# Bounded Fatou Components of Composite Transcendental Entire Functions with Gaps

Authors: Cunji Yang; Shaoming Wang
Journal: Discrete Dynamics in Nature and Society
 (2015)

---

## Abstract

We prove that composite transcendental entire functions with certain gaps have no unbounded Fatou component.


---

## Body

## 1. Introduction

Letf be a transcendental entire function. We write f
1
=
f, and f
n
=
f
∘
f
n
-
1 for n
≥
2 for the nth iterate of f. The Fatou set or set of normality F
(
f
) of f consists of all z in the complex plane C which has a neighborhood U such that the family {
f
n
∣
U
:
n
≥
1
} is a normal family. The Julia set J
(
f
) of f is J
(
f
)
=
C
∖
F
(
f
). For the fundamental results in the iteration theory of rational and entire functions, we refer to the original papers of Fatou [1–3], and Julia [4] and the books of Beardon [5], Carleson and Gamelin [6], Milnor [7], and Ren [8].LetU be a connected component of F
(
f
). Then f
n
(
U
)
⊆
U
n, where U
n is a component of F
(
f
). If there is a smallest positive integer p such that U
p
=
U, then U is periodic of period p. In particular, if p
=
1, then U is called invariant. If for some integer n
≥
1, U
n is periodic, while U is not periodic then U is called preperiodic. If U is periodic and f
n
∣
U
→
∞ then U is called a Baker domain. If all the U
n are disjoint, that is, for n
≠
m, U
n
≠
U
m then U is called a wandering domain. Let f be a transcendental entire function. In 1981, Baker [9] proposed whether every component of F
(
f
) is bounded if the growth of f is sufficiently small. The appropriate growth condition would appear to be of order 1
/
2, minimal type at most. In [9], Baker observed that this condition is best possible in the following sense: for any sufficiently large positive a, the function (1)
f
(
z
)
=
z
-
1
/
2
sin
z
+
z
+
ais of order 1
/
2, mean type, and has an unbounded component U of F
(
f
) containing a segment [
x
0
,
∞
) of the positive real axis, such that f
n
(
z
)
→
∞ as n
→
∞, locally uniformly in U.It is conjectured that if the order off is less than 1
/
2, minimal type, then every component of F
(
f
) is bounded. It is still open for wandering domains, although there are several remarkable results for wandering domains under the assumptions that the growth satisfies, in addition, some regular conditions; see [10–17].Suppose thatf
(
z
)
=
∑
n
=
0
∞
a
n
z
n is an entire function with gaps; that is, some of the a
n are zero, in a certain sense. Then the function has the form f
(
z
)
=
∑
k
=
0
∞
a
k
z
n
k. We say that f
(
z
) has Fabry gaps if n
k
/
k
→
∞ as k
→
∞, and f
(
z
) has Fejér gaps if ∑
k
=
1
∞
(
1
/
n
k
)
<
∞.Wang [18] proved that every component of the Fatou set of an entire function with certain gaps is bounded, by using the properties of the entire functions with such gaps. Wang [18] obtained the following result.Theorem 1.
Letf
(
z
)
=
∑
k
=
0
∞
a
k
z
n
k be an entire function with 0
<
μ
≤
ρ
<
∞. If f
(
z
) has Fabry gaps, then every component of F
(
f
) is bounded.For Fejér gap, Wang [18] proposed the following problem.Letf
(
z
) be an entire function with Fejér gaps, that is, (2)
∑
k
=
1
∞
‍
1
n
k
<
∞
.Is every component of F
(
f
) bounded?For composite of entire function, Qiao [19] proved the following result.Theorem 2.
Leth
(
z
)
=
f
N
∘
f
N
-
1
∘
⋯
∘
f
1 be a transcendental entire function, where f
j
(
j
=
1,2
,
…
,
N
) are entire functions with order ρ
j
<
1
/
2. Then every nonwandering component is bounded.Cao and Wang [20] proved the following result.Theorem 3.
Leth
(
z
)
=
f
N
∘
f
N
-
1
∘
⋯
∘
f
1, where f
j
(
j
=
1,2
,
…
,
N
) are nonconstant holomorphic maps, each having order less than 1/2. If there is a number i
∈
{
1,2
,
…
,
N
} such that the lower order of f
i is greater than 0, then every component of F
(
h
) is bounded.Singh [21] proved the following result.Theorem 4.
LetL be the set of all entire functions f such that, for given ɛ
>
0, (3)
log
⁡
m
(
r
,
f
)
≥
(
1
-
ɛ
)
log
⁡
M
(
r
,
f
)holds for all r outside a set of logarithmic density 0. Let F
=
⋃
K
≥
1
F
K where F
K is the set of all transcendental entire functions f such that (4)
log
log
M
(
r
,
f
)
≥
log
⁡
r
1
/
K
.If h
(
z
)
=
f
N
∘
f
N
-
1
∘
⋯
∘
f
1
(
z
), where f
j
∈
F
⋂
‍
L
(
j
=
1,2
,
…
,
N
), then every component of F
(
h
) is bounded.


## 2. Preliminaries

We use the standard notations for the maximum modulusM
(
r
,
f
), minimum modulus m
(
r
,
f
), order of growth ρ, and lower order of growth μ of a function f; namely, (5)
M
(
r
,
f
)
=
max
⁡
{
f
(
z
)
:
z
=
r
}
,
m
(
r
,
f
)
=
min
⁡
{
f
(
z
)
:
z
=
r
}
,
ρ
=
limsup
⁡
r
→
∞
log
⁡
log
⁡
M
(
r
,
f
)
log
⁡
r
,
μ
=
liminf
⁡
r
→
∞
log
⁡
log
⁡
M
(
r
,
f
)
log
⁡
r
.Briefly, we also denote maximum modulus M
(
r
,
f
) and minimum modulus m
(
r
,
f
) by M
(
r
) and m
(
r
).LetE be a set in C. The logarithmic measure of a set E is defined by ∫
E
‍
(
d
t
/
t
). If E
⊂
[
1
,
∞
), E
(
a
,
b
) denote the part of E in the interval (
a
,
b
), that is, E
(
a
,
b
)
=
E
⋂
‍
(
a
,
b
), then the upper logarithmic density of the set E is defined by (6)
logdens
¯
E
=
limsup
⁡
r
→
∞
1
log
⁡
r
∫
E
(
1
,
r
)
‍
d
t
t
;the lower logarithmic density of the set E is defined by (7)
logdens
_
E
=
liminf
⁡
r
→
∞
1
log
⁡
r
∫
E
(
1
,
r
)
‍
d
t
t
.If the upper and lower logarithmic density are equal, their common value is called the logarithmic density of E.Lemma 5 (see [22]).
Letf be a transcendental entire function. Then there exists R
>
0 such that, for all r
≥
R and all c
>
1, (8)
log
⁡
M
(
r
c
,
f
)
≥
c
log
⁡
M
(
r
,
f
)
.Lemma 6 (see [23]).
Letf be an entire function of finite order with Fabry gaps. Then for given ɛ
>
0, (9)
log
⁡
m
(
r
,
f
)
≥
(
1
-
ɛ
)
log
⁡
M
(
r
,
f
)holds for all r outside a set of logarithmic density 0.Lemma 7 (see [24]).
Letf
=
∑
k
=
0
∞
a
k
z
n
k be an entire function with (10)
n
k
>
k
log
⁡
k
log
log
k
α
a
s
k
⟶
∞for some α
>
2. Then for given ɛ
>
0, (11)
log
⁡
m
(
r
,
f
)
≥
(
1
-
ɛ
)
log
⁡
M
(
r
,
f
)holds for all r outside a set of logarithmic density 0.Lemma 8 (see [25]).
Iff
=
∑
k
=
0
∞
a
k
z
n
k satisfies the gap-condition n
k
>
k
(
log
⁡
k
)
2
+
η, then for given ɛ
>
0, (12)
log
⁡
m
(
r
,
f
)
≥
(
1
-
ɛ
)
log
⁡
M
(
r
,
f
)holds for all r outside a set of finite logarithmic measure.Lemma 9 (see [26]).
For an entire functionf
(
z
) with Fejér gaps and ɛ
>
0, (13)
log
⁡
m
(
r
,
f
)
≥
(
1
-
ɛ
)
log
⁡
M
(
r
,
f
)
.Lemma 10 (see [9]).
LetD be a domain and K a compact subset of D. Let G be the family of all holomorphic functions g on D which omit the values 0, 1 and satisfy the condition that 

g


≥
1 on K. Then there exist constants A and B such that 

g
(
z
′
)


<
A


g
z


B, for every g
∈
G and every z
,
z
′
∈
K.Lemma 11.
Letf be a transcendental entire function of finite order with Fabry gaps. Then there exist L
>
1 and R
>
0 such that, for all r
≥
R, there exists σ satisfying r
≤
σ
≤
r
L and m
(
σ
,
f
)
=
M
(
r
,
f
).Proof.
By Lemma6, given any ɛ
>
0, we have m
(
r
,
f
)
>
M
(
r
,
f
)
1
-
ɛ, for all r outside a set E of logarithmic density 0. Let c
>
1. Then there exists R
0 such that, for all R
>
R
0, there exists s
∈
[
R
,
R
c
] such that m
(
s
,
f
)
>
M
(
s
,
f
)
1
-
ɛ. If not, then there exists a sequence R
j
→
∞ such that m
(
s
,
f
)
≤
M
(
s
,
f
)
1
-
ɛ, for every s
∈
[
R
j
,
R
j
c
]. Thus [
R
j
,
R
j
c
]
⊂
E. So(14)
logdens
_
E
=
liminf
⁡
r
→
∞
1
log
⁡
r
∫
E
(
1
,
r
)
‍
d
t
t
≥
liminf
⁡
j
→
∞
1
log
⁡
R
j
c
∫
E
(
1
,
R
j
c
)
‍
d
t
t
≥
liminf
⁡
j
→
∞
1
c
log
⁡
R
j
∫
R
j
R
j
c
‍
d
t
t
=
1
-
1
c
,contradicting logdense
(
E
)
=
0.
SetR
1
-
2
ɛ
=
r. There exists s
∈
[
r
1
/
(
1
-
2
ɛ
)
,
r
c
/
(
1
-
2
ɛ
)
] such that m
(
s
,
f
)
>
M
(
s
,
f
)
1
-
ɛ. By Lemma 6, (15)
m
s
,
f
>
M
s
,
f
1
-
ɛ
≥
M
r
1
/
1
-
2
ɛ
,
f
1
-
ɛ
≥
M
r
,
f
(
1
-
ɛ
)
/
(
1
-
2
ɛ
)
≥
M
(
r
,
f
)
.Since m
(
r
,
f
)
≤
M
(
r
,
f
), there exists σ
∈
[
r
,
s
]
⊂
[
r
,
r
L
], where L
=
c
/
(
1
-
2
ɛ
) such that m
(
σ
,
f
)
=
M
(
r
,
f
). This completes the proof of Lemma 11.By Lemma9 and the same method in the proof of Lemma 11, we can prove the following result.Lemma 12.
Letf be a transcendental entire function of finite order with Fejér gaps. Then there exist L
>
1 and R
>
0 such that, for all r
≥
R, there exists σ satisfying r
≤
σ
≤
r
L and m
(
σ
,
f
)
=
M
(
r
,
f
).By Lemma7 and the same method in the proof of Lemma 11, we can prove the following result.Lemma 13.
Letf be a transcendental entire function of finite order with (16)
n
k
>
k
log
⁡
k
log
log
k
α
a
s
k
⟶
∞for some α
>
2. Then there exist L
>
1 and R
>
0 such that, for all r
≥
R, there exists σ satisfying r
≤
σ
≤
r
L and m
(
σ
,
f
)
=
M
(
r
,
f
).


## 3. Main Results

In 2012, the authors proved some results on the bounded Fatou components of transcendental entire functions with gaps; see [27]. In this paper, we investigate the iteration of the composite entire functions with gaps and obtain the following results.Theorem 14.
Leth
(
z
)
=
f
N
∘
f
N
-
1
∘
⋯
∘
f
1
(
z
) be a transcendental entire function, where f
j
=
∑
k
=
0
∞
a
j
k
z
n
j
k
(
j
=
1,2
,
…
,
N
) have Fabry gaps of finite order. If (17)
M
(
r
,
f
j
)
≥
exp
⁡
(
exp
⁡
log
⁡
r
p
)
,
0
<
p
<
1
,then F
(
h
) has no unbounded component.Proof.
The proof follows the idea of Theorems3 and 4. Since f
j is a transcendental entire function of finite order with Fabry gaps, by Lemma 11, there exist L
j
>
1 and r
j sufficiently large such that, for r
≥
r
j, there exists σ
(
j
) such that r
≤
σ
(
j
)
≤
r
L
j and(18)
m
(
σ
(
j
)
,
f
j
)
=
M
(
r
,
f
j
)
,
j
=
1,2
,
…
,
N
.Since f
j is a transcendental entire function of finite order, h must be a transcendental entire function. It follows that there exists a number r
0
>
1 such that M
(
r
,
h
)
>
r
2 for all r
>
r
0, and there exists a number t
j
>
1 such that (19)
exp
⁡
(
r
t
j
)
≥
M
r
,
f
j
log
⁡
M
r
,
f
j
1
/
pfor all r sufficiently large. In fact, if there is a sequence {
r
n
} which tend to ∞ such that (20)
exp
⁡
(
r
n
t
)
<
M
r
n
,
f
j
log
⁡
M
r
n
,
f
j
1
/
p
,then (21)
t
log
⁡
r
n
<
1
+
p
p
·
log
⁡
log
⁡
M
(
r
n
,
f
j
)
.So (22)
t
≤
1
+
p
p
lim
n
→
∞
log
⁡
log
⁡
M
(
r
n
,
f
j
)
log
⁡
r
n
≤
1
+
p
p
ρ
.Which gives a contradiction if we take any t
>
(
1
+
p
/
p
)
ρ.
GivenR
1
(
1
)
>
r
0, define inductively (23)
R
n
(
j
+
1
)
=
M
(
R
n
(
j
)
,
f
j
)
,
j
=
1,2
,
…
,
N
-
1
;
R
n
+
1
(
1
)
=
M
(
R
n
(
N
)
,
f
N
)
.It is easy to see that, for all n
∈
N, (24)
R
n
1
2
<
M
(
R
n
(
1
)
,
h
)
≤
R
n
+
1
(
1
)
.Thus (25)
R
n
+
1
(
1
)
>
R
1
1
2
n
⟶
∞
as
n
⟶
∞and R
n
(
j
)
→
∞ as n
→
∞ for all j
=
1,2
,
…
,
N.
Take numbern
0 sufficiently large, for n
≥
n
0, by Lemma 11, there exists σ
n
(
j
) such that (26)
exp
⁡
R
n
j
t
j
≤
σ
n
j
≤
exp
⁡
L
j
R
n
j
t
j
,
m
σ
n
j
,
f
j
=
M
exp
⁡
R
n
j
t
j
,
f
jfor n
≥
n
0. By the hypotheses of Theorem 14, (19), and Lemma 11, we have(27)
m
σ
n
j
,
f
j
=
M
exp
⁡
R
n
j
t
j
,
f
j
≥
M
M
R
n
j
,
f
j
log
⁡
M
R
n
j
,
f
j
1
/
p
,
f
j
≥
exp
⁡
exp
⁡
log
⁡
M
R
n
j
,
f
j
log
⁡
M
R
n
j
,
f
j
1
/
p
p
=
exp
⁡
exp
⁡
log
⁡
M
R
n
j
,
f
j
log
⁡
M
R
n
j
,
f
j
p
=
exp
⁡
M
R
n
j
,
f
j
log
⁡
M
R
n
j
,
f
j
p
≥
exp
⁡
L
j
+
1
M
R
n
j
,
f
j
t
j
+
1
=
exp
⁡
L
j
+
1
R
n
j
+
1
t
j
+
1for j
=
1,2
,
…
,
N
-
1; and(28)
m
σ
n
N
,
f
N
=
M
exp
⁡
R
n
N
t
N
,
f
N
≥
M
M
R
n
N
,
f
N
log
⁡
M
R
n
N
,
f
N
1
/
p
,
f
N
≥
exp
⁡
exp
⁡
log
⁡
M
R
n
N
,
f
N
log
⁡
M
R
n
N
,
f
N
1
/
p
p
=
exp
⁡
exp
⁡
log
⁡
M
R
n
N
,
f
N
log
⁡
M
R
n
N
,
f
N
p
=
exp
⁡
M
R
n
N
,
f
N
log
⁡
M
R
n
N
,
f
N
p
≥
exp
⁡
L
1
M
R
n
N
,
f
N
t
1
=
exp
⁡
L
1
R
n
+
1
1
t
1
.Suppose on contrary that F
(
h
) has an unbounded component D. Without loss of generality we may assume 0, 1 belong to J
(
h
). Hence each map h
n omits the values 0, 1 in D. It follows from the unboundedness and connectivity of D that there exists n
1
≥
n
0
∈
N such that D meets the circles (29)
α
n
(
j
)
=
{
z
:
z
=
R
n
(
j
)
}
,
β
n
(
j
)
=
z
:
z
=
exp
⁡
L
j
R
n
j
t
j
,
γ
n
(
j
)
=
{
z
:
z
=
σ
n
(
j
)
}for all n
≥
n
1,  j
=
1,2
,
…
,
N.
We choose a valuek
∈
N such that k
≥
n
1 and note that D must contain a path Γ joining a point ω
k
(
1
)
∈
α
k
(
1
) to a point τ
k
+
1
(
1
)
∈
β
k
+
1
(
1
). It is clear that Γ contains two subsets Γ
′, Γ
′′ such that Γ
′ joins ω
k
(
1
)
∈
α
k
(
1
) to η
k
(
1
)
∈
β
k
(
1
) and contains ξ
k
(
1
)
∈
γ
k
(
1
) and Γ
′′ joins δ
k
+
1
(
1
)
∈
α
k
+
1
(
1
) to τ
k
+
1
(
1
)
∈
β
k
+
1
(
1
). We know that R
k
(
2
)
=
M
(
R
k
(
1
)
,
f
1
) and so R
k
(
2
)
≥


f
1
(
ω
k
(
1
)
)

. Also m
(
σ
k
(
1
)
,
f
1
)
≥
exp
⁡
(
L
2
(
R
k
(
2
)
)
t
2
) and so 

f
1
(
ξ
k
(
1
)
)


≥
exp
⁡
(
L
2
(
R
k
(
2
)
)
t
2
). Hence f
1
(
Γ
′
) contains an arc joining a point ω
k
(
2
)
∈
α
k
(
2
) to a point η
k
(
2
)
∈
β
k
(
2
). Similarly f
1
(
Γ
′′
) contains an arc joining a point δ
k
+
1
(
2
)
∈
α
k
+
1
(
2
) to a point τ
k
+
2
(
2
)
∈
β
k
+
2
(
2
).
Repeating the process inductively we obtain thath
(
Γ
′
)
=
f
N
∘
f
N
-
1
∘
⋯
∘
f
1
(
Γ
′
) contains an arc joining ω
k
+
1
(
1
)
∈
α
k
+
1
(
1
) to a point η
k
+
1
(
1
)
∈
β
k
+
1
(
1
) and h
(
Γ
′′
) contains an arc joining a point δ
k
+
2
(
1
)
∈
α
k
+
2
(
1
) to a point τ
k
+
2
(
1
)
∈
β
k
+
2
(
1
).
SinceΓ
′ and Γ
′′ are two subsets of Γ, it follows that h
(
Γ
) must contain an arc joining ω
k
+
1
(
1
)
∈
α
k
+
1
(
1
) to the point τ
k
+
2
(
1
)
∈
β
k
+
2
(
1
). By induction it now follows that h
n
(
Γ
) contains an arc joining a point ω
k
+
n
(
1
)
∈
α
k
+
n
(
1
) to the point τ
k
+
n
+
1
(
1
)
∈
β
k
+
n
+
1
(
1
).
Thush
n
D is a component of F
h containing h
n
Γ, and, on Γ, h
n takes a value of modulus at least R
k
+
n
1 and R
k
+
n
1
→
∞ as n
→
∞. Thus we conclude that R
k
+
n
1
→
∞ locally uniformly in D. Hence there exists N
∈
N such that, for all z
∈
Γ, 

h
n
z


>
1 for all n
>
N. Thus the family {
h
n
}
n
>
N satisfies the conditions of Lemma 10 on Γ, and so there exist constants A, B such that 

h
n
(
z
′
)


<
A


h
n
z


B for all n
>
N and for all z
,
z
′
∈
Γ. Choose z
n
,
z
n
′
∈
Γ with n
>
N such that h
n
(
z
n
)
=
ω
k
+
n
(
1
)
∈
α
k
+
n
(
1
) and h
n
(
z
n
′
)
=
τ
k
+
n
+
1
(
1
)
∈
β
k
+
n
+
1
(
1
); we have (30)
M
R
k
+
n
1
,
h
=
M
(
R
k
+
n
(
1
)
,
f
N
∘
f
N
-
1
∘
⋯
∘
f
1
)
≤
M
(
R
k
+
n
(
N
)
,
f
N
)
=
R
k
+
n
+
1
(
1
)
<
exp
⁡
L
1
R
k
+
n
+
1
1
t
1
=
h
n
(
z
n
′
)
<
A
h
n
z
n
B
=
A
R
k
+
n
1
Bfor all n
>
N which contradicts the fact that h is a transcendental entire function and R
k
+
n
(
1
)
→
∞ as n
→
∞. This completes the proof of Theorem 14.Corollary 15 (see [27]).
Letf
(
z
)
=
∑
k
=
0
∞
a
k
z
n
k be a transcendental entire function of finite order with Fabry gaps. If (31)
M
(
r
,
f
)
≥
exp
⁡
(
exp
⁡
log
⁡
r
p
)
,
0
<
p
<
1
,then F
(
f
) has no unbounded component.Remark 16.
Iff satisfies M
(
r
,
f
)
≥
exp
⁡
(
exp
⁡
(
log
⁡
r
)
p
), 0
<
p
<
1, then ρ
≥
0, so Corollary 15 is an extension of Theorem 1.
By Lemma12 and the same method used in the proof of Theorem 14, we can show the following result.Theorem 17.
Leth
(
z
)
=
f
N
∘
f
N
-
1
∘
⋯
∘
f
1
(
z
) be a transcendental entire function, where f
j
=
∑
k
=
0
∞
a
j
k
z
n
j
k
(
j
=
1,2
,
…
,
N
) have Fejér gaps. If (32)
M
(
r
,
f
j
)
≥
exp
⁡
(
exp
⁡
log
⁡
r
p
)
,
0
<
p
<
1
,then F
(
h
) has no unbounded component.By Theorem17, we have the following.Corollary 18 (see [27]).
Letf
(
z
)
=
∑
k
=
0
∞
a
k
z
n
k be a transcendental entire function with Fejér gaps. If (33)
M
(
r
,
f
)
≥
exp
⁡
(
exp
⁡
log
⁡
r
p
)
,
0
<
p
<
1
,then F
(
f
) has no unbounded component.Remark 19.
Corollary18 partly answers the problem of Wang on the Fejér gaps.
By Lemma13 and the same method of the proof of Theorem 14, we can show Theorem 20.Theorem 20.
Leth
(
z
)
=
f
N
∘
f
N
-
1
∘
⋯
∘
f
1
(
z
) be a transcendental entire function, where f
j
=
∑
k
=
0
∞
a
j
k
z
n
j
k
(
j
=
1,2
,
…
,
N
) are entire functions with the gap-conditions (34)
n
j
k
>
k
log
⁡
k
log
log
k
α
a
s
k
⟶
∞
f
o
r
s
o
m
e
α
>
2
.If (35)
M
(
r
,
f
j
)
≥
exp
⁡
(
exp
⁡
log
⁡
r
p
)
,
0
<
p
<
1
,then F
(
h
) has no unbounded component.By Theorem20, we have the following.Corollary 21.
Leth
(
z
)
=
f
N
∘
f
N
-
1
∘
⋯
∘
f
1
(
z
) be a transcendental entire function, where f
j
=
∑
k
=
0
∞
a
j
k
z
n
j
k
(
j
=
1,2
,
…
,
N
) are entire functions with the gap-conditions n
j
k
>
k
(
log
⁡
k
)
2
+
η for some η
>
0. If (36)
M
(
r
,
f
j
)
≥
exp
⁡
(
exp
⁡
log
⁡
r
p
)
,
0
<
p
<
1
,then F
(
h
) has no unbounded component.



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*Source: 149182-2015-04-12.xml*