Growth and representation of analytic and harmonic functions in the unit disc
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irk-123456789-1245402017-09-30T03:03:34Z Growth and representation of analytic and harmonic functions in the unit disc Chyzhykov, I. 2006 Article Growth and representation of analytic and harmonic functions in the unit disc / I. Chyzhykov // Український математичний вісник. — 2006. — Т. 3, № 1. — С. 32-45. — Бібліогр.: 12 назв. — англ. 1810-3200 2000 MSC. 30E20, 20D50. http://dspace.nbuv.gov.ua/handle/123456789/124540 en Український математичний вісник Інститут прикладної математики і механіки НАН України |
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Chyzhykov, I. Growth and representation of analytic and harmonic functions in the unit disc Український математичний вісник |
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Chyzhykov, I. |
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Growth and representation of analytic and harmonic functions in the unit disc |
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Growth and representation of analytic and harmonic functions in the unit disc |
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Growth and representation of analytic and harmonic functions in the unit disc |
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Growth and representation of analytic and harmonic functions in the unit disc |
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Growth and representation of analytic and harmonic functions in the unit disc |
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growth and representation of analytic and harmonic functions in the unit disc |
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Інститут прикладної математики і механіки НАН України |
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Growth and representation of analytic and harmonic functions in the unit disc / I. Chyzhykov // Український математичний вісник. — 2006. — Т. 3, № 1. — С. 32-45. — Бібліогр.: 12 назв. — англ. |
| series |
Український математичний вісник |
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AT chyzhykovi growthandrepresentationofanalyticandharmonicfunctionsintheunitdisc |
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2025-07-09T01:35:38Z |
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1837131313258168320 |
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Український математичний вiсник
Том 3 (2006), № 1, 32 – 45
Growth and representation of analytic and
harmonic functions in the unit disc
Ihor Chyzhykov
(Presented by M. M. Sheremeta)
Abstract. Let u(z) be harmonic in {|z| < 1}, α ≥ 0, 0 < γ ≤ 1. Let
B(r, u) = max{u(z) : |z| ≤ r}, ω(δ, ψ) be the modulus of continuity of a
function ψ defined on [0, 2π]. We prove that u(z) has the form
u(reiϕ) =
1
2π
2π
∫
0
Pα(r, ϕ− t) dψ(t)
where ψ ∈ BV [0, 2π], and ω(δ, ψ) = O(δγ) (δ ↓ 0), if and only if
B(r, u) = O((1 − r)γ−α−1), r ↑ 1 and sup0<r<1
∫ 2π
0
|uα(reiϕ)| dϕ <
+∞. Here uα is α-fractional integral of u(reiϕ), Pα(r, t) = Γ(1 +
α)ℜ( 2
(1−reit)α+1 − 1).
2000 MSC. 30E20, 20D50.
Key words and phrases. Analytic function, harmonic function, frac-
tional integral, growth estimates.
1. Introduction and main results
1.1. Analytic functions in the unit disc
Let D = {z ∈ C : |z| < 1}. Denote by A(D) the class of analytic
functions in D. For f ∈ A(D) let M(r, f) = max{|f(z)| : |z| = r} be
the maximum modulus, T (r, f) = 1
2π
∫ 2π
0 log+ |f(reiθ)| dθ, 0 < r < 1, the
Nevanlinna characteristic, x+ = max{x, 0}.
Usually, the orders of the growth of analytic functions inD are defined
as
ρM [f ] = lim sup
r↑1
log+ log+M(r, f)
− log(1 − r)
, ρT [f ] = lim sup
r↑1
log+ T (r, f)
− log(1 − r)
.
Received 5.08.2004
ISSN 1810 – 3200. c© Iнститут прикладної математики i механiки НАН України
I. Chyzhykov 33
It is well known that
ρT [f ] ≤ ρM [f ] ≤ ρT [f ] + 1, (1.1)
and all cases are possible. This is in contrast to entire functions where
the corresponding orders are equal. We cite a couple of results concerning
(1.1).
Given α > 1, ρ satisfying ρ ≤ α ≤ ρ+1, C. N. Linden [1] constructed
an analytic function in D \ {1} of the form of so-called Naftalevich–Tsuji
product
g(z) =
∞
∏
n=1
E
(1 − |an|
2
1 − ānz
, p
)
,
∑
n
(1 − |an|)
p+1 <∞,
with the property ρT [g] = ρ, ρM [g] = α. Here
E(w, p) = (1 − w) exp{w + w2/2 + · · · + wp/p}, p ∈ Z+,
is the Weierstrass primary factor, an are the zeros of g(z).
Another approach is used in a paper by M. Sheremeta [2], where, in
particular, given α > 0 a class of analytic functions f represented by gap
series (with Hadamard’s gaps) is extracted such that
1
∫
0
(1 − r)1+αT (r, f) dr < +∞ ⇔
1
∫
0
(1 − r)1+α logM(r, f) dr < +∞.
Prof. O. Skaskiv posed the following problem
Problem 1.1. Given 0 ≤ ρ ≤ α ≤ ρ + 1, describe the class of analytic
function in D such that ρT [f ] = ρ, ρM [f ] = α.
In order to solve Problem 1.1 one needs a parametric representation
of functions analytic in D and of finite order of the growth. Such rep-
resentation was obtained [3] in 1960th by M. M. Djrbashian using the
Riemann–Liouville fractional integral.
For α > 0 consider two subclasses of A(D)
Aα : sup
0<r<1
2π
∫
0
( r
∫
0
(r − t)α−1 log |f(teiϕ)| dt
)+
dϕ < +∞,
A∗
α : sup
0<r<1
2π
∫
0
( r
∫
0
(r − t)α−1 log+ |f(teiϕ)| dt
)
dϕ < +∞.
34 Growth and representation...
Obviously, A∗
α ⊂ Aα. Note that f ∈ A∗
α means
∫ 1
0 T (t, f)(1 − t)α−1dt <
+∞, i.e. f belongs to the convergence class of order α.
Throughout this paper by (1 − w)α, w ∈ D, α ∈ R, we mean the
branch of the power function such that (1 − w)α
∣
∣
w=0
= 1.
Theorem A. The class Aα coincides with the class of functions repre-
sented in the form
f(z) = Cλz
λBα(z) exp
{ 2π
∫
0
dψ(θ)
(1 − e−iθz)α+1
}
≡ Cλz
λBα(z) exp{gα(z)}, (1.2)
where ψ ∈ BV [0, 2π], (zk) is the zero sequence of f(z) such that
∑
k(1−
|zk|)
α+1 < +∞; Bα(z) =
∏
k
(
1 − z
zk
)
exp{−Wα(z, zk)} is a Djrbashian
product
Wα(z, ζ) =
∑
k
Γ(α+ k + 1)
Γ(α+ 1)Γ(1 + k)
×
(
(ζ̄z)k
1
∫
|ζ|
(1 − x)α
xk+1
dx−
(z
ζ
)k
|ζ|
∫
0
(1 − x)αxk−1 dx
)
.
In this paper we confine by the case when f(z) has no zeros and of
finite order of the growth. Then f(z) = Cα exp{gα(z)}, for some α ≥ 0.
Radial and non-tangential limits of gα(z) were investigated in
many papers, e.g. D. J. Hallenbeck, T. H. MacGregor [4, 5], and
M. M. Sheremeta [6], even for complex-valued functions ψ of bounded
variation. It turns out that gα(z) admits above estimates in terms of the
modulus of continuity for ψ. We cite a typical result [6].
Let S(θ, γ) be the closed Stolz angle having vertex eiθ and open-
ing γ, i.e. S(θ, γ) = {z ∈ D : | arg(eiθ − z)| < γ/2}. A function g
defined in D is said to have a nontangential limit at eiθ provided that
limz→eiθ,z∈S(θ,γ) g(z) exists for every γ ∈ [0, π).
Theorem B. Let α > −1, θ ∈ [0, 2π], ψ ∈ BV [0, 2π], and ω be a
nonnegative, increasing continuous, semi-additive function on [0,+∞),
and ω(0) = 0. If
1
∫
0
t−α−2ω(t) dt = ∞, |ψ(t) − ψ(θ)| = o
(
ω(|t− θ|)
)
, t→ θ,
I. Chyzhykov 35
and gα is given by (1.2) then
|gα(z)|
/
1
∫
|1−ze−iθ|
t−α−2ω(t) dt
has the nontangential limit zero at eiθ.
Since lower estimates for |gα(z)| are known only in particular cases
(see [7], Theorem D and Remark 1.3 below), it is interesting to obtain
results which give lower estimates for |gα(z)| in the general situation.
The main purpose of this paper is to describe the growth of |gα(z)|
in terms of the modulus of continuity for ψ and find counterparts for
harmonic functions in D.
Problem 1.1 is not solved, but Theorem 3.3 and the corollary describe
large classes of analytic functions f with the property ρT [f ] = ρ, ρM [f ] =
α, 0 ≤ ρ ≤ α ≤ ρ+ 1. Theorem 3.2 yields asymptotic formulas for gα in
Stolz angles when ψ is not continuous.
1.2. Representation and the growth of harmonic functions
We need some definitions. Let Uθ(δ) = {x ∈ [0, 2π] : |x − θ| < δ},
δ > 0. For ψ : [0, 2π] → R define the moduli of continuity ω(δ, θ;ψ) =
sup{|ψ(x) − ψ(y)| : x, y ∈ Uθ(δ)}, ω(δ;ψ) = supθ∈[0,2π] ω(δ, θ;ψ).
Following [8] we say that ψ ∈ Λγ if ω(δ; f) = O(δγ) (δ ↓ 0).
The fractional integral of order α > 0 for h : (0, 1) → R is defined by
the formulas [3]
D−αh(r) =
1
Γ(α)
r
∫
0
(r − x)α−1h(x) dx,
D0h(r) ≡ h(r), Dαh(r) =
dp
drp
{D−(p−α)h(r)}, α ∈ (p− 1; p], p ∈ N.
Let H(D) be the class of harmonic functions in D. We put uα(reiϕ) =
r−αD−αu(reiϕ), where the fractional integral is taken on the variable r.
We define B(r, u) = max{u(z) : |z| ≤ r} for a subharmonic function u
in D.
Let
Sα(z) = Γ(1 + α)
( 2
(1 − z)α+1
− 1
)
, Pα(r, t) = ℜSα(reit).
Remark 1.1. Note that S0(z) is the Schwartz kernel, P0(r, t) is the
Poisson kernel; Pα(r, t) = Dα(rαP0(r, t)).
36 Growth and representation...
Our starting point is the following two theorems
Theorem C (M. Djrbashian). Let u ∈ H(D), α > −1. Then
u(reiϕ) =
2π
∫
0
Pα(r, ϕ− θ) dψ(θ), (1.3)
where ψ ∈ BV [0, 2π], if and only if
sup
0<r<1
2π
∫
0
|uα(reiϕ)| dϕ < +∞.
Remark 1.2. Actually, for α = 0 it is the classical result of Nevanlinna
on representation of log |F (z)| when F belongs to the Nevanlinna classN .
Theorem D (Hardy–Littlewood). Let u ∈ H(D), 0 < γ ≤ 1. Then
u(reiϕ) =
2π
∫
0
P0(r, ϕ− t)v(t) dt (1.4)
for some function v ∈ Λγ, if and only if
B
(
r,
∂u
∂ϕ
)
= O((1 − r)γ−1), r ↑ 1.
Remark 1.3. Theorem D was originally proved [9] for analytic function
(cf. Theorem 1.2).
Applying methods of proofs of Theorems B and C, we prove the fol-
lowing theorem which describes the growth of functions of form (1.3).
Theorem 1.1. Let u ∈ H(D), α ≥ 0, 0 < γ < 1. Then u(z) has form
(1.3) where ψ is of bounded variation on [0, 2π], and ψ ∈ Λγ, if and only
if
B(r, u) = O((1 − r)γ−α−1), r ↑ 1
and
sup
0<r<1
2π
∫
0
|uα(reiϕ)| dϕ < +∞.
Note that Theorem D corresponds to the case when ψ is absolutely
continuous, but in Theorem 1.1 the general situation is considered.
Similar to that as we deduce Theorem 1.1 from the proposition below,
one can deduce the following generalization of Theorem D.
I. Chyzhykov 37
Theorem 1.2. Let u ∈ H(D), 0 < γ < 1, α ≥ 0. Then
u(reiϕ) =
2π
∫
0
Pα(r, ϕ− t)v(t) dt
for some function v ∈ Λγ, if and only if
B
(
r,
∂u
∂ϕ
)
= O((1 − r)γ−α−1), r ↑ 1.
It is not difficult to prove a counterpart of the last theorem for analytic
functions.
Remark 1.4. Similar to [6], one can prove that if u(z) is represented by
(1.3), then
u(reiθ) = O
( 2π
∫
1−r
ω(τ, ϕ;ψ)
τα+2
dτ
)
, r ↑ 1, reiθ ∈ S(ϕ, τ), 0 ≤ τ < π.
Problem 1.2. Obtain necessary and sufficient conditions for local growth
of u ∈ H(D).
2. Proof of Theorem 1.1
2.1. The case α = 0
First, it is suitable to prove an important particular case of Theo-
rem 1.1 in spirit of Theorem D.
Proposition 2.1. Let u ∈ H(D), 0 < γ ≤ 1. Then u(z) has the form
u(reiϕ) =
2π
∫
0
P0(r, ϕ− t) dψ(t) (2.1)
where ψ is of bounded variation on [0, 2π], and ψ ∈ Λγ, if and only if
B(r, u) = O((1 − r)γ−1), r ↑ 1
and
sup
0<r<1
2π
∫
0
|u(reiϕ)| dϕ < +∞.
38 Growth and representation...
In the sequel, the symbol C with indices stands for some positive
constants.
Proof of Proposition 2.1. First, we consider the case γ = 1. Note that
the class Λ1 consists of functions that are integrals of bounded functions.
Thus it is sufficient to apply Theorem (6.3) [8, Ch.IV], which states that
(1.4) holds if and only if B(r, u) is bounded as r ↑ 1.
Consider the case γ ∈ (0, 1).
Necessity. The proof of necessity is standard (cf. [11, Ch.8.2], [6]).
The following estimates of P0(r, t) are well known
∣
∣
∣
∂
∂t
P0(r, t)
∣
∣
∣
≤
2
(1 − r)2
,
∣
∣
∣
∂
∂t
P0(r, t)
∣
∣
∣
≤
π2
t2
, r ≥
1
2
, |t| ≤ π. (2.2)
We extend ψ on R by the formula ψ(t+2π)−ψ(t) = ψ(2π)−ψ(0). Since
P0(r, t) is a periodic and even function in t, we have
u(reiϕ) =
π+ϕ
∫
−π+ϕ
P0(r, θ − ϕ)d(ψ(θ) − ψ(ϕ))
= (ψ(θ) − ψ(ϕ))P0(r, θ − ϕ)
∣
∣
∣
π+ϕ
−π+ϕ
−
π+ϕ
∫
−π+ϕ
∂
∂θ
(P0(r, θ−ϕ))(ψ(θ)−ψ(ϕ)) dθ
= (ψ(2π) − ψ(0))P0(r, π) −
π
∫
−π
∂
∂τ
(P0(r, τ))(ψ(τ + ϕ) − ψ(ϕ)) dτ.
Hence, using (2.2), we obtain
|u(reiϕ)|
≤
C1(ψ)(1 − r)
1 + r
+
(
∫
|τ |≤1−r
+
∫
1−r≤|τ |≤π
)
∣
∣
∣
∂
∂τ
P0(r, τ)
∣
∣
∣
ω(|τ |, ψ;ϕ) dτ
≤ o(1) + 2
∫
|τ |≤1−r
ω(|τ |, ψ;ϕ)
(1 − r)2
dτ +
∫
1−r≤|τ |≤π
π2
τ2
ω(|τ |, ψ;ϕ) dτ
≤ o(1) + 4
ω(1 − r, ψ;ϕ)
1 − r
+ 2π2
∫
1−r≤τ≤π
ω(τ, ψ;ϕ)
τ2
dτ
≤ (2π2 + 4)
∫
1−r≤τ≤π
ω(τ, ψ;ϕ)
τ2
dτ +O(1), r ↑ 1. (2.3)
I. Chyzhykov 39
Here, we have used increasing of the modulus of continuity. Since ψ ∈ Λγ ,
ω(τ, ψ;ϕ) = O(τγ) as τ ↓ 0. Thus, (2.3) yields
B(r, u) ≤ C2(γ)(1 − r)γ−1, r ↑ 1.
Sufficiency. Let u(reiϕ) be harmonic for r < 1, and
∫ 2π
0 |u(reiϕ)| dϕ ≤
C3.
Remark 2.1. By Theorem C, we have (2.1), where ψ ∈ BV [0, 2π], and
one can take ψ such that at any point θ of continuity of ψ for some
sequence (rn) ([3], [10, p. 57]).
ψ(θ) = lim
rn↑1
θ
∫
0
u(rne
iφ) dφ. (2.4)
Let F (z) be an analytic function in D1 such that ℜF (z) = u(z). By
the theorem of Zygmund [8, Th. (2.30), Ch. VII] B(r, u) = O((1−r)γ−1)
implies M(r, F ) = O((1 − r)γ−1) as r ↑ 1.
Define the analytic function Φ(z) =
∫ z
0 F (ζ) dζ, z ∈ D. For any fixed
ϕ ∈ [0, 2π] and 0 < r′ < r′′ < 1, we have
|Φ(r′′eiϕ) − Φ(r′eiϕ)| =
∣
∣
∣
∣
∣
r′′
∫
r′
F (ρeiϕ)eiϕ dρ
∣
∣
∣
∣
∣
≤ C4
r′′
∫
r′
(1 − ρ)γ−1 dρ ≤
C4
γ
(1 − r′)1−γ .
Therefore, by Cauchy’s criterion, there exists limr↑1 Φ(reiϕ) ≡ Φ(eiϕ)
uniformly in ϕ. Consequently, Φ̃(ϕ)
def
= Φ(eiϕ) is a continuous function
on [0, 2π].
Let us prove that Φ̃ ∈ Λγ . Let h ∈ (0, 1), z0 = eiϕ, z1 = (1 − h)eiϕ,
z2 = (1 − h)ei(ϕ+h), z3 = ei(ϕ+h).
Then by Cauchy’s theorem
Φ(z3) − Φ(z0) =
∫
[0,z3]
F (z) dz +
∫
[z0,0]
F (z) dz
=
(
∫
[z0,z1]
+
z2
∫
z1
+
∫
[z2,z3]
)
F (z) dz.
40 Growth and representation...
For sufficiently small h > 0, we have
∣
∣
∣
∣
∣
∫
[z0,z1]
F (z) dz
∣
∣
∣
∣
∣
≤
1
∫
1−r
C4
(1 − h)1−γ
dr =
C4
γ
hγ .
Similarly, |
∫
[z2,z3] F (z) dz| ≤ C4
γ h
γ . It is obvious that |
∫ z2
z1
F (z) dz| ≤
C4h
γ .
Therefore, |Φ(ei(ϕ+h)) − Φ(eiϕ)| ≤ C4(
2
γ + 1)hγ , so Φ̃ ∈ Λγ .
For R ∈ (0, 1) define
λR(θ) =
θ
∫
0
F (Reiσ) dσ =
θ
∫
0
dΦ(Reiσ)
iReiσ
=
Φ(Reiθ)
iReiθ
−
Φ(R)
iR
+
1
R
θ
∫
0
Φ(Reiσ)e−iσ dσ.
Since Φ(z) is continuous in {z : |z| ≤ 1}, Φ(Reiσ)⇉
θ
Φ(eiσ) as R ↑ 1.
Consequently, as R ↑ 1
λR(θ) ⇉ −Φ(eiθ)ie−iθ + iΦ(1) +
θ
∫
0
Φ(eiσ)e−iσ dσ ≡ λ(θ) ∈ C[0, 2π].
Since Φ̃ ∈ Λγ , we have λ ∈ Λγ . On the other hand, u(reiϕ) =
∫ 2π
0 P0(r,
ϕ− t) dψ(t), where, by (2.4) and the definition of λ
ψ(θ) = lim
r↑1
θ
∫
0
ℜF (reiφ) dφ = λ(θ).
Thus, ψ ∈ Λγ .
2.2. The case α > 0
Necessity. Let u has form (1.3), where ψ ∈ BV [0, 2π] ∩ Λγ . This
implies
uα(reiϕ) =
2π
∫
0
P0(r, ϕ− θ) dψ(θ). (2.5)
By the proposition we have supr<1
∫ 2π
0 |uα(reiϕ)| dϕ<+∞ and B(r, uα)=
O((1 − r)γ−1) as r ↑ 1.
I. Chyzhykov 41
We use the following formula [3, Chap. IX, (2.9)]
u(reiϕ) =
1
2π
2π
∫
0
Pα(r/ρ, ϕ− θ)uα(ρeiθ) dθ, 0 ≤ r < ρ < 1.
Taking ρ = (1 + r)/2 and using the estimate B(r, uα) = O((1 − r)γ−1)
(r ↑ 1), we obtain
|u(reiϕ)| = O
( 2π
∫
0
(1 − ρ)γ−1 dθ
|1 − r
ρe
i(ϕ−θ)|1+α
)
= O
(
(1 − r)γ−1(ρ− r)−α
)
= O
(
(1 − r)γ−1−α
)
, r ↑ 1.
The necessity is proved.
Sufficiency. Let
∫ 2π
0 |uα(reiϕ)| dϕ < +∞ uniformly for all r ∈ (0, 1).
Then by Theorem C we have (2.5), where ψ ∈ BV [0, 2π].
We need the following elementary lemma.
Lemma 2.1. Let (∀x ∈ [0, 1)) f ∈ L[0, x), and 0 ≤ η < β, and |f(x)| =
O((1 − x)−β) as x ↑ 1. Then |D−ηf(x)| = O((1 − x)η−β) as x ↑ 1.
Proof. Using the definition of the fractional integral and standard esti-
mates we obtain
|D−ηf(x)| =
∣
∣
∣
∣
∣
1
Γ(η)
x
∫
0
f(t)(x− t)η−1 dt
∣
∣
∣
∣
∣
= O
( x
∫
0
(x− t)η−1
(1 − t)β
dt
)
= O
( x−2(1−x)
∫
0
+
x
∫
x−2(1−x)
)
(x− t)η−1
(1 − t)β
dt
= O
( x−2(1−x)
∫
0
(1 − t)η−1−β dt+ (1 − x)η−1
x
∫
x−2(1−x)
dt
(1 − t)β
)
= O((1 − x)η−β), x ↑ 1.
The lemma is proved.
Since B(r, u) = O((1 − r)γ−1−α) as r ↑ 1, by the lemma B(r, uα) =
O((1 − r)γ−1) as r ↑ 1. Therefore, by Proposition 2.1, ψ ∈ Λγ .
42 Growth and representation...
3. Further results for analytic functions
There is an analogue of Theorem C for analytic functions proved by
M. Djrbashian [3]. The following theorem can be proved as the proposi-
tion and Theorem 1.1 were proved.
Theorem 3.1. Let f(z) be an analytic function in D, α ≥ 0, 0 < γ < 1.
Then f(z) has the form
f(reiϕ) =
2π
∫
0
Sα(r, ϕ− t) dψ(t) + iℑf(0) (3.1)
where ψ is of bounded variation on [0, 2π], and ψ ∈ Λγ, if and only if
B(r, |f |) = O((1 − r)γ−α−1), r ↑ 1
and
sup
0<r<1
∫ 2π
0
|ℜfα(reiϕ)| dϕ < +∞,
where fα(reit) = r−αD−αf(reit).
Theorem 1.1 does not cover the case when ψ ∈ Λ0, in particular,
when ψ is not continuous. Here, following [6] we are able to prove a
more precise result. It looks to be known, but I have not found it in a
literature.
Theorem 3.2. Let f(z) have the form
f(z) =
2π
∫
0
(1 − ze−it)−α dψ(t), z ∈ D
where α > 0, ψ ∈ BV [0, 2π]. If {tk} is the set of the discontinuity points
of ψ with jumps {hk}, then
f(z) =
hk + o(1)
(1 − ze−itk)α
, z → eitk , z ∈ S(tk, τ), τ ∈ [0, π), (3.2)
and
f(z) =
o(1)
(1 − ze−it)α
, z → eit, z ∈ S(t, τ), t 6∈ {tk}, τ ∈ [0, π). (3.3)
I. Chyzhykov 43
Proof. Since ψ ∈ BV [0, 2π], the set {tk} is at most countable. Without
loss of generality we can assume that ψ is continuous from the right.
Then hk = ψ(tk) − ψ(tk − 0). It is sufficient to prove (3.2) for k = 1.
We may assume that t1 ∈ (0, 2π). Let
H1(t) =
{
0, 0 ≤ t < t1,
h1, t1 ≤ t ≤ 2π.
We extend ψ on R by the formula ψ(t + 2π) − ψ(t) = ψ(2π) − ψ(0) as
well as H1. The function g(t)
def
= ψ(t)−H1(t) is continuous at the points
t = t1 + 2πk, k ∈ Z, so ω(δ, t1, g) = o(1) as δ ↓ 0. We have
f(z) =
2π
∫
0
(1 − ze−it)−α dg(t) +
h1
(1 − ze−it1)α
. (3.4)
Let ω1(δ) = max{
√
ω(δ, t1, g), δ
α/2}. It is easy to see that ω1(δ) satisfies
the hypotheses of Theorem B on ω(δ). Applying Theorem B to the
integral from (3.4) we obtain (3.2) from (3.4).
Relationship (3.3) follows directly from Theorem B if we choose
ω2(δ) = max{
√
ω(δ, t, ψ), δα/2}.
For ψ ∈ BV [0, 2π] we define τ [ψ] to be sup γ satisfying ψ ∈ Λγ . In
particular, ω(δ, ψ) ∈ Λτ [ψ]−ε \ Λτ [ψ]+ε.
Theorem 3.3. Let F (z) be analytic in D,
log |F (reiϕ)| =
2π
∫
0
Pα(r, ϕ− t) dψ(t),
where ψ ∈ BV [0, 2π], τ [ψ] = τ ∈ [0, 1). Then ρM [F ] = α+1−τ , ρT [F ] ≤
α. If, in addition, ψ is not absolutely continuous, then ρT [F ] = α.
Corollary 3.1. Suppose that the conditions of Theorem 3.3 hold, and
τ = 0. Then ρM [F ] = ρT [F ] + 1 = α+ 1.
Proof of Theorem 3.3. First, let τ ∈ (0, 1). By Theorem 1.1
sup
r<1
2π
∫
0
|uα(reiϕ)| dϕ < +∞.
Since ω(δ, ψ) ∈ Λτ−ε\Λτ+ε, 0 < ε ≤ min{τ, 1−τ}, applying Theorem 1.1
again, we have
logM(r, F ) = B(r, log |F |) = O((1 − r)τ−α−1−ε),
44 Growth and representation...
logM(r, F ) 6= O((1 − r)τ−α−1+ε), r ↑ 1,
i.e. ρM [F ] = α+ 1 − τ .
Further,
T (r, f) =
1
2π
2π
∫
0
(
1
2π
2π
∫
0
Pα(r, ϕ− t) dψ(t)
)+
dϕ
≤
1
4π2
2π
∫
0
2π
∫
0
P+
α (r, ϕ− t) dψ(t) dϕ
≤
1
4π2
2π
∫
0
dψ(t)
2π
∫
0
2
|1 − reiθ|α+1
dθ
=
{
O((1 − r)−α), α > 0,
O
(
log 1
1−r
)
, α = 0,
r ↑ 1,
i.e. ρT [f ] ≤ α.
In order to complete the proof of Theorem 3.3 we need the following
result of F. A. Shamoian [12], which compares the classes Aα and A∗
α.
Theorem E ([12, Theorem 3]). The function
F (z) = exp
{
1
2π
2π
∫
0
Sα(ze−iθ) dψ(θ)
}
∈ A∗
α
if and only if: 1) ψ is absolutely continuous;
2)
2π
∫
0
2π
∫
0
|ψ(θ + t) − 2ψ(θ) + ψ(θ − t)|
t2
dt dθ < +∞.
As we noted above F ∈ A∗
α if and only if T (r, F ) belongs to the con-
vergence class of order α. Therefore, if ψ is not absolutely continuous, F
has the growth at least of the divergence class of order α, i.e. ρT [F ] ≥ α.
If τ = 0, then similarly one can deduce that ρT [f ] ≤ α. Since
ω(δ;ψ) 6∈ Λε, ε > 0,
logM(r, f) 6= O((1 − r)−α−1+ε), r ↑ 1,
i.e. ρM [F ] ≥ α + 1. Using the inequality ρM [F ] ≤ ρT [F ] + 1, we obtain
ρM [F ] = ρT [F ] + 1 = α+ 1, i.e. the statement of the corollary.
I. Chyzhykov 45
Remark 3.1. The condition τ < 1 in Theorem 3.3 is essential. In fact,
by the Cauchy theorem on residues
2π
∫
0
dθ
(1 − e−iθz)n
= 2π, n ∈ N, z ∈ D.
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Contact information
Ihor Chyzhykov Faculty of Mechanics and Mathematics,
Lviv Ivan Franko National University,
Universytets’ka 1,
79000, Lviv
Ukraine
E-Mail: ichyzh@lviv.farlep.net
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