Simply connected fast escaping Fatou components

0.19 MB·English

by  D. J. Sixsmith

SIMPLY CONNECTED FAST ESCAPING FATOU COMPONENTS
D.J.SIXSMITH
2
1
Abstract. We give an example of a transcendental entire function with a
0
simply connected fast escaping Fatou component, but with no multiply con-
2
nectedFatoucomponents. Wealsogiveanewcriterionforpointstobeinthe

n fastescapingset.
a
J

9

1. Introduction
]
S
Suppose that f :C→C is atranscendentalentire function. The Fatou set F(f)
D
is defined as the set of points z ∈ C such that (fn)n∈N is a normal family in a
.
h neighborhoodofz. TheJulia set J(f)isthe complementofF(f). Anintroduction
t to the properties of these sets was given in [3].
a
m The fast escaping set A(f) was introduced in [5]. We use the definition
[ A(f)={z :there exists ℓ∈N such that |fn+ℓ(z)|≥Mn(R,f), for n∈N},

1 given in [10]. Here, the maximum modulus function M(r,f)= max |f(z)|, for
|z|=r
v r>0, Mn(r,f) denotes repeated iterationof M(r,f) with respect to r, and R>0
6
can be taken to be any value such that M(r,f) > r, for r ≥ R. We write M(r)
2
9 when it is clear from the context which function is being considered.
1 Suppose that U =U0 is a component of F(f). If U ∩A(f)6=∅, then U ⊂A(f)
. [10,Theorem1.2]. WecallaFatoucomponentinA(f)fast escaping. DenotebyU
1 n
0 thecomponentofF(f)containingfn(U). Wesaythatacomponentiswandering if
2 U = U implies that n= m. All fast escaping Fatou components are wandering;
n m
1 [5, Lemma 4] and [10, Corollary 4.2].
:
v For a transcendental entire function, all multiply connected Fatou components
i are fast escaping; [8, Theorem 2] and [10, Theorem 4.4]. The first example of
X
a transcendental entire function with a multiply connected Fatou component was
r
a constructed by Baker in [1]. Other examples are found in, for example, [2], [4], [6]
and [7].
The only known example of a simply connected fast escaping Fatou component
was given by Bergweiler [4], using a quasi-conformal surgery technique from [7].
This function also has multiply connected Fatou components. In fact, in [4], the
properties of the multiply connected Fatou components are used to show that the
simply connected Fatou components are fast escaping.
This prompts the question of whether a transcendentalentire function can have
simply connected fast escaping Fatou components without having multiply con-
nected Fatou components. We answer this in the affirmative, using a direct con-
struction and a recent result on the size of multiply connected Fatou components
[6, Theorem 1.2] to prove the following.
Theorem 1. There is atranscendentalentirefunctionwith asimply connectedfast
escaping Fatou component, and no multiply connected Fatou components.
1
2 D.J.SIXSMITH
To prove Theorem 1 we require a new sufficient condition for points to be in
A(f), which may be of independent interest.
Theorem 2. Supposethat f is atranscendentalentirefunction. Supposealso that,
for R >0, ǫ:[R ,∞)→(0,1) is a nonincreasing function, such that
0 0
(1.1) ǫ(Mn(r))≥ǫ(r)n+1, for r ≥R and n∈N.
0
Define η(r)=ǫ(r)M(r), for r ≥R . Then there exists R ≥R such that
0 1 0
A(f)={z : there exists ℓ∈N such that |fn+ℓ(z)|≥ηn(R′), for n∈N},
for R′ ≥R .
1
Note that this is a generalisation of [10, Theorem 2.7], which is obtained from
Theorem 2 when ǫ is constant.
2. The definition of the function
In this section we define a transcendental entire function, f, which has all the
properties defined in Theorem 1. Since f is very complicated, we first outline
informally the construction of f, starting with simpler functions which only have
some of these properties. We then give the full construction. A detailed proof of
Theorem 1 is given in subsequent sections.
Consider first a transcendental entire function defined by a power series;
∞ 2
z
g(z)=z 1+ , 0<a <a <··· .
1 2
a
k=1(cid:18) k(cid:19)
Y
The sequence (an)n∈N can be chosen so that the following holds: we can define
another sequence, (bn)n∈N, such that bn is approximately equal to an, −bn is close
to a critical point of g, and g(−b ) is close to −b . It can then be shown that a
n n+1
small disc centred at −b is mapped by g into a small disc centred at −b . By
n n+1
Montel’s theorem, these discs must be in the Fatou set of g. Moreover,these discs
cannotbe inmultiply connectedFatoucomponentsofg since, by [6, Theorem1.2],
any open set contained in a multiply connected Fatou component of g must, after
afinite number of iterationsofg, coveranannulus surroundingthe origin. Finally,
it can be shown, by comparing |g(−b )| to M(b ,g) = g(b ), that these discs are
n n n
contained in fast escaping Fatou components of g.
However,g does nothaveallthe propertiesdefined inTheorem1. Inparticular,
byconsideringthebehaviourofg inlargeannuliwhichomitthezerosofg,itcanbe
shown that g has multiply connected Fatou components. Thus g has very similar
properties to the example in [4].
WenotethatnozeroofgcanbeinamultiplyconnectedFatoucomponent,since
0 is a fixed point. In order to prevent the existence of multiply connected Fatou
components, we add further zeros to the function, along the negative real axis.
This requires some care. The addition of too many zeros – for example, spaced
linearly along the negative real axis – leads to a breakdown of other parts of the
construction. The addition of a zero with modulus insufficiently distant from a
n
leads to a similar breakdown.
Weuse[6,Theorem1.2]toshowthatonlyarelativelysmallnumberofadditional
zerosarerequired. Inparticular,supposethath isatranscendentalentirefunction
withh(0)=0andwithzerosofmodulusr <r <r <···. Then,by[6,Theorem
0 1 2
SIMPLY CONNECTED FAST ESCAPING FATOU COMPONENTS 3
1.2],hhasnomultiplyconnectedFatoucomponentsiflim logr /logr exists
k→∞ k+1 k
and is equal to 1.
To use this result, we need to understand the behaviour of loga /loga , for
n+1 n
large n. From the recursive definition that we use to ensure that the sequences
(an)n∈N and (bn)n∈N have the required properties, (see (3.15)), we find that, for
large n, loga /loga is close to n3. See (3.9) for a precise statement of how the
n+1 n
term n3 arises here.
This suggests the following. Define µ = n3/n, for n ∈ N. To simplify some
n
displays we set µ = µm, and observe that µ = 1 and µ = n3, for n ∈ N.
n,m n n,0 n,n
We now define a more complicated transcendental entire function;
∞ k−1 2
z
h(z)=z 1+ , 0<a <a <··· .
aµk,l 1 2
k=1l=0(cid:18) k (cid:19)
Y Y
The sequence (an)n∈N in this definition is not the same as in the definition of g,
but serves the same purpose, and is chosen similarly. This function has zeros of
modulus
··· ,a , aµn, aµn,2, ··· , aµn,n−1,a , aµn+1··· .
n n n n n+1 n+1
Sinceitisreadilyseenthatµ →1asn→∞,thisfunctiondoesnothavemulti-
n
ply connected Fatou components. However, two further adjustments are required.
Firstly, the zero of modulus aµn,n−1 is sufficiently close to the zero of modulus
n
a that the original construction breaks down. We resolve this by omitting this
n+1
zero. Secondly, the value of loga /loga is not close enough to n3, for large
n+1 n
n, to ensure that lim loga /logaµk,k−2 = 1. We resolve this by adding one
k→∞ k+1 k
additional zero, which serves no other purpose in the construction. This zero is
defined using two additional sequences, (αn)n∈N and (βn)n∈N, which we choose to
keep loga /loga sufficiently close to n3.
n+1 n
Now we are able to indicate the form of the function f in Theorem 1. Let f be
the transcendental entire function;
∞ 2αk k−2 2
z z
(2.1) f(z)=z 1+ 1+ ,
kY=3 aβkk! Yl=0(cid:18) aµkk,l(cid:19) 
where 0 < a3 < a4 < ··· , αn ∈ {0,1,2,...}, βn ∈ R, for n ∈N. Again, the se-
quence (an)n∈N in this definition is notthe same asthat in the definitionof g or h,
butservesthesamepurpose,andischosensimilarly. Therelatedsequence(bn)n∈N,
discussedafterthedefinitionofg,isdefinedforf by(3.14). Thesequences(αn)n∈N
and(βn)n∈N arethetwosequencesmentionedattheendofthepreviousparagraph.
ThestructureoftheproofofTheorem1isasfollows. First,inSection3,wegive
thedefinitionofthevarioussequencesin(2.1),andweproveanumberofestimates
on the modulus of the zeros of f. Next, in Section 4, we show that there are no
multiply connected Fatou components of f. In Section 5 we show that there are
intervals on the negative real axis each contained in some Fatou component of f.
Finally, in Section 6 we prove Theorem 2 and then use this to show that these
Fatou components of f are fast escaping. It is clear that Theorem 1 follows from
these results.
4 D.J.SIXSMITH
Remark: Rippon and Stallard asked [10, Question 1] if there can be unbounded
fast escaping Fatou components of a transcendental entire function. It can be
shown that the Fatou components of the function f are all bounded. Indeed, it is
straightforward to prove that the number of zeros of f in the disc {z : |z| < r}
is O(logr), and hence that logM(r,f) = O((logr)2). It follows that f has no
unbounded Fatou components [10, Theorem 1.9(b)], and, moreover, that the set
A(f) has a structure known as a spider’s web. For more details we refer to [10].
3. Defining the sequences
In this section we first define the sequences (αn)n∈N and (βn)n∈N, and then de-
fine the sequences (an)n∈N and (bn)n∈N.
Recall from Section 2 that µ =n3/n and µ =µm; we also define, for n≥3,
n n,m n
n−2
µ −1
n,n−2
(3.1) σ = µ =µ .
n n,l n
µ −1
n
l=1
X
We define (αn)n∈N to be a sequence of integers and (βn)n∈N to be a sequence of
real numbers. Assume that N is even and chosen sufficiently large for subsequent
0
estimates to hold. Set
0, for n<N ,
0
N3+2N2+6N +2, for n=N ,
(3.2) 2α = 0 0 0 0
n 3n2+n+6, for n>N0, n even,
3n2+n+4, for n>N , n odd.
0
Note that αn is an integer, for n∈N. Set
0, for n<N ,
0
β = 1 (n4−σ ), for n≥N , n even,
n αn n 0
 1 (n3(2n−1) −σ ), for n≥N , n odd.
αn 2 n 0
We observe that thesechoices imply that
2
(3.3) τ = (α β +σ )
n n3 n n n
satisfies
2n, for n≥N , n even,
0
(3.4) τ =
n
(2n−1, for n≥N0, n odd,
and
(3.5) 2α =3n2+n+3+τ −τ , for n>N .
n n n−1 0
We also define a sequence of integers (Tn)n∈N by
n3+2n−3, for n even,
(3.6) T =
n (n3+2n−2, for n odd.
Next we prove a result which gives various relationships between these sequences.
SIMPLY CONNECTED FAST ESCAPING FATOU COMPONENTS 5
Lemma 3.1. The following all hold for the choice of sequences above:
3 2
(3.7) α ∼ n2, β ∼ n2, as n→∞,
n n
2 3
n
(3.8) 2 α =n3+2n2+4n+2+τ , for n≥N ,
k n 0
k=3
X
n−1 n−1k−2
(3.9) 1+2 α + 2=n3+τ =T , for n≥N ,
k n−1 n 0
k=3 k=3l=0
X XX
and
(3.10) µ <β <µ , for large n.
n,2 n n,n−3
Proof. The first half of (3.7) is immediate from (3.2). Now, by (3.2), (3.3) and
(3.4),
2n σ
n
(3.11) β ∼ n− , as n→∞.
n 3 n3
(cid:16) (cid:17)
We have that
x 1
(3.12) ≤log(1+x)≤x, for 0<x< .
2 2
Putting x=µ −1, we obtain
n
3 6
(3.13) logn≤µ −1≤ logn, for large n.
n
n n
Hence, by (3.1),
σ µ µ −1 1 n
n n n,n−2
= ∼ =O as n→∞,
n3 µ µ −1 µ (µ −1) logn
n,n n n n (cid:18) (cid:19)
and the second half of (3.7) follows by (3.11).
We can see that (3.8) holds by induction. For, it is immediately satisfied when
n=N . When n=m>N we have, by (3.5) and (3.8) with n=m−1,
0 0
m m−1
2 α =2α +2 α =m3+2m2+4m+2+τ .
k m k m
k=3 k=3
X X
Finally (3.9) follows from (3.4), (3.6), and (3.8), and (3.10) follows from (3.7). (cid:3)
Next we define the sequence (an)n∈N recursively,and for each n∈N put
2 T
n
(3.14) b =a − a = a .
n n n n
T +2 T +2
n n
Choose a and N large, and set a = an3, for 3≤ n <N . We assume that a
3 1 n+1 n 1 3
andN arechosensufficiently largefor variousestimates inthe sequelto hold. For
1
n≥N , we define
1
(3.15)
2n−1 2αk k−2 2
(T +2) b b b
n+1 n n n
a = b 1− 1− 1− .
n+1 Tn+1 n(cid:18) an(cid:19) kY=3 aβkk! Yl=0(cid:18) aµkk,l(cid:19) 
Finally in this section we prove a set of inequalities which concern the growth
 
of the sequence (a ), and the ratios of these numbers to the modulus of the other
n
6 D.J.SIXSMITH
zeros of f. Note that (3.7) and (3.17) imply that the product in (2.1) is locally
uniformly convergentin C.
Lemma 3.2. The following inequalities hold for the sequence (a ) defined above.
n
For n≥3,
(3.16) an3−2/n ≤a ≤an3+2/n,
n n+1 n
(3.17) a >exp(en),
n
and, for large n,
a α a
(3.18) n ≤exp(−en/2), n n ≤exp(−en/2),
aµnn aβnn
(3.19) aµnn,n−2 ≤exp(−en/2), αnaβnn ≤exp(−en/2).
a a
n+1 n+1
Proof. First, assume that (3.16) holds for 3 ≤ n ≤ m. Equation (3.17) follows for
3≤n≤m by a simple induction. Hence, for sufficiently large m, by (3.7), (3.13),
(3.16) and (3.17):
a
m =a1−µm ≤exp(em(1−µ ))≤exp(−em/2);
aµm m m
m
α a
m m ≤3m2a1−m2/2 ≤exp(−em/2);
aβm m
m
aµm,m−2 2
ma ≤(aµmm,m−2)1−µm,2+m2 ≤exp(em(1−µm,2+ m))≤exp(−em/2);
m+1
αmamβm ≤3m2am2−m3+2/m ≤exp(−em/2).
a m
m+1
It remains to prove (3.16). We can assume, by taking N sufficiently large, that
1
(3.16) holds for 3 ≤ n ≤ m−1, for some large m. We can assume also that m is
sufficientlylargethat(3.9),(3.10)andvariousotherestimatesusedinthefollowing
hold. We need to prove that (3.16) holds for n=m. Now, by (3.15) and (3.3),
2
(T +2) b L
a = m+1 bκm 1− m 1
m+1 Tm+1 m (cid:18) am(cid:19) mk=−31 a2kαkβk lk=−02a2kµk,l
=k aκmm QL1 n Q o
ma2αm−1βm−1 m−3a2µm−1,l L2
m−1 l=1 m−1
(3.20) =k aκmm QL1,
ma(m−1)3τm−1 L2
m−1
where, by (3.14),
(T +2) T κm 2 2
m+1 m
k = ,
m
T T +2 T +2
m+1 (cid:18) m (cid:19) (cid:18) m (cid:19)
by (3.9),
m−1 m−1k−2
κ =1+2 α + 2=m3+τ =T ,
m k m−1 m
k=3 k=3 l=0
X X X
SIMPLY CONNECTED FAST ESCAPING FATOU COMPONENTS 7
m−1 aβk 2αkk−2 aµk,l 2
L = 1− k 1− k ,
1  b b 
kY=3  m ! Yl=0(cid:18) m (cid:19) 
and, by (3.3) again,
 
m−2 k−2 m−2 m−2
L =a2 a2αkβk a2µk,l =a2 a2(1+αkβk+Plk=−12µk,l) =a2 a2+k3τk.
2 m−1 k k m−1 k m−1 k
( )
k=3 l=0 k=3 k=3
Y Y Y Y
We now estimate the terms in this equality. Firstly, by (3.6), and noting that
(T /(T +2))Tm >1/e2, we obtain
m m
1
m−6 <k <8m−6.
m
8
Secondly, by (3.16), with n=m−1,
a(m−1)3 ≤a(1−(m−21)4)−1 ≤a1+(m−41)4,
m−1 m m
and so, by (3.4),
aκmm ≥aκm−τm−1−(4mτm−−1)14 =am3−(4mτm−−1)14 ≥am3−m1 .
a(m−1)3τm−1 m m m
m−1
Similarly, by (3.4) and (3.16),
aκmm ≤am3+m1 .
a(m−1)3τm−1 m
m−1
Thirdly, we consider L . Noting (3.10), and by (3.14) and (3.19), we see that
1
the smallest term in this product is
aµm−1,m−3 2 2aµm−1,m−3 2
1> 1− m−1 > 1− m−1 >1−exp(−em4 ).
b a
m ! m !
There are fewer than m terms in L of the form (1−ap /b )q, p ∈ R, q ∈ N,
1 m−1 m
andsoofcomparablesizeto this term. The othertermsinthe productforL tend
1
to 1 sufficiently quickly, by (3.16), that 1/2<L <1.
1
Finally, we consider L . Observe that, by (3.4) and (3.16), the largest term in
2
this product is
a2 a2+(m−2)3τm−2 <a2 a4(m−2)4 <a2 a8(m−1) <a16/m2.
m−1 m−2 m−1 m−2 m−1 m−1 m
By (3.16), all other terms in the product for L decrease sufficiently quickly that
2
1 < L < a32/m2.
2 m
Thus, by (3.20), for sufficiently large m,
a ≥ 1 m−6am3−m1−m322 ≥am3−m2 ,
m+1 m m
16
and similarly,
a ≤8m−6am3+m1 ≤am3+m2 .
m+1 m m
This completes the proof of Lemma 3.2. (cid:3)
8 D.J.SIXSMITH
4. There are no multiply connected Fatou components
In this section we prove the following result.
Lemma4.1. Thetranscendentalentirefunctionf doesnothavemultiplyconnected
Fatou components.
We use the definition of an annulus
A(r ,r )={z :r <|z|<r }, for 0<r <r .
1 2 1 2 1 2
We need the following, which is part of [6, Theorem 1.2].
Theorem A. Suppose that g is a transcendental entire function with a multiply
connected Fatou component U. For each z ∈ U there exists α > 0 such that, for
0
sufficiently large n∈N,
gn(U)⊃A(|gn(z )|1−α,|gn(z )|1+α).
0 0
Proof of Lemma 4.1. Observe that, for large n, in the closed annulus A(a ,a )
n n+1
there are zeros of f on the negative real axis of modulus an,aµnn,aµnn,2,...,anµn,n−2
and a . Note also that 0 is a fixed point of f, and so no zero of f can be in a
n+1
multiply connected Fatou component of f. Now, by (3.16),
a ≤an3+2/n <(aµn,n−2)µn,2+2/n.
n+1 n n
Hence,forlargen,thereis atleastonezerooff inanyannulusA(r,rµn,2+2/n), for
a ≤r ≤a . Note that µ +2/n→1 as n→∞.
n n+1 n,2
Now,byTheoremA,iff hasamultiplyconnectedFatoucomponent,thenthere
isac>1,anda sequence(ri)i∈N,tending to infinity,suchthatthe annuliA(ri,ric)
arecontainedinmultiplyconnectedFatoucomponentsoff. Thisisincontradiction
to the observations aboveregardingthe distribution of zerosof f and the fact that
these zeros do not lie in multiply connected Fatou components. Hence there can
be no multiply connected Fatou components of f. (cid:3)
5. There are simply connected Fatou components
Next we show that f has simply connected Fatou components.
Lemma 5.1. Define B = {z : |z +b | < δ b }, where δ = n−15. Then, for
n n n n n
large n, we have f(B )⊂ B , and B is contained in a simply connected Fatou
n n+1 n
component of f.
Proof. Suppose that z ∈ B , in which case z = −b +wb where |w| < δ . We
n n n n
assume throughout this section that n is sufficiently large for various estimates to
hold. By (2.1), (3.14) and (3.15),
f(z)
=I I ,
1 2
−b
n+1
where
(5.1)
2n−1 2αkk−2 2
b b b
n n n
I =(1−w) 1+w 1+w 1+w ,
1 (cid:18) an−bn(cid:19) kY=3 aβkk −bn! Yl=0(cid:18) aµkk,l −bn(cid:19) 
 
SIMPLY CONNECTED FAST ESCAPING FATOU COMPONENTS 9
and
n−2 z 2 ∞ z 2αk ∞ k−2 z 2
I = 1+ 1+ 1+ .
2 l=1(cid:18) aµnn,l(cid:19) k=n aβkk! k=n+1l=0(cid:18) aµkk,l(cid:19)
Y Y Y Y
First consider I . This is a polynomial in w of degree T +2, by (3.9). Write
1 n
Tn+2
I =1+ η wj, for η ∈C.
1 j j
j=1
X
Then
(5.2) |I −1|≤|η w|+|η w2|+···+|η wTn+2|.
1 1 2 Tn+2
We consider η . We have, by (3.2), (3.9), (3.14), (3.16), (3.17) and (3.19),
1
n−1 k−2
b b b
n n n
|η |= 1+2 +2 α +
1 (cid:12)(cid:12) bn−an Xk=3( kbn−aβkk Xl=0 bn−aµkk,l)(cid:12)(cid:12)
(cid:12) (cid:12)
(cid:12) n−1 aβk k−2 aµ(cid:12)k,l
(5.3) =(cid:12)1−T +2 α 1+ k + 1+ k(cid:12)
(cid:12)(cid:12) n Xk=3( k bn−aβkk! Xl=0(cid:18) bn−aµkk,l(cid:19))(cid:12)(cid:12)
(cid:12) (cid:12)
(cid:12) n−1 aβk k−2 aµk,l (cid:12)
=2(cid:12) α k + k (cid:12)
(cid:12)(cid:12)Xk=3( kbn−aβkk Xl=0 bn−aµkk,l)(cid:12)(cid:12)
(cid:12)(cid:12) aβn−1 aµn−1,n−3 (cid:12)(cid:12)
≤4(cid:12)nα n−1 +n2 n−1 (cid:12)
n−1
(cid:12) an an (cid:12)
(cid:12) (cid:12)
≤ex(cid:12)(cid:12)p(−en/4). (cid:12)(cid:12)
(cid:12) (cid:12)
Note that the cancellation in (5.3) occurs because, due to the choice of b and T ,
n n
−b is very close to a critical point of f.
n
Next consider η , for k > 1. Note that the coefficients of w in the product for
k
I have modulus at most b /(a −b )=T /2<n3. Moreover, the degree of I is
1 n n n n 1
less than 2n3, and so the expansion of (5.1) contains less than (2n3)k terms in wk.
Hence
|η |<(n3)k.(2n3)k <n7k, for large n.
k
Hence |η w2|<n−16. The other terms in (5.2) decrease sufficiently quickly that
2
(5.4) |I −1|<2n−16, for large n.
1
Now we consider I . Observe that each term in the product for I has modulus
2 2
less than 1. Hence, since −log(1−x) ≤log(1+2x), for 0<x<1/2, we have, by
(3.12), (3.16), (3.17) and (3.18),
n−2 ∞ ∞ k−2
2|z| 2|z| 2|z|
0<−log|I |≤2 log 1+ + α log 1+ + log 1+
2 l=1 (cid:18) aµnn,l(cid:19) k=n k aβkk! k=n+1l=0 (cid:18) aµkk,l(cid:19)!
X X X X
n−2 ∞ ∞ k−2
a a a
n n n
≤8 + α +
l=1 aµnn,l k=n kaβkk k=n+1l=0 aµkk,l!
X X X X
≤16 exp(−en/2)+exp(−en/2)+exp(−en/2)
≤exp(cid:16)(−en/4). (cid:17)
10 D.J.SIXSMITH
Thus,1−2exp(−en/4)≤|I |≤1.This,togetherwith(5.4),establishesthefirstpart
2
of the lemma. It follows from Montel’s theorem that, for large n, B is contained
n
in a Fatou component, which must be simply connected by Lemma 4.1. (cid:3)
Remark. Let V be the Fatou component containing B . These Fatou com-
n n
ponents are distinct. For, suppose that V = V with m 6= n. Because all the
m n
coefficients of z in (2.1) are real, the Fatou set F(f) must be invariant under re-
flection in the real axis. Hence, all points on the negative real axis between B
n
and B must be in V , as otherwise V would be multiply connected. This is a
m m m
contradiction since these points include the zeros of f.
6. The simply connected Fatou components are fast escaping
In this section we first prove Theorem 2, and then we use this result to prove
the following.
Lemma 6.1. Let V , n∈N, be the simply connected Fatou components defined at
n
the end of Section 5. Then V ⊂A(f), for large n.
n
Proof of Theorem 2. We need two facts about the maximum modulus function,
M(ρ), of a transcendental entire function. The first is well known, and the second
is from [9, Lemma 2.2]:
logM(ρ)
(6.1) →∞ as ρ→∞,
logρ
and there exists R>0 such that
(6.2) M(ρc)≥M(ρ)c, for ρ≥R, c>1.
Fix r ≥ R such that M(r) > r, for r ≥ r . Whenever r ≥ r there is a unique
0 0 0 0
n∈N suchthat Mn−1(r )≤r<Mn(r ). Hence, since ǫ is nonincreasing,by (1.1)
0 0
and (6.1)
ǫ(r)r ≥ǫ(Mn(r ))Mn−1(r )≥ǫ(r )n+1Mn−1(r )→∞ as n→∞.
0 0 0 0
Hence
(6.3) ǫ(r)r →∞ as r →∞.
By (6.1) and (6.3) we see that, given k > 0, we can ensure that
logM(ǫ(r)R)
(6.4) >k, for large r, R≥r.
log(ǫ(r)R)
A little algebra shows that this is equivalent to
(6.5) M(ǫ(r)R)l−ogl(oǫg(rǫ)(Rr)) >ǫ(r)−k, for large r, R≥r.
In (6.2) we replace c with logR/log(ǫ(r)R), and replace ρ with ǫ(r)R. We obtain,
using (6.5) with k =3, that there exists R ≥R such that
1 0
(6.6) M(R)≥ǫ(r)−3M(ǫ(r)R), for R≥r≥R .
1
Note that we can assume that R is sufficiently large that M(r) > r, for r ≥
1
R /ǫ(r), and also, by (6.3), that η(r)>r, for r≥R .
1 1
We claim next that we have
(6.7) ηk(r)≥ǫ(r)−k−1Mk(ǫ(r)r)>Mk(ǫ(r)r), for r ≥R , k ∈N.
1