1. No category

Bounded Fatou components of composite
transcendental entire functions with gaps
Cunji Yang1 , Shaoming Wang
Department of Mathematics and Computer Sciences, Dali University, Dali, Yunnan
671003, P.R. China
Abstract
We prove that composite transcendental entire functions with certain gaps
have no unbounded Fatou component.
Keywords:
Transcendental entire function, gap power series, composite, Fatou
component
2010 MSC: Primary: 30D05; Secondary: 37F10.
1. Introduction
Let f 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 it 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].
Let U be a connected component of F (f ). Then f n (U ) ⊆ Un , where Un
is a component of F (f ). If there is a smallest positive integer p such that
Up = U , then U is periodic of period p. In particular, if p = 1, then U is called
invariant. If for some integer n ≥ 1, Un 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 Un are disjoint, that is, for n 6= m, Un 6= Um then
1
Corresponding author: Email addresses: [email protected] (Cunji Yang)
Preprint submitted to Discrete Dynamics in Nature and Society
October 10, 2014
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 should be of order 21 , minimal type at most. In [9], Baker observed
that this condition is best possible in following sense: for any sufficiently
large positive a, the function
√
1
f (z) = z − 2 sin z + z + a
is of order 21 , mean type, and has an unbounded component U of F (f )
containing a segment [x0 , ∞) of the positive real axis, such that f n (z) → ∞
as n → ∞, locally uniformly in U .
The conjecture that if the order of f is less than 21 , minimal type, then
every component of F (f ) is bounded 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].
∞
P
Suppose that f (z) =
an z n is an entire function with gaps, i. e.,
n=0
some of the an are zero, in a certain sense. Then the function has the form
∞
P
ak z nk . We say that f (z) has Fabry gaps if nkk → ∞ as k → ∞,
f (z) =
k=0
and f (z) has Fej´er gaps if
∞
P
k=1
1
nk
< ∞.
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 1.1([18]). Let f (z) =
∞
P
ak z nk be an entire function with 0 < µ ≤
k=0
ρ < ∞. If f (z) has Fabry gaps, then every component of F (f ) is bounded.
For Fej´er gap, Wang [18] proposed the following problem.
Problem 1.2([18]). Let f (z) be an entire function with Fej´er gaps, i.e.,
∞
X
1
< ∞.
n
k
k=1
Is every component of F (f ) bounded ?
2
For composite of entire function, Qiao [19] proved the following result.
Theorem 1.3([19]). Let h(z) = fN ◦ fN −1 ◦ · · · ◦ f1 be a transcendental entire
function, where fj (j = 1, 2, . . . , N ) are entire functions with order ρj < 12 .
Then every non-wandering component is bounded.
Cao and Wang [20] proved the following result.
Theorem 1.4([20]). Let h(z) = fN ◦fN −1 ◦· · ·◦f1 , where fj (j = 1, 2, . . . , N )
are non-constant holomorphic maps, each having order less than 12 . If there
is a number i ∈ {1, 2, . . . , N } such that the lower order of fi is greater than
0, then every component of F (h) is bounded.
Singh [21] proved the following result.
Theorem 1.5([21]). Let L be the set of all entire functions f such that for
given ε > 0,
log m(r, f ) ≥ (1 − ε) log M (r, f )
S
holds for all r outside a set of logarithmic density 0. Let F =
FK where
K≥1
FK is the set of all transcendental entire functions f such that
1
log log M (r, f ) ≥ (log r) K .
T
If h(z) = fN ◦ fN −1 ◦ · · · ◦ f1 (z), where fj ∈ F L (j = 1, 2, . . . , N ), then
every component of F (h) is bounded.
2. Preliminaries
We use the standard notations for the maximum modulus M (r, f ), minimum modulus m(r, f ), order of growth ρ and lower order of growth µ of a
function f, namely,
M (r, f ) = max{|f (z)| : |z| = r},
m(r, f ) = min{|f (z)| : |z| = r},
ρ = lim sup
r→∞
log log M (r, f )
,
log r
3
µ = lim inf
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).
R dtLet E be a set in C. The logarithmic measure of a set E is defined by
. If E ⊂ [1,
E t
T∞), E(a, b) denote the part of E in the interval (a, b),
i.e., E(a, b) = E (a, b), then the upper logarithmic density of the set E is
defined by
Z
1
dt
logdensE = lim sup
;
r→∞ log r E(1,r) t
The lower logarithmic density of the set E is defined by
Z
1
dt
.
logdensE = lim inf
r→∞ log r E(1,r) t
If the upper and lower logarithmic density are equal, their common value is
called the logarithmic density of E.
Lemma 2.1([22]). Let f be a transcendental entire function. Then there
exists R > 0 such that, for all r ≥ R and all c > 1,
log M (rc , f ) ≥ c log M (r, f ).
Lemma 2.2([23]). Let f be an entire functions of finite order with Fabry
gaps. Then for given ε > 0,
log m(r, f ) ≥ (1 − ε) log M (r, f )
holds for all r outside a set of logarithmic density 0.
Lemma 2.3([24]). Let f =
∞
P
ak z nk be an entire function with
k=0
nk > k log k(log log k)α as k → ∞
for some α > 2. Then for given ε > 0,
log m(r, f ) ≥ (1 − ε) log M (r, f )
holds for all r outside a set of logarithmic density 0.
4
Lemma 2.4([25]). If f =
∞
P
ak z nk satisfies the gap condition nk > k(log k)2+η ,
k=0
then for given ε > 0,
log m(r, f ) ≥ (1 − ε) log M (r, f )
holds for all r outside a set of finite logarithmic measure.
Lemma 2.5([26]). For an entire function f (z) with Fej´er gaps and ε > 0,
log m(r, f ) ≥ (1 − ε) log M (r, f ).
Lemma 2.6([9]). Let D be a domain and K be 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 0 )| < A|g(z)|B , for every g ∈ G and every z, z 0 ∈ K.
Lemma 2.7. Let f be a transcendental entire function of finite order with
Fabry gaps. Then there exists L > 1 and R > 0 such that, for all r ≥ R,
there exists σ satisfying r ≤ σ ≤ rL and m(σ, f ) = M (r, f ).
Proof. By Lemma 2.2, given any ε > 0, we have m(r, f ) > M (r, f )1−ε , for
all r outside a set E of logarithmic density 0. Let c > 1. Then there exists
R0 such that for all R > R0 , there exists s ∈ [R, Rc ] such that m(s, f ) >
M (s, f )1−ε . If not, then there exists a sequence Rj → ∞ such that m(s, f ) ≤
M (s, f )1−ε , for every s ∈ [Rj , Rjc ]. Thus [Rj , Rjc ] ⊂ E. So
Z
1
dt
logdensE = lim inf
r→∞ log r E(1,r) t
Z
1
dt
≥ lim inf
c
j→∞ log R
j E(1,Rjc ) t
Z Rjc
1
dt
≥ lim inf
j→∞ c log Rj R
t
j
1
=1− .
c
contradicting logdense(E) = 0.
1
c
Set R1−2ε = r. There exists s ∈ [r 1−2ε , r 1−2ε ] such that m(s, f ) >
M (s, f )1−ε . By Lemma 2.2,
1−ε
1
m(s, f ) > M (s, f )1−ε ≥ M (r 1−2ε , f )1−ε ≥ M (r, f ) 1−2ε ≥ M (r, f ).
5
c
such
Since m(r, f ) ≤ M (r, f ), there exists σ ∈ [r, s] ⊂ [r, rL ], where L = 1−2ε
that m(σ, f ) = M (r, f ). This completes the proof of Lemma 2.7. By Lemma 2.5 and the same method in the proof of Lemma 2.7, we can
prove the following result.
Lemma 2.8. Let f be a transcendental entire function of finite order with
Fej´er gaps. Then there exists L > 1 and R > 0 such that, for all r ≥ R,
there exists σ satisfying r ≤ σ ≤ rL and m(σ, f ) = M (r, f ).
By Lemma 2.3 and the same method in the proof of Lemma 2.7, we can
prove the following result.
Lemma 2.9. Let f be a transcendental entire function of finite order with
nk > k log k(log log k)α as k → ∞
for some α > 2. Then there exists L > 1 and R > 0 such that, for all
r ≥ R, there exists σ satisfying r ≤ σ ≤ rL 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 3.1. Let h(z) = fN ◦ fN −1 ◦ · · · ◦ f1 (z) be a transcendental entire
∞
P
functions, where fj =
ajk z njk (j = 1, 2, . . . , N ) have Fabry gaps of finite
k=0
order. If
M (r, fj ) ≥ exp(exp(log r)p ), 0 < p < 1,
then F (h) has no unbounded component.
Proof. The proof follows the idea of Theorem 1.4 and Theorem 1.5. Since fj
is a transcendental entire function of finite order with Fabry gaps, by Lemma
2.7, there exists Lj > 1 and rj sufficiently large such that for r ≥ rj , there
exists σ (j) such that
r ≤ σ (j) ≤ rLj and m(σ (j) , fj ) = M (r, fj )
6
for j = 1, 2, . . . , N. Since fj is a transcendental entire function of finite
order, h must be a transcendental entire function. It follows that there exists
a number r0 > 1 such that M (r, h) > r2 for all r > r0 , and there exists a
number tj > 1 such that
exp(rtj ) ≥ M (r, fj )(log M (r,fj ))
1
p
(1)
for all r sufficiently large. In fact, if there is a sequence {rn } tend to ∞ such
that
1
p
exp(rnt ) < M (rn , fj )(log M (rn ,fj )) ,
then
t log rn <
So
t≤
1+p
· log log M (rn , fj ).
p
log log M (rn , fj )
1+p
1+p
lim
≤
ρ.
n→∞
p
log rn
p
Which gives a contradiction if we take any t >
1+p
ρ.
p
(1)
Given R1 > r0 . Define inductively
(1)
Rn(j+1) = M (Rn(j) , fj ), j = 1, 2, . . . , N − 1; Rn+1 = M (Rn(N ) , fN ).
It is easy to see that for all n ∈ N,
(1)
(Rn(1) )2 < M (Rn(1) , h) ≤ Rn+1 .
Thus
(1)
(1)
n
Rn+1 > (R1 )2 → ∞ as n → ∞
(j)
and Rn → ∞ as n → ∞ for all j = 1, 2, . . . , N .
Take number n0 sufficiently large, for n ≥ n0 , by Lemma 2.7, there exists
(j)
σn such that
exp((Rn(j) )tj ) ≤ σn(j) ≤ exp(Lj (Rn(j) )tj )
and
m(σn(j) , fj ) = M (exp((Rn(j) )tj ), fj )
for n ≥ n0 . By the hypotheses of Theorem 3.1, (1) and Lemma 2.7, we have
7
(j)
(j)
m(σn , fj ) = M exp((Rn )tj ), fj
≥M
(j)
1
p
M (Rn(j) , fj )(log M (Rn ,fj )) , fj
≥ exp exp
1
(j)
p
log M (Rn(j) , fj )(log M (Rn ,fj ))
p = exp exp((log M (Rn(j) , fj ))(log M (Rn(j) , fj ))p )
(j)
p
= exp M (Rn(j) , fj )(log M (Rn ,fj ))
≥ exp Lj+1 M (Rn(j) , fj )tj+1
= exp Lj+1 (Rn(j+1) )tj+1
for j = 1, 2, . . . , N −1; and
(N )
(N )
m(σn , fN ) = M exp((Rn )tN ), fN
1
(N )
(N )
(log M (Rn ,fN )) p
≥ M M (Rn , fN )
, fN
≥ exp exp
(N )
log M (Rn(N ) , fN )(log M (Rn
1
,fN )) p
p = exp exp((log M (Rn(N ) , fN ))(log M (Rn(N ) , fN ))p )
(N )
p
= exp M (Rn(N ) , fN )(log M (Rn ,fN ))
≥ exp L1 M (Rn(N ) , fN )t1
(1)
= exp L1 (Rn+1 )t1 .
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 hn
omits the values 0, 1 in D. It follow from the unboundedness and connectivity
of D that there exists n1 ≥ n0 ∈ N such that D meets the circles
αn(j) = {z : |z| = Rn(j) }, βn(j) = {z : |z| = exp(Lj (Rn(j) )tj )}, γn(j) = {z : |z| = σn(j) }
for all n ≥ n1 , j = 1, 2, . . . , N .
8
We choose a value k ∈ N such that k ≥ n1 and note that D must contain
(1)
(1)
(1)
(1)
a path Γ joining a point ωk ∈ αk to a point τk+1 ∈ βk+1 . It is clear that
(1)
(1)
(1)
(1)
Γ contains two subsets Γ0 , Γ00 such that Γ0 joining ωk ∈ αk to ηk ∈ βk
(1)
(1)
(1)
(1)
(1)
(1)
and contains ξk ∈ γk , and Γ00 joining δk+1 ∈ αk+1 to τk+1 ∈ βk+1 . We
(2)
(1)
(2)
(1)
(1)
know that Rk = M (Rk , f1 ) and so Rk ≥ |f1 (ωk )|. Also m(σk , f1 ) ≥
(2)
(1)
(2)
exp(L2 (Rk )t2 ) and so |f1 (ξk )| ≥ exp(L2 (Rk )t2 ). Hence f1 (Γ0 ) contains an
(2)
(2)
(2)
(2)
arc joining a point ωk ∈ αk to a point ηk ∈ βk . Similarly f1 (Γ00 ) contains
(2)
(2)
(2)
(2)
an arc joining a point δk+1 ∈ αk+1 to a point τk+2 ∈ βk+2 .
Repeating the process inductively we obtain that h(Γ0 ) = fN ◦ fN −1 ◦ · · · ◦
(1)
(1)
(1)
(1)
f1 (Γ0 ) contains an arc joining ωk+1 ∈ αk+1 to a point ηk+1 ∈ βk+1 and h(Γ00 )
(1)
(1)
(1)
(1)
contains an arc joining a point δk+2 ∈ αk+2 to a point τk+2 ∈ βk+2 .
Since Γ0 and Γ00 are two subsets of Γ, it follows that h(Γ) must contain
(1)
(1)
(1)
(1)
an arc joining ωk+1 ∈ αk+1 to the point τk+2 ∈ βk+2 . By induction it now
(1)
(1)
follows that hn (Γ) contains an arc joining a point ωk+n ∈ αk+n to the point
(1)
(1)
τk+n+1 ∈ βk+n+1 .
Thus hn (D) is a component of F (h) containing hn (Γ), and on Γ, hn takes a
(1)
(1)
value of modulus at least Rk+n and Rk+n → ∞ as n → ∞. Thus we conclude
(1)
that Rk+n → ∞ locally uniformly in D. Hence there exists N ∈ N such that
for all z ∈ Γ, |hn (z)| > 1 for all n > N . Thus the family {hn }n>N satisfy the
conditions of Lemma 2.6 on Γ, and so there exists constants A, B such that
|hn (z 0 )| < A|hn (z)|B for all n > N and for all z, z 0 ∈ Γ. Choose zn , zn0 ∈ Γ
(1)
(1)
(1)
(1)
with n > N such that hn (zn ) = ωk+n ∈ αk+n and hn (zn0 ) = τk+n+1 ∈ βk+n+1 ,
we have
(1)
(1)
(N )
(1)
M (Rk+n , h) = M (Rk+n , fN ◦ fN −1 ◦ · · · ◦ f1 ) ≤ M (Rk+n , fN ) = Rk+n+1
(1)
< exp L1 (Rk+n+1 )t1 = |hn (zn0 )|
(1)
< A|hn (zn )|B = A|Rk+n |B
for all n > N which contradicts h is a transcendental entire function and
(1)
Rk+n → ∞ as n → ∞. This completes the proof of Theorem 3.1. Corollary 3.2([27]). Let f (z) =
∞
P
ak z nk be a transcendental entire function
k=0
of finite order with Fabry gaps. If
M (r, f ) ≥ exp(exp(log r)p ), 0 < p < 1,
9
then F (f ) has no unbounded component.
Remark 3.3. If f satisfies M (r, f ) ≥ exp(exp(log r)p ), 0 < p < 1, then
ρ ≥ 0, so Corollary 3.2 is an extension of Theorem 1.1.
By Lemma 2.8 and the same method using in the proof of Theorem 3.1, we
can show following result.
Theorem 3.4. Let h(z) = fN ◦ fN −1 ◦ · · · ◦ f1 (z) be a transcendental entire
∞
P
function, where fj =
ajk z njk (j = 1, 2, . . . , N ) have Fej´er gaps. If
k=0
M (r, fj ) ≥ exp(exp(log r)p ), 0 < p < 1,
then F (h) has no unbounded component.
By Theorem 3.4, we have
Corollary 3.5([27]). Let f (z) =
∞
P
ak z nk be a transcendental entire function
k=0
with Fej´er gaps. If
M (r, f ) ≥ exp(exp(log r)p ), 0 < p < 1,
then F (f ) has no unbounded component.
Remark 3.6. Corollary 3.5 is the partly answer the problem of Wang on
the Fej´er gaps.
By Lemma 2.9 and the same method of the proof of Theorem 3.1, we can
show Theorem 3.7.
Theorem 3.7. Let h(z) = fN ◦ fN −1 ◦ · · · ◦ f1 (z) be a transcendental entire
∞
P
functions, where fj =
ajk z njk (j = 1, 2, . . . , N ) have entire functions with
k=0
the gap-conditions
njk > k log k(log log k)α as k → ∞ f or some α > 2.
If
M (r, fj ) ≥ exp(exp(log r)p ), 0 < p < 1,
10
then F (h) has no unbounded component.
By Theorem 3.7, we have
Corollary 3.8.([27]). Let h(z) = fN ◦ fN −1 ◦ · · · ◦ f1 (z) be a transcenden∞
P
tal entire functions, where fj =
ajk z njk (j = 1, 2, . . . , N ) have entire
k=0
functions with the gap-conditions njk > k(log k)2+η for some η > 0. If
M (r, fj ) ≥ exp(exp(log r)p ), 0 < p < 1,
then F (h) has no unbounded component.
Conflict of Interests
The author declares that there is no conflict of interests regarding the publication of this paper.
Acknowledgements
This work was Supported by the National Natural Science Foundation of
China (Grant No. 11261002), the Natural Science Foundation of Yunnan
Province of China (Grant No.2012FZ167 ) and the Educational Commission
of Yunnan Province of China (Grant No. 2012z121).
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