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.
Proof.
The proof follows the idea of Theorems 3 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
/
p
for 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
.
Given
R
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 number
n
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
j
for
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
+
1
for
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 value
k
∈
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 that
h
(
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
)
.
Since
Γ
′
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
)
.
Thus
h
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
B
for 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.