j�Rn+p
�2(1� j�j)
1� jzj
�2
d~�(2)(�)
� 4
(1� jzj)2
1X
k=n+p
(1�Rk)
2
Z
�Ak
d~�(2)(�) � 52
(1� q)(1� jzj)2
1X
k=n+p
(1�Rk)
=
52(1 �Rn+p)
(1� q)2(1� jzj)2 �
C2(q)
1�Rn
:
Thus, u2 is represented by the integral of subharmonic function log jEj of z, and
the integral converges uniformly on compact subsets in D , and so u2 2 SH(D ).
Since 1� j�j2 � 3=4 for � 2 supp ~�(2), using (2.6) and (2.3) we have
ju2(0)j �
Z
D
j log jE(1 � j�j2; 1)jj d~�(2)(�) �
Z
D
2(1 � j�j2)2d~�(2)(�)
� 8
1X
k=0
Z
�Ak
(1� j�j)2d~�(2)(�) � 104
1� q
1X
k=0
(1�Rk) = C3(q): (2.7)
Let us estimate T (r; u2)
def
= 1
2�
R 2�
0
u+2 (re
i�) d� for r � Rn, where u
+ = maxfu; 0g.
Note that for j�j � Rn+2, jzj � Rn we have
1�j�j2
j1���zj � 2. Thus
log
���E�1� j�j2
1� ��z
; 1)
���� 12e
1 � j�j2
j1 � ��zj
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2 217
I.E. Chyzhykov
in this case. Using the latter estimate, (2.6), (2.3), and the lemma [10, Ch. 5.10,
p. 226], we get
T (r; u2) � 1
2�
2�Z
0
�n+1X
k=0
Z
�Ak
12e
1� j�j2
j1 � ��rei�j d�
(2)
k (�)
�
d�
+
1
2�
2�Z
0
� 1X
k=n+2
Z
�Ak
6e
(1� j�j2)2
j1� ��rei�j2 d�
(2)
k (�)
�
d�e
� C4(q)
�n+1X
k=0
Z
�Ak
(1� j�j2) log 1
1� r
d�
(2)
k (�) +
1X
k=n+2
Z
�Ak
(1� j�j2)2
1� r
d�
(2)
k (�)
�
� C5(q)
�n+1X
k=0
log
1
1� r
+
1X
k=n+2
1�Rk
1� r
�
� C6(q)n log
1
1� r
� C7(q) log
2 1
1� r
:
Finally, by the First main theorem for subharmonic functions [1, Ch. 3.9]
m(r; u2)
def
=
1
2�
2�Z
0
u�2 (re
i�)d�
= T (r; u2)�
rZ
0
n(t; u2)
t
dt� u2(0) � T (r; u2) + C3(q):
Therefore
R 2�
0
ju2(rei�)jd� � 4�T (r; u2) + C8(q):
Consequently,
Z
jzj�1
ju2(z)j dm(z) � 4�
1Z
0
T (r; u2) dr + C8(q)
� C9(q)
1Z
0
log2
1
1� r
dr � C10(q):
Lemma 1 is proved.
2.2. Approximation of ~�1
The following theorem plays a key role in approximation of u.
218 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2
Approximation of Subharmonic Functions in the Unit Disk
Theorem E. Let � be a measure in R2 with compact support, supp� � �, and
�(�) 2 N, where � is a rectangle with the ratio of side lengths l0 � 1. Suppose,
in addition, that for any line L, parallel to a side of �, there is at most one point
p 2 L such that
0 < �(fpg)(< 1) while always �(L n fpg) = 0; (2.8)
Then there exists a system of rectangles �k � � with sides parallel to the sides of
�, and measures �k with the following properties:
1) supp�k � �k;
2) �k(�k) = 1,
P
k �k = �;
3) the interiors of the convex hulls of the supports of �k are pairwise disjoint;
4) the ratio of the side lengths of rectangles �k lies in the interval [1=l; l], where
l = maxfl0; 3g;
5) each point of the plane belongs to the interiors of at most 4 rectangles �k.
Theorem E was proved by R.S. Yulmukhametov [2, Th. 1] for absolutely con-
tinuous measures (i.e., � such that m(E) = 0 ) �(E) = 0) and l0 = 1. In this
case condition (2.8) is ful�lled automatically. In [8, Th. 2.1] D. Drasin showed
that Yulmukhametov's proof works if the condition of continuity is replaced by
condition (2.8). We can drop condition (2.8) rotating the initial square [8]. One
can also consider Th. E as a formal consequence of Th. 3 [4]. Here l0 plays role
for a �nite set of rectangles corresponding to small k's, but in [4] it plays the
principal role in the proof.
R e m a r k 2.1. In the proof of Th. E [8] the rectangles �k are obtained by
splitting the given rectangles, starting with �, into smaller ones in the following
way. The length of the smaller side of initial rectangle coincides with that of the
side of the rectangle obtained in the �rst generation, and the length of the other
side of new rectangle is between one third and two thirds of the length of the
other side of initial rectangle. Thus we can start with a rectangle instead of a
square and l = maxfl0; 3g.
Let u1(z) = u(z) � u2(z) � u3(z). Then �u1 = ~�(1), �
(1)
n;m( �An;m) 2 2Z+,
n 2 Z+, 0 � m �Mn � 1.
Let
Pn;m = logAn;m
=
n
s = � + it : logRn � � � logRn+1;
2�m
Mn
� t � 2�(m+ 1)
Mn
o
:
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2 219
I.E. Chyzhykov
According to (2.4) the ratio of the sides of Pn;m is
log Rn+1
Rn
2�=[ 2�
(1�q)(1�Rn) ]
! 1; n!1: (2.9)
Let d�n;m(s)
def
= d�
(1)
n;m(es), s 2 Pn;m, (i. e., �n;m(S) = �
(1)
n;m(expS) for every
Borel set S � C ). By our assumptions the conditions of Th. E are satis�ed for
� = Pn;m and � = �n;m=2, and all admissible n;m. By Theorem E there exists a
system (Pnkm, �nmk) of rectangles and measures, k � Nnm, 0 � m �Mn�1 with
the properties: 1) �nmk(Pnmk) = 1; 2) supp �nmk � Pnmk; 3) 2
P
k �nmk = �n;m;
4) every point s such that Re s < 0, 0 � Im s < 2� belongs to the interiors of at
most four rectangles Pnmk; 5) the ratio of the side lengths lies between two positive
constants. Indexing the new system (Pnmk; 2�nmk) with the natural numbers, we
obtain a system (P (l); �(l)) with �(l)(P (l)) = 2, supp �(l) � P (l), etc.
Let the measure �(l) de�ned on D be such that d�(l)(es)
def
= d�(l)(s), Re s < 0,
0 � Im s < 2�, Q(l) = expP (l). Let
�l
def
=
1
2
Z
Q(l)
�d�(l)(�) (2.10)
be the center of mass of Q(l), l 2 N.
We de�ne �
(1)
l , �
(2)
l as solutions of the system8>>>>><
>>>>>:
�
(1)
l + �
(2)
l =
Z
Q(l)
�d�(l)(�);
(�
(1)
l )2 + (�
(2)
l )2 =
Z
Q(l)
�2d�(l)(�):
(2.11)
From (2.11) and (2.10) it follows that (see [3, 4] or Lem. 3 below)
j�(j)l � �lj � diamQ(l) � dl; j 2 f1; 2g:
Consequently, we obtain
max
�2Q(l)
j� � �
(j)
l j � 2dl; j 2 f1; 2g; sup
�2Q(l)
j� � �lj � dl: (2.12)
We write
�l(z)
def
=
Z
Q(l)
�
log
��� z � �
1� z��
���� 1
2
log
��� z � �
(1)
l
1� z��
(1)
l
���� 1
2
log
��� z � �
(2)
l
1� z��
(2)
l
���� d�(l)(�);
V (z)
def
=
X
l
�l(z):
220 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2
Approximation of Subharmonic Functions in the Unit Disk
Fix a su�ciently large m (in particular, m � 13) and z 2 Am. Let L+ be the
set of l's such that Q(l) � U(0; Rm�13), and L� the set of l's with Q(l) � f� :
Rm+13 � j�j < 1g, L0 = N n (L� [ L+).
Lemma 2. There exists l� 2 N such that �1, �2 2 U(Q(l); 2dl), l 2 L+ [ L�,
l > l� imply
1
16
jz � �2j � jz � �1j � 16jz � �2j:
P r o o f o f L e m m a 2. First, let l 2 L+, i.e., z 2 Am, Q
(l) � Ap,
p � m� 13. In view of (2.9), Q(l) = expP (l) is �almost a square�. More precisely,
there exists l� 2 N such that for all l > l�
diamQ(l) = dl <
3
2
(Rp+1 �Rp); Q(l) � Ap:
Since �1, �2 2 U(Q(l); 2dl), we have
Rp � 3(Rp+1 �Rp) � j�2j � Rp+1 + 3(Rp+1 �Rp); (2.13)
jz � �1j � jz � �2j � j�2 � �1j � jz � �2j � 5dl � jz � �2j � 15
2
(Rp+1 �Rp):
(2.14)
On the other hand, by the choice of q (see (2.2)) and (2.13)
jz � �2j � Rm �Rp+1 � Rp+13 �Rp+1 � 3(Rp+1 �Rp)
=
12X
s=1
(Rp+s+1 �Rp+s)� 3(Rp+1 �Rp)
=
� 12X
s=1
qs � 3
�
(Rp+1 �Rp) > 8(Rp+1 �Rp):
The latter inequality and (2.14) yield
jz � �1j � jz � �2j � 15
2
(Rp+1 �Rp) > jz � �2j � 15
16
jz � �2j = 1
16
jz � �2j:
For l 2 L�, Q(l) � fRm+13 � j�j < 1g we have p � m+13, and the inequality
(2.14) still holds.
Similarly, by the choice of q and (2.13)
jz � �2j � Rp�3 �Rm+1 � Rp�3 �Rp�12 � 9q4(Rp+1 �Rp) > 8(Rp+1 �Rp);
that together with (2.14) implies the required inequality in this case. Lemma 2 is
proved.
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2 221
I.E. Chyzhykov
Let l 2 L� [ L+. For � 2 Q(l), we de�ne L(�) = Ll(�) = log
�
z��
1��z�
�
, where
logw is an arbitrary branch of Logw in w(Q(l)), w(�) = z��
1��z� . Then L(�) is
analytic in Q(l). We will use the following identities:
L(�)� L(�
(1)
l ) =
�Z
�
(1)
l
L0(s) ds = L0(�(1)l )(� � �
(1)
l ) +
�Z
�
(1)
l
L00(s)(� � s) ds
= L0(�(1)l )(� � �
(1)
l ) +
1
2
L00(�(1)l )(� � �
(1)
l )2 +
1
2
�Z
�
(1)
l
L000(s)(� � s)2 ds: (2.15)
Elementary geometric arguments show that j1
�z
� �j�1 � jz� �j�1 for z; � 2 D .
Since L0(�) = 1
��z +
�z
1��z�
, we have
jL0(�)j � 2
j� � zj ; jL00(�)j � 2
j� � zj2 ; jL000(�)j � 4
j� � zj3 : (2.16)
Now we estimate j�l(z)j for l 2 L+ [ L�. By the de�nitions of L(�), �l(z),
(2.15) and (2.11) we have
j�l(z)j =
���Re
Z
Q(l)
�
L(�)� L(�
(1)
l )� 1
2
(L(�
(2)
l )� L(�
(1)
l )
�
d�(l)(�)
���
=
���Re
Z
Q(l)
�
L0(�(1)l )
�
� � 1
2
(�
(1)
l + �
(2)
l )
�
+
�Z
�
(1)
l
L00(s)(� � s)ds� 1
2
�
(2)
lZ
�
(1)
l
L00(s)(�(2)l � s)ds
�
d�(l)(�)
���
=
���Re
Z
Q(l)
� �Z
�
(1)
l
L00(s)(� � s)ds� 1
2
�
(2)
lZ
�
(1)
l
L00(s)(�(2)l � s)ds
�
d�(l)(�)
���: (2.17)
Using (2.17), (2.16) and (2.12), we obtain
j�l(z)j �
Z
Q(l)
�Z
�
(1)
l
2j� � sj
js� zj2 jdsj d�
(l)(�) +
1
2
Z
Q(l)
�
(2)
lZ
�
(1)
l
2j�(2)l � sjjdsj
js� zj2 d�(l)(�)
� 12d2l max
s2Bl
1
js� zj2 ; (2.18)
222 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2
Approximation of Subharmonic Functions in the Unit Disk
where Bl = U(Q(l); 2dl). Applying Lem. 2, we have (z 2 �Am)
X
l2L�
j�l(z)j � 12
X
l2L�
d2l max
s2Bl
1
js� zj2 � C11
X
l2L�
Z
Q(l)
dm(z)
jz � �j2
� 4C11
Z
Rm+13�j�j<1
dm(z)
jz � �j2 � C12
1Z
Rm+13
d�
�� jzj � C13(q): (2.19)
Similarly,
X
l2L+
j�l(z)j � 12
X
l2L+
d2l max
s2Bl
1
js� zj2 � 4C11
X
l2L+
Z
j�j�Rm�13
dm(z)
jz � �j2
� C12
Rm�13Z
0
d�
jzj � �
� C14(q) log
1
1� jzj : (2.20)
Hence Z
jzj�Rn
X
l2L+[L�
j�l(z)j dm(z) < C15(q): (2.21)
It remains to estimate
R
jzj�Rn
P
l2L0 j�l(z)j dm(z). Here we follow the ar-
guments from [3, e.-g.]. If dist (z;Q(l)) > 10dl, similarly to (2.17), from (2.15),
(2.11), (2.16) and (2.12) we deduce
j�l(z)j =
���Re
Z
Q(l)
�
L0(�(1)l )
�
� � 1
2
(�
(1)
l + �
(2)
l )
�
+
L00(�(1)l )
2
�
�2 � (�
(1)
l )2 + (�
(2)
l )2
2
+ �
(1)
l (�
(1)
l + �
(2)
l � 2�)
�
+
1
2
�Z
�
(1)
l
L000(s)(� � s)2ds� 1
4
�
(2)
lZ
�
(1)
l
L000(s)(�(2)l � s)2ds
�
d�(l)(�)
���
=
����Re
Z
Q(l)
�
1
2
�Z
�
(1)
l
L000(s)(� � s)2 ds� 1
4
�
(2)
lZ
�
(1)
l
L000(s)(� � s)2 ds
�
d�(l)(�)
����
� 6d3l max
s2Bl
1
js� zj3 �
6d3l
j�(1)l � zj3
max
s2Bl
�
1 +
j�(1)l � sj
js� zj
�3
� 26d3l
j�(1)l � zj3
: (2.22)
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2 223
I.E. Chyzhykov
Since L0 depends only on m when z 2 Am, we have
Z
Am
X
l2L0
j�l(z)j dm(z) �
X
l2L0
� Z
AmnU(�(1)l
;10dl)
+
Z
U(�
(1)
l
;10dl)
�
j�l(z)j dm(z)
�
X
l2L0
� Z
AmnU(�(1)l
;10dl)
26d3l
jz � �
(1)
l j3
dm(z) +
Z
U(�
(1)
l
;10dl)
j�l(z)j dm(z)
�
: (2.23)
For the �rst sum we obtain
X
l2L0
26d3l
Z
AmnU(�(1)l
;10dl)
1
jz � �
(1)
l j3
dm(z) �
X
l2L0
52�d3l
2Z
10dl
tdt
t3
� 6�
X
l2L0
d2l � C16
X
l2L0
m(Q(l)): (2.24)
We now estimate the second sum. By the de�nition of �l(z)
�l(z) =
Z
Q(l)
�
log
���z � �
10dl
���� 1
2
log
���z � �
(1)
l
10dl
���� 1
2
log
���z � �
(2)
l
10dl
���� d�(l)(�)
�
Z
Q(l)
�
log j1� z�j � 1
2
log j1� z�
(1)
l j � 1
2
log j1� z�
(2)
l j
�
d�(l)(�) � I1 + I2:
The integral
R jI1j dm(z) is estimated in [3, g.], [4, p. 232]. We have
Z
U(�
(1)
l
;10dl)
jI1j dm(z) � C17m(Q(l)): (2.25)
To estimate jI2j we note that for l su�ciently large, jz � �j � 15dl, � 2
U(Q(l); 2dl), z 2 D , we have j arg z � arg �j � 16dl � 16(1 � jzj)j by the choice
of q. Hence,
j1
z
� ��j � 1
jzj � 1 + 1� j�j+ j�jj1� ei(arg ��arg z)j � C 0
17(1� jzj):
Thus, j1=z � ��j � 1� jzj. Therefore
jI2j �
Z
Q(l)
1
2
��� log j1z � ��j2
j1z � �
(1)
l jj1z � �
(2)
l j
���d�(l)(�) � C18:
224 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2
Approximation of Subharmonic Functions in the Unit Disk
Thus, Z
U(�
(1)
l
;10dl)
jI2j dm(z) � C19(q)m(Q(l)): (2.26)
Finally, using (2.24)�(2.26) we deduce
Z
�Am
X
l2L0
j�l(z)j dm(z)
� C20
X
l2L0
m(Q(l)) � 4�C20(R
2
m+13 �R2
m�13) � C21(q)(Rm+1 �Rm):
Hence,
R
jzj�Rn
P
l2L0 j�l(z)jdm(z) � C20(q), and this with (2.21) yields that
Z
jzj�Rn
jV (z)jdm(z) � C22(q); n! +1: (2.27)
Now we construct the function f1 approximating u1.
Let Kn(z) = u1(z) �
P
Q(l)�U(0;Rn)�l(z); K(z) = u1(z) � V (z). By the
de�nition of �l(z), Kn 2 SH(D ) and
�Kn
��
U(0;Rn)
(z) =
nX
l=1
�
Æ(z � �
(1)
l ) + Æ(z � �
(2)
l )
�
;
where Æ(�) is the unit mass supported at u = 0. For jzj � Rn, j � N � n+14 as
in (2.19) we have
jKj(z)�K(z)j �
X
Q(l)�fj�j�RN+1g
j�l(z)j
� C23
Z
RN+1�j�j<1
dm(z)
jz � �j2 � C24
1�RN+1
RN+1 � jzj ! 0; N ! +1:
Therefore Kn(z) � K(z) on the compact sets in D as n ! +1, and �K
���
D
=P
l(Æ(z � �(1)l ) + Æ(z � �(2)l ). Hence, K(z) = log jf1(z)j, where f1 is analytic in D .
2.3. Approximation of u3
Let u3 be de�ned by (2.1),
N = 2
�
n(1=2; u3)=2
�
; �0 = inffr � 0 : n(r; u3) � Ng:
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2 225
I.E. Chyzhykov
We represent �u3 = �1 + �2, where �1 and �2 are measures such that
supp�1 � U(0; �0); supp�2 � U
�
0;
1
2
�
n U(0; �0);
�1
�
U
�
0;
1
2
��
= N; 0 � �2
�
U
�
0;
1
2
��
< 2:
Let v2(z) =
R
U(0; 1
2
)
log jz � �j d�2(�). Then, using the last estimate,
Z
D
jv2(z)j dm(z) �
Z
U(0;1=2)
Z
D
j log jz � �jj dm(z) d�2(�)
�
Z
U(0;1=2)
Z
U(�;2)
j log jz � �jj dm(z) d�2(�) � C25n
�1
2
; v2
�
� 2C25:
If N = 0, there remains nothing to prove. Otherwise, we have to approximate
v1(z) = u3(z)� v2(z) =
Z
U(0;�0)
log jz � �j d�1(�): (2.28)
In this connection we recall the question of Sodin (Question 2 in [9, p. 315]).
Given a Borel measure � we de�ne the logarithmic potential of � by the
equality
U�(z) =
Z
log jz � �j d�(�):
Question. Let � be a probability measure supported by the square Q = fz =
x + iy : jxj � 1
2 ; jyj � 1
2g. Is it possible to �nd a sequence of polynomials Pn,
degPn = n, such thatZZ
jxj�1
jyj�1
jnU�(z) � log jPn(z)jj dxdy = O(1) (n! +1)?
We should say that the solution is given essentially in [3], but not asserted.
Hence we prove the following
Proposition. Let � be a measure supported by the square Q, and �(Q) =
N 2 N. Then there is an absolute constant C and a polynomial PN such thatZZ
�
jU�(z) � log jPN (z)jj dxdy < C;
where � = fz = x+ iy : jxj � 1; jyj � 1g.
226 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2
Approximation of Subharmonic Functions in the Unit Disk
P r o o f o f t h e p r o p o s i t i o n. As in the proof of Th. 1, if there
are points p 2 Q such that �(fpg) � 1, we represent � = � + ~� where for any
p 2 Q we have �(fpg) < 1, and ~� is a �nite (at most N summand) sum of the
Dirac measures. Then U~� = log
Q
k jz � pkj, so it remains to approximate U� .
By Lemma 2.4 [8] there exists a rotation to the system of orthogonal coordinates
such that if L is any line parallel to either of the coordinate axes, there is at most
one point p 2 L with �(fpg) > 0, while always �(L n fpg) = 0. After rotation the
support of new measure, which is still denoted by �, is contained in
p
2Q.
If ! is a probability measure supported on Q, then RR� jU!(z)jdm(z) is uni-
formly bounded. Therefore we can assume that N 2 2N.
By Theorem E, there exists a system (Pl; �l) of rectangles and measures 1 �
l �M� with the properties: 1) �l(Pl) = 2; 2) supp �l � Pl; 3)
P
l �l = �; 4) every
point s 2 Q belongs to the interiors of at most four rectangles Pl; 5) the ratio of
side lengths lays between 1/3 and 3.
Let
�l =
1
2
Z
Pl
�d�l(�) (2.29)
be the center of mass of Pl, 1 � l �M� .
We de�ne �
(1)
l , �
(2)
l as solutions of the system
8>>>>><
>>>>>:
�
(1)
l + �
(2)
l =
Z
Pl
�d�l(�);
(�
(1)
l )2 + (�
(2)
l )2 =
Z
Pl
�2d�l(�):
We have
j�(j)l � �lj � diamPl � Dl; j 2 f1; 2g;
max
�2Pl
j� � �
(j)
l j � 2Dl; j 2 f1; 2g; sup
�2Pl
j� � �lj � Dl: (2.30)
We write
(z) =
X
l
Z
Pl
�
log
���z � �
���� 1
2
log
���z � �
(1)
l
���� 1
2
log
���z � �
(2)
l
���� d�l(�)
�
X
l
Æl(z): (2.31)
Since we have rotated the system of coordinate, it is su�cient to prove thatR
U(0;
p
2)
j
(z)j dm(z) is bounded by an absolute constant.
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2 227
I.E. Chyzhykov
For � 2 Pl, z 62 Pl we de�ne �(�) = �l(�) = log
�
z � �
�
, where log(z � �) is an
arbitrary branch of Log(z � �) in z � Pl. Then �(�) is analytic in Pl.
We have
j�000(�)j � 2
j� � zj3 : (2.32)
As in subsection 2.2, we have
jÆl(z)j �
���Re
Z
Pl
�
�(�)� �(�
(1)
l )� 1
2
(�(�
(2)
l )� �(�
(1)
l )
�
d�l(�)
���
�
����Re
Z
Pl
�
1
2
�Z
�
(1)
l
�000(s)(� � s)2 ds� 1
4
�
(2)
lZ
�
(1)
l
�000(s)(� � s)2 ds
�
d�l(�)
����: (2.33)
If dist (z; Pl) > 10Dl, the last estimate and (2.32) yield
jÆl(z)j � 24D3
l max
s2El
1
js� zj3 �
24D3
l
j�(1)l � zj3
max
s2El
�
1 +
j�(1)l � sj
js� zj
�
� 103D3
l
j�(1)l � zj3
;
where El = U(Pl; 2Dl).
Then
Z
U(0;
p
2)nU(�(1)
l
;10Dl)
103D3
l
jz � �
(1)
l j3
dm(z) � 206�D3
l
2Z
10Dl
tdt
t3
� 21�D2
l � C26m(Pl):
On the other hand, by the de�nition of Æl(z)Z
U(�
(1)
l
;10Dl)
Æl(z)dm(z) =
Z
U(�
(1)
l
;10Dl)
Z
Pl
�
log
���z � �
10Dl
���
�1
2
log
���z � �
(1)
l
10Dl
���� 1
2
log
���z � �
(2)
l
10Dl
���� d�l(�)dm(z) � C27m(Pl):
From (2.31) and the latter estimates, it follows thatZ
U(0;
p
2)
j
(z)jdm(z) �
X
l
Z
U(0;
p
2)
Æl(z)dm(z)
� C28
X
l
m(Pl) � 4C28m(
p
2Q) = C29: (2.34)
228 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2
Approximation of Subharmonic Functions in the Unit Disk
Thus, P(z) =Q
l(z � �(1)l )(z � �(2)l ) is a required polynomial. This completes the
proof of the proposition.
Finally, let f = f1P. By Lemma 1, (2.27), and (2.34) we have (n! +1)
Z
jzj�Rn
j log jf(z)j � u(z)jj dm(z) �
Z
jzj�Rn
(jK(z)� u1(z)jj+ ju2(z)j
+j log jPj � u3(z)j) dm(z) �
Z
jzj�Rn
(jV (z)j + j
(z)j) dm(z) + C10(q) � C30(q):
Fixing any q satisfying (2.2) we �nish the proof of Th. 1.
3. Uniform Approximation
In this section we prove some counterparts of results due to Yu. Lyubarskii
and Eu. Malinnikova [3]. We start with the counterparts of notions introduced in
[3], which re�ect regularity properties of measures.
De�nition 1. Let b : [0; 1)! (0;+1) be such that b(r) � 1� r,
b(r1) � b(r2) as 1� r1 � 1� r2; r1 " 1: (3.1)
A measure � on D admits a partition of slow variation with the function b if
there exist the integers N , p and the sequences (Q(l)) of subsets of D and (�(l)) of
measures with the following properties:
i) supp�(l) � Q(l), �(l)(Q(l)) = p;
ii) supp (��Pl �
(l)) � D , (��Pl �
(l))(D ) < +1;
iii) 1 � dist (0; Q(l)) � K(p) diamQ(l), and each z 2 D belongs to at most N
various Q(l)'s;
iv) for each l the set logQ(l) is a rectangle with the sides parallel to coordi-
nate axes, and the ratio of sides lengths lies between two positive constants
independent of l;
v) diamQ(l) � b(dist (Q(l); 0)).
R e ma r k 3.1. This is similar to [3], except we have introduced the parameter
p (p = 2 in [3]). Property iii) is adapted for D .
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2 229
I.E. Chyzhykov
De�nition 2. Given a function b satisfying (3.1), we say that a measure �
is locally regular with respect to (w.r.t.) b if
b(jzj)Z
0
�(U(z; t))
t
dt = O(1); r0 < jzj < 1;
for some constant r0 2 (0; 1).
Theorem 3. Let u 2 SH(D), b : [0; 1)! (0;+1) satisfy (3.1). Let �u admits
a partition of slow variation, assume that �u is locally regular w.r.t. b, and, with
p from above, that
1Z
0
bp�1(t)
(1� t)p
dt < +1: (3.2)
Then there exists an analytic function f in D such that 8" > 0 9r1 2 (0; 1)
log jf(z)j � u(z) = O(1); r1 < jzj < 1; z 62 E";
where E" = fz 2 D : dist (z; Zf ) � "b(jzj)g, and for some constant C > 0
log jf(z)j � u(z) < C; z 2 D : (3.3)
Moreover, Zf � U(supp�u;K1(p)b(jzj)), K1(p) is a positive constant, and
T (r; u) � T (r; log jf j) = O(1); r " 1: (3.4)
R e m a r k 3.2. The author does not know whether condition (3.2) is
necessary. But if b(t) = O((1 � t) log��(1 � t), � > 0, t " 1 (3.2) holds for
su�ciently large p. On the other hand, in view of v) the condition b(t) = O(1� t)
as t " 1 is natural.
P r o o f o f T h e o r e m 3. We follow [3] and also use the arguments and
notation from the proof of Th. 1.
Let ~� = �u�
P
l �
(l). Since
��� z��
1���z
���! 1 as jzj " 1 for �xed � 2 D , ~�(D ) < +1,
~u1(z) =
Z
D
log
��� z � �
1� ��z
��� d~�(�)
is a subharmonic function in D and j~u1(z)j < C for r1 < jzj < 1, r1 2 (0; 1). So
we can assume that �u =
P
l �
(l), where �(l) are from Def. 1.
230 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2
Approximation of Subharmonic Functions in the Unit Disk
Fix a partition of slow variation. Instead of points �
(1)
l and �
(2)
l satisfying
(2.11) we de�ne �
(l)
1 , . . . , �
(l)
p from the system
8>>>><
>>>>:
�1 + � � �+ �p =
R
Q(l) �d�
(l)(�);
�21 + � � � + �2p =
R
Q(l) �
2d�(l)(�);
...
�
p
1 + � � �+ �
p
p =
R
Q(l) �
pd�(l)(�):
(3.5)
Lemma 3 is a modi�cation of the estimates in(2.12).
Lemma 3. Let � be a set in C , � is a measure on �, �(�) = p 2 N, diam� =
d. Then for any solution (�1; : : : ; �p) of (3.5) we have j�j � �0j � K1(p)d, where
K1(p) is a constant, �0 is the center of mass of �.
P r o o f o f L e m m a 3. Let �0 =
1
p
R
�
�d�(�) be the center of mass of �.
By induction, it is easy to prove that (3.5) is equivalent to the system
8>>>><
>>>>:
w1+ � � �+ wp = 0;
w2
1+ � � �+ w2
p = J2;
...
w
p
1+ � � �+ w
p
p = Jp;
(3.6)
where wk = �k � �0, Jk =
R
�
(� � �0)
k d�(�), 1 � k � p. Note that
jJkj �
Z
�
j� � �0jk d�(�) � pdk:
From algebra it is well known that the symmetric polynomials
X
1�i1<���<ik�m
wi1 � � �wik ;
1 � k � m, can be obtained from the polynomials
Pm
j=1w
k
j using only �nite
number of operations of addition and multiplication. Therefore (3.6) yields
8>>>>><
>>>>>:
w1 + � � �+ wp = 0;P
1�i1<i2�p
wi1wi2 = b2;
...
w1 � � �wp = bp;
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2 231
I.E. Chyzhykov
where bk =
P
l alk(J1)
s
(k)
1l � � � (Jm)s
(k)
ml , alk = alk(p), s
(k)
jl are nonnegative integers,
and
Pp
j=1 s
(k)
jl j = k. The last equality follows from homogeneousity. Hence, there
exists a constant K1(p) � 2 such that jbkj � K1(p)d
k, 1 � k � p. By Vieta's
formulas ([11, ��51, 52]) wj , 1 � j � p, satisfy the equation
wp + b2w
p�2 � b3w
p�3 + � � �+ (�1)pbp = 0: (3.7)
For jwj = K1(p)d we have
jwp + b2w
p�2 � b3w
p�3 + � � �+ (�1)pbpj � K1(p)(d
2jwjp�2 + � � � + dp)
= K1(p)d
p(Kp�2
1 +K
p�1
1 + � � �+ 1) < 2Kp�1
1 (p)dp � K
p
1 (p)d
p = jwjp:
By Rouch�e's theorem all p roots of (3.7) lay in the disk jwj � K1(p)d, i.e., j�j �
�0j � K1(p)d. Consequently, dist (�j ;�) � K1(p)d.
Applying Lem. 3 to Q(l) we obtain that j�(j)l ��lj � K1(p)dl, 1 � j � p, where
�l =
1
p
R
Q(l) �d�
(l)(�):
Consider
V (z) =
X
l
jl(z)
def
=
X
l
Z
Q(l)
�
log
��� z � �
1� �z�
���� 1
p
pX
j=1
log
��� z � �
(j)
l
1� �z�
(j)
l
����d�(l)(�):
For Rn = 1� 2�n, z 2 Am, m is �xed, we de�ne the sets of indices L+, L� and
L0 as in the proof of Th. 1.
The estimate of
P
l2L�
jl(z) repeats that of
P
l2L�
�l(z), so
X
l2L�
jjl(z)j � C31: (3.8)
Following [3], we estimate
P
l2L0 jl(z). Let bm = b(Rm). Note that dl � bm for
l 2 L0 by condition v). As in (2.18), we have
jjl(z)j � C32d
3
l max
s2U(Q(l);K1(p)dl)
js� zj�3 � C 0
32
d3l
j�(1)l � zj3
; (3.9)
provided that dist (z;Q(l)) � 3K1(p)dl. Then����
X
l2L0
Q(l)\U(z;3K1(p)dl)=?
jl(z)
���� � C32
X
l2L0
d3l
j�(1)l � zj3
� C33bm
Z
jz��j>C34bm
dm(�)
jz � �j3 � C35
bm
bm
= C35: (3.10)
232 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2
Approximation of Subharmonic Functions in the Unit Disk
Let now l be such that Q(l) \ U(z; 3K1(p)dl) 6= ?. Since dl � bm, the number of
these l is bounded uniformly in l. For z 62 E" we have (1 � k � p)
log jz � �
(k)
l j = log bm + log
jz � �
(k)
l j
bm
= log b(jzj) +O(1): (3.11)
Therefore
jl(z) =
Z
Q(l)
�
log jz � �j � 1
p
pX
k=1
log jz � �
(k)
l j
�
d�(l)(�)
�1
p
Z
Q(l)
log
j1� �z�jpQp
k=1 j1� �z�
(k)
l j
d�(l)(�) = J3 + J4:
As in the proof of the proposition (see the estimate of I2), one can show that
j1�z � �j � 1� jzj � j1�z � �
(j)
l j. Hence, we have J4 = O(1).
Let �z(t) = �(U(z; t)). Further, using (3.11),
J3 =
Z
Q(l)nU(z;b(jzj))
log jz � �jd�(l)(�) +
Z
U(z;b(jzj))
log jz � �j d�(l)(�)
�p log b(jzj) +O(1) = �(l)(Q(l) n U(z; b(jzj)) log b(jzj) +O(1)
+
b(jzj)Z
0
log t d�(l)z (t)� p log b(jzj) = �(l)(Q(l) n U(z; b(jzj)) log b(jzj)
+O(1) + �(l)(U(z; b(jzj)) log b(jzj)�
b(jzj)Z
0
�
(l)
z (t)
t
dt
�p log b(jzj) = �
b(jzj)Z
0
�
(l)
z (t)
t
dt+O(1) = O(1) (3.12)
by the regularity of �u w.r.t b(t). Together with (3.10) it yields
X
l2L0
jjl(z)j = O(1); z 62 E": (3.13)
Now we estimate
P
l2L+ jl(z). Integration by parts gives us
L(�)�L(�(1)l ) =
mX
k=1
1
k!
L(k)(�
(1)
l )(���(1)l )k+
1
m!
�Z
�
(1)
l
L(m+1)(s)(��s)m ds; (3.14)
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2 233
I.E. Chyzhykov
where L(�) = log z��
1��z�
,
jL(k)(�)j � 2(k � 1)!
jz � �jk : (3.15)
The de�nition of �
(k)
l , 1 � k � p, allows us to cancel the �rst p moments. There-
fore, similarly to (2.18) and (2.22), we have
jjl(z)j � C28d
p+1
l max
s2U(Q(l);K1(p)dl)
js� zj�p�1: (3.16)
Then (z 2 Am)
X
l2L+
jjl(z)j � C28
X
l2L+
d
p+1
l
jz � �
(1)
l jp+1
� C29
X
l2L+
d
p�1
l
Z
Q(l)
dm(z)
jz � �jp+1
� C30(N; p; q)
X
n�m�12
bp�1(Rn)
Z
�An
dm(z)
jz � �jp+1
� C31
Z
j�j�Rm�12
bp�1(j�j)dm(�)
jz � �jp+1
� C32
Rm�12Z
0
bp�1(�)
(jzj � �)p
d� � C33
1Z
0
b(�)p�1
(1� �)p
d� < +1:
Using the latter inequality, (3.13) and (3.8) we obtain jV (z)j = O(1) for z 62 E".
The construction of f is similar to that of Th. 1. It remains to prove (3.3) for
z 2 E".
By (3.10) it is su�cient to consider l with Q(l) \ U(z; 3K1(p)dl) 6= ?. For all
su�ciently large l 2 L0 we have
���
Z
Q(l)
log jz � �jd�(l)(�)
��� �
Z
U(z;4K1(p)dl)
log
1
jz � �jd�
(l)(�)
�
4K1(p)dlZ
0
log
1
t
d�(l)z (t) = log
1
4K1(p)dl
�(l)z (4K1(p)dl) +
4K1(p)dlZ
0
�
(l)
z (t)
t
dt = O(1):
(3.17)
Then we have
log jf(z)j � u(z) = O(1) +
X
l2L0
� pX
k=1
log jz � �
(k)
l j �
Z
Q(l)
log jz � �jd�(l)(�)
�
< C;
because jz � �
(j)
l j = O(b(jzj)) < 1 for l � l0 and (3.3) is proved.
234 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2
Approximation of Subharmonic Functions in the Unit Disk
Finally, in order to prove (3.4) we note that for z 2 E", in view of (3.8), (3.10),
(3.17), we have
log jf(z)j � u(z) =
mX
j=1
log jz � �j j+O(1);
where �j 2 Zf , and m are uniformly bounded. Then T (r; u� log jf j) is bounded,
and consequently
T (r; u) = T (r; log jf j) + T (r; u� log jf j) +O(1) = T (r; log jf j) +O(1):
P r o o f o f T h e o r e m 2. Let �j = �u
���
[
j ]
. By the assumptions of the
theorem we have �u =
Pm
j=1 �j . We can write u =
Pm
j=1 uj, where uj 2 SH(D ),
and �uj = �j . Therefore, it is su�cient to approximate each uj, 1 � j � m,
separately.
We write R(r) = (1�r)�1 andW (R) = R�(R). Then �j(U(0; r)) = �jW (R(r)).
Put b(t) = (1 � t)=W (R(r)). Then condition (3.2) is satis�ed. We are going to
prove that �j admits a partition of slow variation and is locally regular w.r.t. b(t).
We de�ne a sequence (rn) from the relation �jW (R(rn)) = 2n, n 2 N. Then,
using the theorem on the inverse function and the properties of proximate order
[7, Ch. 1, �12], we have (r0 2 (rn; rn+1))
rn+1 � rn =
2(1 � r0)2
�jW 0(R(r0))
=
(2 + o(1))R(r0)(1 � r)02
�j�W (R(r0))
=
2 + o(1)
�j�
b(r0) � b(rn):
Let Q(n) = fz : rn � jzj � rn+1; '
�
n � � � '+
n g, where '�n = �j(rn) �K(rn+1 �
rn), '
+
n = �j(rn) +K(rn+1 � rn). Since j�0j(t)j � K, we have �j(r) 2 ['�n ; '
+
n ],
rn � r � rn+1. Let �
(n) = �j
���
Q(n)
. Then, by the de�nition of rn, �
(n)(Q(n)) = 2.
Therefore conditions i) and iv) in the de�nition of partition of slow growth are
satis�ed. Condition ii) is trivial. Since diamQ(n) � b(rn) � (1�rn)1+�(rn), � > 0,
conditions iii) and v) are valid. Therefore, � admits a partition of slow growth
w.r.t. b, N = p = 2.
Finally, we check the local regularity of �j w.r.t. b(t). For jzj = r, � � b(r)
we have
�j(U(z; �)) � �jW (R(r + �))��jW (R(r � �)) =W 0(R(r�))
2��j
(1 � r�)2
= (2 + o(1))�j��
W (R(r�))
1� r�
� 3���j
b(r)
:
Then
R b(r)
0
�(U(z;�))
� d� � 3��j , as required.
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2 235
I.E. Chyzhykov
Applying Th. 3 we obtain (1.8), (1.9), and (1.10) for some analytic function
fj in D .
Finally, we de�ne f =
Qm
j=1 fj.
The theorem is proved.
I would like to thank Prof. O. Skaskiv, who read the paper and made valuable
suggestions, as well as other participants of the Lviv seminar on the theory of an-
alytic functions for their valuable comments that contributed to the improvement
of the initial version of the paper. I wish also to thank the anonymous referee for
a number of useful comments and corrections.
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