Математичнi Студiї. Т.13, №2
Matematychni Studii. V.13, No.2
УДК 517.53
M. V. Zаbоlоtsкii
AN EXAMPLE OF ENTIRE FUNCTION OF STRONGLY
REGULAR GROWTH
M. V. Zаbоlоtsкii. An example of entire function of strongly regular growth, Matematychni
Studii, 13 (2000) 145–148.
We construct an entire function of zero order of strongly regular growth such that its zeros
do not have any angular density with respect to the function v(r) = rλ(r) , where λ(r) is
a proximate order of the counting function n(r).
Н. В. Заболоцкий. Пример целой функции сильно регулярного роста // Математичнi Студiї. – 2000. – Т.13, №2. – C.145–148.
Построена целая функция нулевого порядка сильно регулярного роста такая, что её
нули не имеют угловой плотности относительно функции v(r) = rλ(r) , где λ(r) — уточненный порядок считающей функции n(r).
Let f be an entire function of positive order and completely regular growth (c.r.g.) by
Levin-Pfluger [1, p. 182]. The necessary and sufficient condition of c.r.g. of functions f
of non-entire order is the existence of the angular density ∆(ψ) of its zeros relatively the
comparison function V (r) = rρ(r) , ρ(r) be the proximate order of function f .
In the case of zero order the proximate order ρ(r) of function f will not be the proximate
order of the counting function n(r) of its zeros. Then n(r) = o(V (r)) as r → ∞. Hence zeros
of an entire function of zero order always have the angular density ∆(ψ) = 0. Therefore
in case of zero order we consider the comparison function v(r) = rλ(r) , where λ(r) is the
proximate order of the function n(r).
In [2] it was introduced a new notion of regular growth for entire functions of zero order.
Let f be an entire function, f (0) = 1, n(r, α, β) be the number of zeros an = |an |eiαn of the
function f in theS
sector {z : |z| ≤ r, α < arg z ≤ β}, ln f is a univalent branch of Ln f in the
iαn
: r ≥ |an |} such that ln f (0) = 0. The ray arg z = θ is ordinary,
domain D = C \ ∞
n=1 {re
if
lim lim n(r, θ − h, θ + h)/v(r) = 0.
h→0 r→∞
We shall say that a function f has a strongly regular growth (s.r.g.) if for all ordinary rays
arg z = θ we have
ln f (reiθ ) − N (r)
lim
= Hf (θ),
r→∞,r∈E
/
v(r)
where ER is some set of zero relative measure, i.e., mes (E ∩ [0, r]) = o(r) (r → ∞), and
r
N (r) = 0 n(t)/tdt.
2000 Mathematics Subject Classification: 30D15.
c M. V. Zbltsii, 2000
146
M. V. ZАBОLОTSКII
Without loss of generality suppose that the ray arg z = −π is ordinary for zeros of the
functions. We say that the set of roots of an entire function has angular density if the limit
∆(ψ) = limr→∞ n(r, −π, ψ)/v(r) exists for all ψ with the exception, perhaps, of ψ belonging
to some countable set.
Theorem A [2]. Let f be an entire function which zeros have angular density ∆(ψ). Then
f has a s.r.g., moreover for all ordinary rays arg z = θ we have
Z π
arg−π ei(θ−ψ−π) d∆(ψ), −π ≤ θ < π,
Hf (θ) = i
−π
where arg−π z is the value of arg z ∈ [−π, π). On the contrary, let f be an entire function
whose zeros lie on a finite system of rays arg z = θj , j = 1, ..., k. and f has a s.r.g. Then
zeros f have angular density ∆(ψ), moreover
X
1
∆(ψ) =
Im
(Hf (θj − 0) − Hf (θj + 0)) .
2π
−π≤θ <ψ
j
The following theorem implies that s.r.g. of entire function f is not a sufficient condition
for existence of angular density of its zeros. Thus, this indicates that the condition of zeros
placement on a finite system of rays is essential in Theorem A.
Theorem. Let α(r) be a continuous positive increasing to +∞ on [0, +∞) function. Then
for arbitrary increasing to +∞ on [0, +∞) function β(r) there exist an entire function
f, f (0) = 1, and a set E of zero relative measure such that for all θ, 0 ≤ θ < 2π, with
the exception of θ belonging to a countable set, the following holds
ln f (reiθ ) = N (r) + o(β(r)), r → ∞, r ∈
/ E,
n(r)
lim
=1
r→∞ α(r)
(1)
(2)
and zeros of f do not have any angular density with respect to the function α(r).
Proof. Without loss of generality, we shall assume that β(r) ≥ 0 and β(r) < α(r) for all
r > 0. We choose two increasing sequences (rk ) of positive numbers and nk of positive
integers by the induction. Let r1 ≥ 1 be such that α(r1 ) ∈ N, n1 = α(r1 ), and assume that
r2 , r3 , ..., rm and n2 , n3 , ..., nm are already chosen. Then we choose rm+1 such that
α(rm )
1
< ,
α(rm+1 )
m
α(rm+1 ) ∈ N, β(rm+1 ) > π(m + 2)2 ,
rm+1 > 4rm ,
(3)
(4)
and nm+1 = α(rm+1 ) − α(rm ).
Q nk
We consider the entire function f (z) = +∞
1
−
(z/r
)
. Let rm ≤ r < rm+1 . Then
k
k=1
n(r) =
X
rk ≤r
nk = α(r1 ) +
X
α(rk ) − α(rk−1 ) = α(rm ) ≤ α(r),
rk ≤r
whence (2) follows. Further, if 0 ≤ α < β < 2π, then
β−α X
β−α
n(r, α, β) = (1 + o(1))
nk = (1 + o(1))
α(rm ),
2π r ≤r
2π
k
r → ∞,
AN EXAMPLE OF ENTIRE FUNCTION OF STRONGLY REGULAR GROWTH
147
and from (3) we have that the limit limr→∞ n(r)/α(r) does not exist.
Let’s show that f satisfies condition (1). Let z = reiθ , θ 6= 2πj/nk , k ∈ N, j = 0, 1, . . . , nk .
Then
Z z
+∞
+∞
z nk Z z f 0 (ζ)
X
X
−ζ nk −1 dζ
iθ
=
=
dζ =
nk
ln f (re ) =
ln 1 −
nk
nk
rk
0 f (ζ)
0 rk − ζ
k=1
k=1
Z r
+∞
+∞
X
X
tnk −1 dt
=
nk
=
Jk .
(5)
nk − r nk e−ink θ
0 t
k
k=1
k=1
Set τ = tnk , ϕk = −nk θ, Rk = rknk , R = rnk . We have
Z R
Z R
dτ
R
sin
ϕ
dτ
k
k
=
=
|Im Jk | = Im
iϕ
iϕ
2
k
k| 0 τ − Rk e
0 |τ − Rk e
R Z R
τ
−
R
cos
ϕ
dτ
k
k = arctg
= Rk sin ϕk
< π,
2
2
Rk sin ϕk (τ − R cos ϕ )2 + R sin ϕ
k
0
k
k
k
0
and owing to (4)
m+1
X
|Im Jk | ≤ (m + 1)π <
k=1
β(rm )
.
m+1
Let rm ≤ r < rm+1 . Then from (3) for k ≥ m+2 we obtain
2x for 0 ≤ x ≤ 1/2, we have
Z
|Jk | ≤
0
r
(6)
r
≤ 4m+1−k . Since − ln(1−x) ≤
rk
k−m−1
r nk r nk
nk tnk −1 dt
1
.
= − ln 1 −
≤2
≤2
nk
n
rk − t k
rk
rk
4
Therefore
+∞
X
+∞
X
1
= 1,
|Jk | ≤
2k
k=m+2
k=1
(7)
arg f (reiθ ) ≤ β(r) + 1 = o(β(r)), r → ∞.
m+1
(8)
and from (5)–(7) we have
Find the asymptotics of ln |f (reiθ )|. We have (R = rnk )
r nk
R
Rk e−iϕk r
k
ink θ Re Jk = ln
+ ln 1 −
= nk ln + ln 1 −
e .
Rk
R rk
r
Let rm +
−2
rm
nm
r nk
k
r
≤ r ≤ rm+1 +
rm+1
.
nm+1
(9)
Then for k ≤ m − 1 it holds
r nk r nk
r nk r nk
k
k
k
k
≤ ln 1 −
≤ ln 1 −
eink θ ≤ ln 1 +
≤
. (10)
r
r
r
r
Further,
r nm r nm
r nm m
m
m
ln 1 −
≤ Re Jm = ln 1 −
einm θ ≤ ln 1 +
≤ ln 2.
r
r
r
(11)
148
M. V. ZАBОLОTSКII
and
ln 1 −
r nm m
r
≥ ln 1 −
rm
rm + rm /nm
nm ≥
1 −nm
2
≥ ln 1 − 1 +
≥ ln 1 −
. (12)
nm
e
Analogously, we obtain
Re Jm+1
nm+1
nm+1 r
r
inm+1 θ ≤ ln 1 −
e
≤ ln 2,
≤ ln 1 + rm+1
rm+1
(13)
and
Re Jm+1 ≥ ln 1 −
nm+1 r
≥ ln 1 − 1 −
rm+1
1
nm+1 nm+1
2
≥ ln 1 −
e
.
(14)
S
From (7) and (9)–(14) for r ∈
/ E, E = +∞
k=1 (rk − rk /nk , rk + rk /nk ), we obtain
m−1
m
X
X
r
nk ln ≤
|Re Jk | + |Re Jm |+
ln |f (reiθ )| −
rk r=1
k=1
+|Re Jm+1 | +
+∞
X
|Jk | ≤ 2
m−1
X
k=1
k=m+2
1
+ O(1) = O(1), r → ∞.
4nk
(15)
From (8) and (15) we have (1), since the relative measure of the set E is zero. Indeed, for
rm ≤ r ≤ rm+1 we obtain
mes (E ∩ [0, r]) ≤
2rm
2r1 2r2
+
+ ... +
≤
n1
n2
nm
m
X
1
2rm
2rm
2rm
2rm
+
= 2rm
,
≤ m−1 + m−2 + ... +
m−k
4
n1 4
n2
4nm−1
nm
4
nk
k=1
and by the Töplitz theorem, we have
mes (E ∩ [0, r]) = o(rm ) = o(r), r → ∞.
REFERENCES
1. Левин Б. Я. Распределение корней целых функций, М.: Гостехиздат, 1956.
2. Заболоцкий Н. В. Сильно регулярный рост целых функций нулевого порядка, Мат. заметки 63
(1998), №2, 196–208.
Lviv National University, Faculty of Mechanics and Mathematics
Received 27.01.2000