# iterates of meromorphic functions on escaping Fatou components

Zheng Jian-Hua and Wu Cheng-Fa  Department of Mathematical Sciences, Tsinghua
University, Beijing, 100084, P. R. China<EMAIL_ADDRESS>Institute for Advanced Study, Shenzhen University, Shenzhen, 518060, P. R.
China<EMAIL_ADDRESS>
###### Abstract.

In this paper, we prove that the ratio of the modulus of the iterates of two
points in an escaping Fatou component may be bounded even if the orbit of the
component contains an infinite modulus annulus sequence and this case cannot
happen when the maximal modulus of the meromorphic function is large enough.
Therefore, we extend the related results for entire functions to ones for
meromorphic functions with infinitely many poles. And we investigate the fast
escaping Fatou components of meromorphic functions defined in [44] in terms of
the Nevanlinna characteristic instead of the maximal modulus in [12] and show
that the multiply-connected wandering domain is a part of the fast escaping
set under a growth condition of the maximal modulus. Finally we give examples
of wandering domains escaping at arbitrary fast rate and slow rate.

###### Key words and phrases:

Escaping Sets, Meromorphic Functions, Fatou set, Julia set

###### 2010 Mathematics Subject Classification:

37F10 (primary), 30D05 (secondary)

Supported by the grant (No. 12071239) of NSF of China.

##  1\. Introduction and Main Results

Let $f$ be a meromorphic function which is not a $M\ddot{o}bius$
transformation and let $f^{n}$, $n\in\mathbb{N}$, denote the $n$th iterate of
$f$. The Fatou set $F(f)$ of $f$ is defined as set of points,
$z\in\hat{\mathbb{C}}$, such that $\\{f^{n}\\}_{n\in\mathbb{N}}$ is well-
defined and forms a normal family in some neighborhood of $z$. Therefore,
$F(f)$ is an open set. Since $F(f)$ is completely invariant under $f$, i.e.,
$f(z)\in F(f)$ if and only if $z\in F(f)$, for a component $U$ of $F(f)$,
$f(U)$ is contained in a component of $F(f)$. We write $U_{n}$ for the
component of $F(f)$ such that $f^{n}(U)\subseteq U_{n}$. The complement
$J(f)=\hat{\mathbb{C}}\setminus F(f)$ of $F(f)$ is called Julia set of $f$.
Then $\bigcup_{n=0}^{\infty}f^{-n}(\infty)\subset J(f)$ and if
$\bigcup_{n=0}^{\infty}f^{-n}(\infty)$ contains three distinct points in the
extended complex plane $\hat{\mathbb{C}}$, then
$\overline{\bigcup_{n=0}^{\infty}f^{-n}(\infty)}=J(f)$. Thus every $f^{n}$ is
analytic in $F(f)$ for $n\geq 1$.

In this paper, we investigate the Fatou component $U$ of $f$ with
$f^{n}|_{U}\to\infty(n\to\infty)$. For convenience of statements, we call such
an $U$ an escaping Fatou component of $f$. Let us first determine what kinds
of Fatou components will be escaping to $\infty$ under iterates.

A periodic Fatou component $W$ of $f$ with period $p\geq 1$, i.e.,
$f^{p}(W)\subseteq W=W_{p}$ and $p$ is the minimal positive integer such that
the inclusion holds, is known as a Baker domain if $f^{np}|_{W}\to
a(n\to\infty)$, but $f^{p}$ is not defined at $a$. Then $\infty$ must be the
limit value of $f^{np+j}$ in $W$ for some $0\leq j\leq p-1$ and so $W_{j}$ is
unbounded. If an escaping Fatou component $U$ of $f$ is periodic, i.e., for
some positive integer $p$, $f^{p}(U)\subseteq U=U_{p}$, then $U$ is a Baker
domain of $f$. The periodic Fatou components are classified into five possible
types: Attracting domains, Parabolic domains, Siegel disks, Herman rings and
Baker domains; see [10], Theorem 6.

A Fatou component $U$ is called a wandering domain of $f$, if for any pair of
positive integers $m\not=n$, $U_{m}\not=U_{n}$ and actually $U_{m}\cap
U_{n}=\emptyset$. We know that the escaping Fatou component $U$ is either a
wandering domain, a Baker domain or a preimage of Baker domain. Many examples
of such Baker domains and wandering domains have been revealed; see [26] for a
survey about Baker domains and [2, 3, 8, 21, 18, 22, 30, 11, 36, 15, 14, 19,
23], here we mention some of many papers, for wandering domains; the existence
of wandering domains with some special behavior of geometry, analysis or
dynamics and/or for some special class of meromorphic functions continues to
be an important topic and keeps to attract many interests.

Let $U$ be an escaping Fatou component of a meromorphic function $f$.
Generally for any two distinct points $a$ and $b$ in $U$ there exist $M>1$ and
a positive integer $N$ such that for $\forall\ n\geq N$,

(1.1) $|f^{n}(b)|^{1/M}\leq|f^{n}(a)|\leq|f^{n}(b)|^{M}$

and if $\hat{\mathbb{C}}\setminus U$ contains an unbounded component, we have
a more precise inequality

(1.2) $M^{-1}|f^{n}(a)|\leq|f^{n}(b)|\leq M|f^{n}(a)|,\ \forall\ n\geq N.$

(see [1], Lemma 5 or [10], Lemma 7). Therefore, if $U$ is an unbounded or
simply-connected wandering domain, then (1.2) holds. As we know, the escaping
Baker domain is unbounded and may be either simply-connected or multiply-
connected. However, it was shown in [43] and [25] that a Baker domain is
always such that (1.2) holds. Therefore, an escaping Fatou component $U$ not
satisfying (1.2) must be bounded, multiply-connected and wandering.

The first entire function with multiply connected Fatou component was
constructed by I. N. Baker [2, 3, 4] and every multiply-connected Fatou
component $U$ of a transcendental entire function was shown in Töpfer [40] and
Baker [5] to be a bounded escaping wandering domain and in [42] and [13] that
for all sufficiently large $n$, $f^{n}(U)$ contains a round annulus centered
at the origin with the large modulus tending to $\infty$ as $n\to\infty$. But
it is not this case for a meromorphic function with poles. A meromorphic
function may have the Herman rings, multiply connected attracting domains,
parabolic domains and Baker domains; see [16] for examples of meromorphic
functions with only one pole that have invariant multiply connected Fatou
component. A meromorphic function was constructed by Baker, Kotus and Lü [8]
to have a multiply connected wandering domain $U$ of preassigned connectivity
such that the limit set of $\\{f^{n}|_{U}\\}$ is an infinite set including
$\infty$. A planar region $E$ is said to have connectivity $m$ if
$\hat{\mathbb{C}}\setminus E$ has $m$ components. Such a wandering domain of a
transcendental entire function must be simply connected and such an example of
entire function was constructed by Eremenko and Lyubich [18]. Some examples of
meromorphic functions with finitely or infinitely many poles which have
simply, doubly or infinitely connected wandering domains and which change
connectivity of wandering domains under iterates can be found in [30].

There exists a meromorphic function with bounded, multiply-connected, escaping
and wandering domains separating the origin and $\infty$ where (1.2) holds;
see Example 5.2 of [45]. As we know, there exist the escaping wandering
domains such that (1.1) holds but (1.2) does not hold. In fact, an entire
function does not satisfy (1.2) in its multiply connected Fatou component and,
however, (1.1) is best possible; see Theorem 1.1 in [13]. Furthermore, if
(1.2) does not hold, then the $U$ contains two points $a$ and $b$ such that

(1.3) $|f^{n_{k}}(a)|/|f^{n_{k}}(b)|\to\infty(k\to\infty).$

We have shown in [45] that in this case, $\bigcup_{n=0}^{\infty}f^{n}(U)$
contains a sequence of annuli $A(r_{m},R_{m})$ with
$R_{m}/r_{m}\to\infty(m\to\infty)$ and $r_{m}\to\infty(m\to\infty)$; we call
such a sequence of annuli an infinite modulus annulus sequence; Here and
henceforth, for $R>r>0$ we denote round annulus $\\{z:\ r<|z|<R\\}$ by
$A(r,R)$. And for all sufficiently large $n_{k}$, $U_{n_{k}}=f^{n_{k}}(U)$
separates the origin and $\infty$. A condition was given in [45] to ensure the
establishment of that for all sufficiently large $n$, $U_{n}=f^{n}(U)$
surrounds the origin and contains a large round annulus centered at the origin
which form an infinite modulus annulus sequence. We can construct meromorphic
functions which have wandering domain $U$ such that $f^{2n}(U)$ surrounds the
origin and contains a large round annulus centered at the origin with (1.3),
but $f^{2n-1}(U)$ does not surround the origin by suitably modifying the
construction of Example 5.4 in [45]. Here we omit the constructing details
which are routine but lengthy.

This leads us to beg the following

Question $\mathcal{A}$: For an escaping Fatou component $U$ of $f$, if
$\bigcup_{n=0}^{\infty}f^{n}(U)$ contains an infinite modulus annulus
sequence, are there some pair of points $a$ and $b$ in $U$ such that (1.2)
does not hold ?

An equivalent statement of Question $\mathcal{A}$ is that if (1.2) is
satisfied in an escaping Fatou component $U$ of $f$, then whether may
$\bigcup_{n=0}^{\infty}f^{n}(U)$ contain an infinite modulus annulus sequence?
It turns out that it is this case. There exist wandering domains whose orbit
contains an infinite modulus annulus sequence and (1.2) holds. This is the
first result of this paper.

######  Theorem 1.

There exists a transcendental meromorphic function $f$ which has an escaping
wandering domain $U$ such that for all sufficiently large $n$,
$f^{n}(U)\supset A(r_{n},R_{n})$ with $R_{n}/r_{n}\to\infty(n\to\infty)$ and
$r_{n}\to\infty(n\to\infty)$, but for any two points $a,b\in U$, there exists
an $M>1$ such that for all sufficiently large $n$,

$M^{-1}|f^{n}(a)|\leq|f^{n}(b)|\leq M|f^{n}(a)|.$

In another preparing paper studying the Baker domains, we construct a
meromorphic function which has a Baker domain containing an infinite modulus
annulus sequence and not essentially conjugating to a Möbius transformation.

However, the answer to Question $\mathcal{A}$ is positive for a transcendental
entire function; Indeed, (1.2) does not hold for an entire function in its
multiply-connected Fatou component, as we mentioned earlier; see Theorem 1.1
in [13]. In this paper, we establish a more general result allowed to deal
with infinitely many poles.

In order to clearly state our second result of this paper, we describe the
notations appearing often below. Here and below $M(r,f)$ ($\hat{m}(r,f)$,
respectively) denotes the maximal (minimal, respectively) modulus of $f$ on
the circle $\\{|z|=r\\}$, i.e.,

$M(r,f)=\max\\{|f(z)|:\ |z|=r\\},\ \hat{m}(r,f)=\min\\{|f(z)|:\ |z|=r\\}.$

If $f$ has a pole in $\\{|z|=r\\}$, then $M(r,f)=+\infty$. $T(r,f)$ is the
characteristic of Nevanlinna of $f$; $m(r,f)$ is the proximity function of
$f$; $N(r,f)$ is the integrated counting function of $f$ in Nevanlinna theory
(see [20] and [41]). Please see Section 2 for their exact definitions.

For an escaping Fatou component $U$ and $c\in U$, define

$h_{n}(z):=\frac{\log|f^{n}(z)|}{\log|f^{n}(c)|},\ \forall\ z\in U,$
$\overline{h}(z):=\limsup_{n\to\infty}h_{n}(z),\
\underline{h}(z):=\liminf_{n\to\infty}h_{n}(z),\ \forall\ z\in U.$

In view of (1.1) we easily see that $\overline{h}(z)$ and $\underline{h}(z)$
exist on $U$. In [13], Bergweiler, Rippon and Stallard first introduced and
proved that $h(z):=\overline{h}(z)=\underline{h}(z)$ on a multiply connected
Fatou component $U$ of an entire function $f$ and applied $h$ successfully to
describe the geometric structure of $U$; This work leads to $h$ becoming a
significant function.

######  Theorem 2.

Assume that for some $C\geq 1$, $c>0$ and $r_{0}>0$, we have

(1.4) $\log M(Cr,f)\geq\left(1-\frac{c}{\log r}\right)T(r,f),\ \forall\
r>r_{0}.$

Then there exist constants $D>1$, $d>1$ and $R_{0}>0$ such that for a Fatou
component $U$ of $f$, if for some $m>0$ and some $r\geq R_{0}$, we have

(1.5) $A(D^{-d}r,D^{d}r)\subseteq f^{m}(U),$

then $U$ is an escaping wandering domain,
$h(z):=\overline{h}(z)=\underline{h}(z)$ exists on $U$ and $h(z)$ is a non-
constant positive harmonic function on $U$. Therefore for any point $z_{0}\in
U$ and any neighborhood $V$ of $z_{0}$ in $U$, we have for some constant
$0<\alpha<1$ and all sufficiently large $n$

$f^{n}(V)\supset A(|f^{n}(z_{0})|^{1-\alpha},|f^{n}(z_{0})|^{1+\alpha});$

moreover, there exist two points $a$ and $b$ in $V$ and $\tau>1$ such that

$|f^{n}(a)|^{\tau}\leq|f^{n}(b)|$

and

$A_{n}=A(r_{n},R_{n})\subseteq f^{n}(V)$

with $R_{n}\geq r_{n}^{\tau}\to\infty(n\to\infty)$ and $A_{n+1}\subseteq
f(A_{n})$.

We can give effective conditions to determine $D,d$ and $R_{0}$ in Theorem 2.
We note that the condition which $D,d$ and $R_{0}$ satisfy is actually for the
result of Lemma 6 to hold. Therefore from the proof of Theorem 1.1 in [45], a
rough calculation shows that $D,d$ and $R_{0}$ satisfying the following
inequalities are available for Theorem 2:

(1.6) $D>2Ce^{c+1},\ \frac{\pi}{d\cos\frac{\pi}{2d}}\left(1+\frac{\pi}{2\log
D}\right)<\frac{1}{6},\ R_{0}\geq D^{d}r_{0}$

and

(1.7) $\frac{T(r,f)}{\log r}>d\log D,\ \forall\ r\geq R_{0}.$

It is easy to see that that if $\bigcup_{n=0}^{\infty}f^{n}(U)$ contains an
infinite modulus annulus sequence for an escaping Fatou component $U$ of $f$,
then we have (1.5) with $D,d$ and $R_{0}$ such that (1.6) and (1.7) hold.

Let us make remarks on (1.4). Theorem 1 tells us that (1.4) cannot be removed
in Theorem 2. In Example 5.3 in [45], we constructed a meromorphic function
which has an escaping wandering domain $U$ in which there exist two points $a$
and $b$ such that $|f^{n}(b)|/|f^{n}(a)|\to\infty(n\to\infty)$ but
$h_{U}(z)\equiv 1$ on $U$.

If $f$ only has at most finitely many poles, then (1.4) immediately holds. So
Theorem 2 extends the related results in [13] to a meromorphic function with
infinitely many poles. If

(1.8) $T(r,f)\geq N(r,f)\log r,\ \forall\ r\geq r_{0},$

then we have

$m(r,f)=T(r,f)-N(r,f)\geq\left(1-\frac{1}{\log r}\right)T(r,f),\ \forall\
r\geq r_{0}.$

Hence, (1.8) implies (1.4) and that the Nevanlinna deficiency
$\delta(\infty,f)$ of $f$ of poles is $1$. It is easy to find a meromorphic
function $f$ satisfying (1.4) and $\delta(\infty,f)<1$. Consider the function
$f(z)=\dfrac{e^{z}-1}{e^{-z}+1}$. Then

$m(r,f)\sim\frac{r}{\pi},\ \ N(r,f)\sim\frac{r}{\pi}\ (r\to\infty)$

and $\delta(\infty,f)=\frac{1}{2}$. And

$\log M(r,f)\geq\log\frac{e^{r}-1}{e^{-r}+1}\sim r\sim\pi T(r,f)\
(r\to\infty).$

Then we obtain (1.4) for $C=1$.

Finally, we mention that Theorem 2 is still true even if (1.4) is satisfied
for all possibly except a set of $r$ with finite logarithmic measure.

The existence of large annuli in the orbit of an escaping Fatou component
guarantees the stability of the annuli in the sense that the suitable large
annuli still exist under relatively small changes to the function in question.

######  Theorem 3.

Let $f$ be a meromorphic function satisfying the conditions in Theorem 2. If
$f$ has an escaping Fatou component $U$ such that
$\bigcup_{n=0}^{\infty}f^{n}(U)$ contains an infinite modulus annulus
sequence, then for any meromorphic function $g$ which is analytic on
$\bigcup_{n=N}^{\infty}f^{n}(U)$ and such that for some $N$ and for
$0<\delta<1$

(1.9) $M(r,f-g)\leq M(r,f)^{\delta},\text{for}\
\\{|z|=r\\}\subset\bigcup_{n=N}^{\infty}f^{n}(U),$

$g$ has an escaping Fatou component $V$ on which the results in Theorem 2 also
hold for $g$.

Theorem 3 was proved in [13] for the case when $f$ and $g$ are entire. Theorem
3 gives us a strong argument to be able to find a meromorphic function $f$
with $\delta(\infty,f)=0$ which has an escaping wandering domain $U$ with
properties stated in Theorem 2 under $f$. This is stated as follows.

######  Theorem 4.

There exists a meromorphic function $f$ with $\delta(\infty,f)=0$ which has a
wandering domain $W$ such that the results in Theorem 2 hold for $f$ and $W$.

Furthermore, we can find a meromorphic function that has the results in
Theorem 2, but that does not satisfy (1.4) in a set of $r$ with the infinite
logarithmic measure.

Next we investigate escaping speed of points in an escaping Fatou component
under iterates. A fast escaping set of a transcendental entire function $f$,
denoted by $A(f)$, was introduced by Bergweiler and Hinkkanen [12] and can be
defined [29] by

$A(f)=\\{z\in\mathbb{C}:\ {\rm there\ exists}\ L\in\mathbb{N}\ {\rm such\
that}\ |f^{n+L}(z)|\geq M_{n}(R,f),\ {\rm for}\ n\in\mathbb{N}\\},$

where $R>\min\limits_{z\in J(f)}|z|$ and $M_{n}(r,f)$ is the $n$th iterate of
$M(r,f)$ with respect to $r$. $A(f)$ is independent of the choice of $R$. It
turns out that the fast escaping set $A(f)$ possesses many interesting
dynamical behaviors and attracts a lot of attentions; see [29] for the detail;
for example, every component of $A(f)$ is unbounded, which is related to a
conjecture of Eremenko ([17]). This has led to a lot of recent researches,
here we mention some of them: [24, 37, 38, 34, 33].

The maximal modulus $M(r,f)$ of an entire function has proved to be powerful
in the study of iterate theory, but it is inferior for a meromorphic function
with infinitely many poles. Although we can define the maximal modulus of a
meromorphic function, basically $f^{n}$ with $n>1$ may not be a meromorphic
function. Hence we cannot consider the fast escaping set of a meromorphic
function with the help of maximal modulus. Rippon and Stallard in [27]
mentioned how to define the fast escaping set for a meromorphic function. In
[41] we used the Nevanlinna characteristic in the place of the maximal modulus
to define it.

For a meromorphic function $f$, set $\hat{T}(r,f)=\exp T(r,f)$ and
$\hat{T}_{n}(r,f)$ is the $n$th iterate of $\hat{T}(r,f)$ with respect to $r$.
Define

$M(f)=\\{z\in\mathbb{C}:\ {\rm there\ exists}\ L\in\mathbb{N}\ {\rm such\
that}\ |f^{n+L}(z)|\geq\hat{T}_{n}(R,f),\ {\rm for}\ n\in\mathbb{N}\\}.$

Other equivalent definitions of $A(f)$ (cf.[12] and [29]) can be defined with
the Nevanlinna characteristic for an entire function $f$, but not for a
meromorphic function with poles. In [41], we showed that

(1) $M(f)$ is non-empty and completely invariant;

(2) $J(f)=\partial M(f)$;

(3) if $U$ is a Fatou component of $f$ and $U\cap M(f)\not=\emptyset$, then
$\overline{U}\subset M(f)$ and $U$ is an escaping wandering domain of $f$.

Let us compare $M(f)$ to $A(f)$ for an entire function and establish the
following

######  Theorem 5.

Let $f$ be a transcendental entire function. Then $M(f)=A(f)$.

Closure of a multiply connected Fatou component of an entire function must be
in the fast escaping set $A(f)$, proved in [32]. But basically, this
conclusion does not hold for a meromorphic function. In following, we give a
sufficient condition.

######  Theorem 6.

Let $f$ be a transcendental meromorphic function with (1.4). Then for
$D>e^{3c}$ and $R_{0}\geq r_{0}$ with property that for $r\geq R_{0}$,

(1.10) $\frac{T(r,f)}{\log r}>\frac{22.56(\kappa+\log D)}{\log D-3c},$

if a Fatou component $U$ of $f$ satisfies that for some $m>0$ and some
$r>CR_{0}$,

(1.11) $A(r,Dr)\subset f^{m}(U),$

we have $\overline{U}\subset M(f)$.

As we know, a simply connected wandering domain of an entire function may not
be in $A(f)$. Are there any simply connected wandering domain in $A(f)$, which
is asked in [28]. The existence of such simply connected wandering domains has
been shown by Bergweiler [11] and Sixsmith [36].

A point $z_{0}$ in an escaping Baker domain has escaping rate

$\log|f^{n}(z_{0})|=O(n)\ (n\to\infty);$

see [6, 10, 43, 25]. In another paper we find a Baker domain where
$|f^{n}(z)|=o(n)(n\to\infty)$ and $f^{n}(z)$ goes forward to $\infty$ in a
slow spiral. For the case of escaping wandering domains, we can construct
escaping wandering domains with any escaping rate under iterates, that is to
say, given a sequence $\\{a_{n}\\}$ of positive numbers tending to $\infty$,
we can find a wandering domain $U$ such that for any point $z_{0}\in U$, we
have

(1.12) $|f^{n}(z_{0})|\sim a_{n}(n\to\infty).$

######  Theorem 7.

Let $\\{a_{n}\\}$ be arbitrary sequence of positive numbers with
$1<a_{n}<a_{n+1}\to\infty(n\to\infty)$. For $0<\rho\leq\infty$, there exists
an entire function $f$ with order at least $\rho$ which has an escaping
wandering domain $U$ such that for some point $z_{0}$, $|f^{n}(z_{0})|\sim
a_{n}(n\to\infty)$. In terms of (1.2), for any point $a\in U$, there exists
$L>1$ such that $\frac{a_{n}}{L}\leq|f^{n}(a)|\leq La_{n}$.

If it is allowed that $f$ is meromorphic in Theorem 7, then we can obtain
(1.12) for any point $z_{0}\in U$. Theorem 7 would not be new, but we could
not find any references about it. It was shown in [31] that every Fatou
component of $z+\sin z+2\pi$ is a bounded wandering domain where $f^{n}(z)\sim
2\pi n(n\to\infty)$; every Fatou component of $z+e^{-z}+2\pi i$ is an
unbounded wandering domain where $f^{n}(z)\sim 2\pi n(n\to\infty)$.

If $f$ is entire, then $a_{n}\leq M(2|z_{0}|,f^{n})\leq M_{n}(2|z_{0}|,f)$,
where $M_{n}(r,f)$ stands for the $n$th iterate of $M(r,f)$ with respect to
$r$. However from Theorem LABEL:thm7 we cannot estimate $a_{n}$ in terms of
$f$ and $z_{0}$ when $f$ is meromorphic with infinitely many poles.

The organization of this paper is the following. We will collect the
preliminary results in Section 2 applied often to the proofs of theorems. We
will prove Theorems 1-4 in Section 3 and Theorems 5-7 in Section 4. We will
make further discussions or remarks after the proofs of theorems.

##  2\. Preliminary Lemmas

Firstly we need a covering lemma which comes from the hyperbolic metric. Let
$\Omega$ be a hyperbolic domain in the extended complex plane
$\hat{\mathbb{C}}$, that is, $\hat{\mathbb{C}}\backslash\Omega$ contains at
least three points. Then there exists on $\Omega$ the hyperbolic metric
$\lambda_{\Omega}(z)|{\rm d}z|$ with Gaussian curvature $-1$ and
$\lambda_{\Omega}(z)$ is the hyperbolic density of $\Omega$ at $z\in\Omega.$
Then

$\lambda_{D}(z)=\frac{1}{|z|\log|z|},\ D=\\{z:|z|>1\\}$

and

(2.1) $\lambda_{A}(z)=\frac{\pi}{\left|z\right|{\rm
mod}(A)\sin(\pi\log(R/\left|z\right|)/{\rm mod}(A))},\forall\ z\in A=A(r,R),$

where $A(r,R)=\\{z:\ r<|z|<R\\}$ and ${\rm mod}(A)=\log R/r$ is the modulus of
$A$. The hyperbolic distance $d_{\Omega}(u,v)$ of two points $u,v\in\Omega$ is
defined by

$d_{\Omega}(u,v)=\inf_{\alpha}\int_{\alpha}\lambda_{\Omega}(z)\left|dz\right|,$

where the infimum is taken over all piecewise smooth paths $\alpha$ in
$\Omega$ joining $u$ and $v$.

######  Lemma 1.

([45]) Let $f$ be analytic on a hyperbolic domain $U$ with $0\not\in f(U)$. If
there exist two distinct points $z_{1}$ and $z_{2}$ in $U$ such that
$|f(z_{1})|>e^{\kappa\delta}|f(z_{2})|$, where $\delta=d_{U}(z_{1},z_{2})$ and
$\kappa=\Gamma(1/4)^{4}/(4\pi^{2})$, then there exists a point $\hat{z}\in U$
such that $|f(z_{2})|\leq|f(\hat{z})|\leq|f(z_{1})|$ and

(2.2) $f(U)\supset
A\left(e^{\kappa}\left(\frac{|f(z_{2})|}{|f(z_{1})|}\right)^{1/\delta}|f(\hat{z})|,\
e^{-\kappa}\left(\frac{|f(z_{1})|}{|f(z_{2})|}\right)^{1/\delta}|f(\hat{z})|\right).$

If $|f(z_{1})|\geq\exp\left(\frac{\kappa\delta}{1-\delta}\right)|f(z_{2})|$
and $0<\delta<1$, then

(2.3) $f(U)\supset A(|f(z_{2})|,|f(z_{1})|).$

In particular, for $\delta\leq\frac{1}{6}$ and $|f(z_{1})|\geq e|f(z_{2})|$,
we have (2.3).

In order to be able to cover an effective annulus, we are forced to calculate
carefully the hyperbolic distance in an annulus.

######  Lemma 2.

Set $A=A(r,R)$ with $0<r<R$. Then for any two points $z_{1},z_{2}\in A$ with
$|z_{2}|\leq|z_{1}|$ we have

$\max\left\\{\frac{\pi}{{\rm
mod}(A)}\log\frac{|z_{1}|}{|z_{2}|},\log\frac{\log\frac{R}{|z_{2}|}}{\log\frac{R}{|z_{1}|}}\right\\}\leq
d_{A}(z_{1},z_{2})\leq\frac{\pi^{2}}{{\rm mod}(A)\hat{K}_{0}}+\frac{\pi}{{\rm
mod}(A)\hat{K}_{1}}\log\frac{|z_{1}|}{|z_{2}|},$

where $\hat{K}_{0}=\max\limits_{j=1,2}\sin(\pi\log(R/|z_{j}|)/{\rm mod}(A))$
and $\hat{K}_{1}=\min\limits_{j=1,2}\sin(\pi\log(R/|z_{j}|)/{\rm mod}(A))$. In
particular, for $z_{1},z_{2}$ with
$\frac{R}{|z_{1}|}=\left(\frac{R}{r}\right)^{\sigma},\frac{R}{|z_{2}|}=\left(\frac{R}{r}\right)^{\tau}$
and $0<\sigma\leq\tau<1$, we have

(2.4) $\max\left\\{(\tau-\sigma)\pi,\log\frac{\tau}{\sigma}\right\\}\leq
d_{A}(z_{1},z_{2})\leq\frac{\pi^{2}}{\hat{K}_{0}{\rm
mod}(A)}+\frac{(\tau-\sigma)\pi}{\hat{K}_{1}},$

with $\hat{K}_{0}=\max\\{\sin(\sigma\pi),\sin(\tau\pi)\\}$ and
$\hat{K}_{1}=\min\\{\sin(\sigma\pi),\sin(\tau\pi)\\}$.

###### Proof.

We prove the first inequality. In view of (2.1), for all $z\in A$ we have

$\lambda_{A}(z)=\frac{\pi}{|z|{\rm mod}(A)\sin(\pi\log(R/|z|)/{\rm
mod}(A))}\geq\frac{1}{|z|\log\frac{R}{|z|}}$

and

$\lambda_{A}(z)\geq\frac{\pi}{|z|{\rm mod}(A)}.$

Therefore

$d_{A}(z_{1},z_{2})\geq\int_{\gamma}\frac{|{\rm
d}z|}{|z|\log\frac{R}{|z|}}\geq\log\frac{\log\frac{R}{|z_{2}|}}{\log\frac{R}{|z_{1}|}}$

and

$d_{A}(z_{1},z_{2})\geq\frac{\pi}{{\rm mod}(A)}\log\frac{|z_{1}|}{|z_{2}|},$

where $\gamma$ is the geodesic curve in $A$ connecting $z_{1}$ and $z_{2}$.

Next we prove the second inequality. For all $z$ with
$|z_{2}|\leq|z|\leq|z_{1}|$ we have

$\sin(\pi\log(R/|z|)/{\rm mod}(A))\geq\min_{j=1,2}\sin(\pi\log(R/|z_{j}|)/{\rm
mod}(A))=\hat{K}_{1}.$

Without loss of generality assume that $\hat{K}_{0}$ attains the maximal value
at $z_{2}$. Then

$\displaystyle d_{A}(z_{1},z_{2})$ $\displaystyle\leq$
$\displaystyle\frac{\pi}{{\rm mod}(A)\hat{K}_{0}}\int_{\gamma_{0}}\frac{|{\rm
d}z|}{|z|}+\frac{\pi}{{\rm mod}(A)\hat{K}_{1}}\int_{\gamma_{1}}\frac{|{\rm
d}z|}{|z|}$ $\displaystyle\leq$ $\displaystyle\frac{\pi^{2}}{{\rm
mod}(A)\hat{K}_{0}}+\frac{\pi}{{\rm
mod}(A)\hat{K}_{1}}\log\frac{|z_{1}|}{|z_{2}|},$

where $\gamma_{0}$ is the shortest arc from $|z_{2}|e^{i\arg z_{1}}$ to
$z_{2}$ and $\gamma_{1}=\\{z=te^{i\arg z_{1}}:\ |z_{1}|\leq t\leq|z_{2}|\\}$.
∎

Please pay attention! The bounds of $d_{A}(z_{1},z_{2})$ only depend on ${\rm
mod}(A)$, but independent of the size of $R$ and $r$.

######  Lemma 3.

([45]) Let $h(z)$ be analytic on the annulus $B=A(r,R)$ with $0<r<R<+\infty$
such that $|h(z)|>1$ on $B$. Then

(2.5) $\displaystyle\log\hat{m}(\rho,h)$ $\displaystyle\geq$
$\displaystyle\exp\left(-\frac{\pi^{2}}{2}\max\left\\{\frac{1}{\log\frac{R}{\rho}},\frac{1}{\log\frac{\rho}{r}}\right\\}\right)\log
M(\rho,h)$ $\displaystyle\geq$
$\displaystyle\min\left\\{\frac{\log\frac{\rho}{r}-\pi}{\log\frac{\rho}{r}+\pi},\frac{\log\frac{R}{\rho}-\pi}{\log\frac{R}{\rho}+\pi}\right\\}\log
M(\rho,h),$

where $\rho\in(r,R)$ and $\hat{m}(\rho,h)=\min\\{|h(z)|:|z|=\rho\\}.$

Secondly we need the basic notations and results in Nevanlinna theory of
meromorphic functions ([20] or [41]). Set $\log^{+}x=\log\max\\{1,x\\}.$ Let
$f$ be a meromorphic function. Define

$m(r,f):=\frac{1}{2\pi}\int_{0}^{2\pi}\log^{+}|f(re^{i\theta})|{\rm d}\theta,$
$N(r,f):=\int_{0}^{r}\frac{n(t,f)-n(0,f)}{t}{\rm d}t+n(0,f)\log r,$

where $n(t,f)$ denotes the number of poles of $f$ counted according to their
multiplicities in $\\{z:|z|<t\\}$, sometimes we write $n(t,\infty)$ for
$n(t,f)$ and $n(t,0)$ for $n\left(t,\frac{1}{f}\right)$ when $f$ is clear in
the context, and the Nevanlinna characteristic of $f$ by

$T(r,f):=m(r,f)+N(r,f).$

Then the deficiency of Nevanlinna of $f$ of poles is

$\delta(\infty,f)=\liminf_{r\to\infty}\frac{m(r,f)}{T(r,f)}=1-\limsup_{r\to\infty}\frac{N(r,f)}{T(r,f)}.$

We write the conclusions about the Nevanlinna characteristic we shall use
below as a lemma, which also holds for $\log M(r,f)$ of a transcendental
entire function $f$; see Theorem 2.2 of [13].

######  Lemma 4.

Let $f$ be a transcendental meromorphic function. Then

(1) $\frac{T(r,f)}{\log r}\rightarrow\infty(r\rightarrow\infty)$;

(2) There exists $r_{0}>0$ such that for $r_{0}\leq r<R$,

(2.6) $T(R,f)\geq\frac{\log R}{\log r}T(r,f);$

(3) There exists $r_{0}>0$ such that for $r_{0}\leq r<R$ and any positive
integer $n$, we have

(2.7) $\hat{T}_{n}(R,f)\geq\hat{T}_{n}(r,f)^{\frac{\log R}{\log r}},$

where $\hat{T}(r,f)=e^{T(r,f)}$ and
$\hat{T}_{n}(r,f)=\hat{T}(\hat{T}_{n-1}(r,f),f),\hat{T}_{0}(r,f)=r.$

Since $T(r,f)$ is a logarithmic convex function, the result (2) in Lemma 4
follows from the result (1) in Lemma 4 and further (3) in Lemma 4 follows (see
[45]).

##  3\. Proofs of Theorems 1-4

We need Runge Theorem to complete the proofs of Theorems 1.

Runge Theorem(cf. [35]). Let $W$ be a compact set on the complex plane and let
$f(z)$ be analytic on $W$. Assume that $E$ is a set which intersects every
component of $\mathbb{C}\setminus W$. Then for any $\varepsilon>0$, there
exists a rational function $R(z)$ such that all poles of $R(z)$ lie in $E$ and

$|f(z)-R(z)|<\varepsilon,\ \forall z\in W.$

Now let us go to the proof of Theorem 1.

Take four sequences $\\{r_{n}\\},\ \\{R_{n}\\},\ \\{r^{\prime}_{n}\\}$ and
$\\{R^{\prime}_{n}\\}$ such that
$10<r_{n}<r_{n}^{\prime}<R_{n}^{\prime}<R_{n}$, $2\leq
r^{\prime}_{n}/r_{n}\leq 3$, $2\leq R_{n}/R^{\prime}_{n}\leq 3$,
$9R_{n}<r_{n+1}$ and $R_{n}/r_{n}=R_{n+1}^{\prime}/r_{n+1}^{\prime}$. Thus
$\frac{R_{n}}{r_{n}}=\frac{R_{n+1}^{\prime}}{r_{n+1}^{\prime}}\leq\frac{1}{4}\frac{R_{n+1}}{r_{n+1}}$,
and $\frac{R_{n}}{r_{n}}\geq 4^{n-1}\frac{R_{1}}{r_{1}}\to\infty(n\to\infty)$.

Define

$T_{n}(z)=\frac{r^{\prime}_{n+1}}{r_{n}}z:\ A(r_{n},R_{n})\to
A(r^{\prime}_{n+1},R^{\prime}_{n+1}).$

Take $a_{n}$ and $b_{n}$ such that $b_{n}>R_{n}+n$, $\frac{b_{n}}{R_{n}}\to
1(n\to\infty)$ and $a_{n}<r_{n}-n$, $\frac{a_{n}}{r_{n}}\to
1\break(n\to\infty)$. Set $A_{n}=A(r_{n},R_{n}),\ C_{n}=\\{z:|z|=b_{n}\
\text{or}\ a_{n}\\}$ and $B_{n}=B(0,r_{n}/4)$ with $r_{1}>4$.

Take a sequence of positive numbers $\\{\varepsilon_{n}\\}$ such that
$\varepsilon_{n+1}<\frac{1}{2}\varepsilon_{n}$ and
$\varepsilon_{1}<\frac{1}{2}$. In view of the Runge’s Theorem, we have a
rational function $f_{1}(z)$ such that

$|f_{1}(z)-T_{1}(z)|<\varepsilon_{1},\forall\ z\in A_{1};\
|f_{1}(z)|<\varepsilon_{1},\forall\ z\in B_{1}$ $|f_{1}(z)|<\varepsilon_{1},\
\forall\ z\in C_{1}$

and inductively, we have rational function $f_{n+1}(z)$ such that

$\left|\sum_{k=1}^{n+1}f_{k}(z)-T_{n+1}(z)\right|<\varepsilon_{n+1},\forall\
z\in A_{n+1};\ |f_{n+1}(z)|<\varepsilon_{n+1},\forall\ z\in B_{n+1}$

and

$\left|\sum_{k=1}^{n+1}f_{k}(z)\right|<\varepsilon_{n+1},\forall\ z\in
C_{n+1}.$

Write $f(z)=\sum_{n=1}^{\infty}f_{n}(z)$. Since the series is uniformly
convergent on any compact subset of $\mathbb{C}$, $f(z)$ is a meromorphic
function on $\mathbb{C}$.

For $z\in B_{1}$, we have

$|f(z)|\leq\sum_{n=1}^{\infty}|f_{n}(z)|<\sum_{n=1}^{\infty}\varepsilon_{n}<1,$

that is to say, $f(B_{1})\subset B(0,1)$ and so $B_{1}$ is contained in an
invariant Fatou component of $f$. For $z\in C_{n}$, we have, by noting that
$C_{n}\subset B_{m}$ for $m>n$,

$|f(z)|\leq\left|\sum_{k=1}^{n}f_{k}(z)\right|+\sum_{k=n+1}^{\infty}|f_{k}(z)|<\varepsilon_{n}+\sum_{k=n+1}^{\infty}|f_{k}(z)|<\sum_{k=n}^{\infty}\varepsilon_{k}<1$

and so $f(C_{n})\subset B(0,1)$ and $C_{n}$ is contained in a preperiodic
Fatou component of $f$. Since for $z\in A_{n}=A(r_{n},R_{n})\subset B_{n+1}$,

(3.1)
$|f(z)-T_{n}(z)|\leq\left|\sum_{k=1}^{n}f_{k}(z)-T_{n}(z)\right|+\left|\sum_{k=n+1}^{\infty}f_{k}(z)\right|<\sum_{k=n}^{\infty}\varepsilon_{k}=\varepsilon^{\prime}_{n}\
(say),$

we have

(3.2) $f(A(r_{n},R_{n}))\subset
A\left(r^{\prime}_{n+1}-\varepsilon^{\prime}_{n},\
R^{\prime}_{n+1}+\varepsilon^{\prime}_{n}\right)\subset A(r_{n+1},R_{n+1}).$

Therefore, $A_{n}$ is contained in a wandering domain $U_{n}$ of $f$ and
$f:U_{n}\to U_{n+1}$ is proper. Since $f$ is univalent in $A_{n}$ by the
Rouché Theorem, $U_{n}$ is not doubly connected. Each of $U_{n}$ has no
isolated boundary points. If $U_{n}$ is finitely connected, then in view of
the Riemann-Hurwitz Theorem, $f$ is univalent in $U_{n}$, but the modulus of
annulus $A_{m}$ tends to infinity as $m$ does, which contracts to the
conformal invariant property of annulus modulus. Therefore, $U_{n}$ must be
infinitely connected.

For a point $a\in A_{n}$, it follows from (3.1) that

$\frac{r^{\prime}_{n+1}}{r_{n}}|a|-\varepsilon^{\prime}_{n}\leq|f(a)|\leq\frac{r^{\prime}_{n+1}}{r_{n}}|a|+\varepsilon^{\prime}_{n}$

and

$\frac{r^{\prime}_{n+1}-\varepsilon^{\prime}_{n}}{r_{n}}\leq\frac{|f(a)|}{|a|}\leq\frac{r^{\prime}_{n+1}+\varepsilon^{\prime}_{n}}{r_{n}}.$

Inductively, from (3.2) we have

$\prod_{k=1}^{m}\frac{r^{\prime}_{n+k}-\varepsilon^{\prime}_{n+k-1}}{r_{n+k-1}}\leq\frac{|f^{m}(a)|}{|a|}\leq\prod_{k=1}^{m}\frac{r^{\prime}_{n+k}+\varepsilon^{\prime}_{n+k-1}}{r_{n+k-1}}.$

We note that

$\displaystyle\prod_{k=1}^{\infty}\frac{r^{\prime}_{n+k}+\varepsilon^{\prime}_{n+k-1}}{r^{\prime}_{n+k}-\varepsilon^{\prime}_{n+k-1}}$
$\displaystyle=$
$\displaystyle\prod_{k=1}^{\infty}\left(1+\frac{2\varepsilon^{\prime}_{n+k-1}}{r^{\prime}_{n+k}-\varepsilon^{\prime}_{n+k-1}}\right)$
$\displaystyle<$
$\displaystyle\exp\sum_{k=1}^{\infty}\frac{2\varepsilon^{\prime}_{n+k-1}}{r^{\prime}_{n+k}-\varepsilon^{\prime}_{n+k-1}}<e^{2}.$

Thus for two points $a$ and $b$ in $A_{n}$, we have

(3.3)
$\frac{|f^{m}(a)|}{|f^{m}(b)|}\leq\frac{|a|}{|b|}\prod_{k=1}^{m}\frac{r^{\prime}_{n+k}+\varepsilon^{\prime}_{n+k-1}}{r^{\prime}_{n+k}-\varepsilon^{\prime}_{n+k-1}}<\frac{|a|}{|b|}e^{2},\
\forall m\in\mathbb{N}.$

Now we need to treat two cases:

Case (I): $|f^{m}(b)|\leq R^{\prime}_{m}<R_{m}\leq|f^{m}(a)|$;

Case (II): $|f^{m}(b)|\leq r_{m}<r^{\prime}_{m}\leq|f^{m}(a)|$.

Below we only prove that Case (I) would be impossible. The same method yields
the same assertion to Case (II). Set $E_{m}=A(a_{m},b_{m}).$ In view of the
Principle of Hyperbolic Metric (Shwarz-Pick Lemma), we calculate

$\displaystyle\ d_{U}(a,b)$ $\displaystyle\geq$ $\displaystyle
d_{U_{m}}(f^{m}(a),f^{m}(b))$ $\displaystyle\geq$ $\displaystyle
d_{E_{m}}(f^{m}(b),f^{m}(a))$ $\displaystyle\geq$ $\displaystyle
d_{E_{m}}(R^{\prime}_{m},R_{m})$ $\displaystyle=$
$\displaystyle\int_{R^{\prime}_{m}}^{R_{m}}\lambda_{E_{m}}(z)|{\rm d}z|$
$\displaystyle=$ $\displaystyle\int_{R^{\prime}_{m}}^{R_{m}}\frac{\pi}{t\ {\rm
mod}(E_{m})\sin(\pi\log\frac{b_{m}}{t}/{\rm mod}(E_{m}))}{\rm d}t$
$\displaystyle\geq$ $\displaystyle\int_{R^{\prime}_{m}}^{R_{m}}\frac{{\rm
d}t}{t\log\frac{b_{m}}{t}}$ $\displaystyle=$
$\displaystyle\log\frac{\log\frac{b_{m}}{R^{\prime}_{m}}}{\log\frac{b_{m}}{R_{m}}}$
$\displaystyle=$
$\displaystyle\log\left(1+\frac{\log\frac{R_{m}}{R^{\prime}_{m}}}{\log\frac{b_{m}}{R_{m}}}\right)$
$\displaystyle\geq$ $\displaystyle\log\left(1+\frac{\log
2}{\log\frac{b_{m}}{R_{m}}}\right)\to\infty(m\to\infty).$

A contradiction is derived. Then for all sufficiently large $m>0$, $f^{m}(a)$
and $f^{m}(b)$ lie in $A_{m}$ or $A(a_{m},r^{\prime}_{m})$ or
$A(R^{\prime}_{m},b_{m})$ at the same time.

By noting that $2\leq r^{\prime}_{n}/a_{n}\leq 4$ and $2\leq
b_{n}/R^{\prime}_{n}\leq 4$ together with (3.3), we deduce that for any pair
of points $a,b\in U$ and all sufficiently large $n$

$\frac{|f^{n}(a)|}{|f^{n}(b)|}\leq\max\left\\{4,\frac{|a|}{|b|}e^{2}\right\\}.$

∎

For the proof of Theorem 2, we first establish two lemmas.

######  Lemma 5.

Let $f$ be a meromorphic function on $\\{z:\ |z|\leq R\\}$ and $f$ be analytic
on $\overline{A}(r,R)$ with $0<r<R/4$. Then for $r+1<r^{\prime}<R^{\prime}\leq
R$, on $\\{|z|=r^{\prime}\\}$, we have

(3.4)
$\log^{+}|f(z)|\leq\left(\frac{R^{\prime}+r^{\prime}}{R^{\prime}-r^{\prime}}+\left(\log\frac{R}{r}\right)^{-1}\log\frac{R^{\prime}}{r^{\prime}-r}\right)T(R,f).$

###### Proof.

In view of the Poisson formula, for $z\in D=\\{z:\ |z|<R^{\prime}\\}$ with
$f(z)\not=0,\ \infty$, we have

(3.5) $\ \log|f(z)|\leq m(D,z,f)+N(D,z,f),$

where

$m(D,z,f)=\frac{1}{2\pi}\int_{\partial D}\log^{+}|f(\zeta)|\frac{\partial
G_{D}(\zeta,z)}{\partial\vec{n}}{\rm d}s,$ $N(D,z,f)=\sum\limits_{b_{n}\in
D}G_{D}(b_{n},z),$

$G_{D}$ is the Green function for $D$ and all $b_{n}$ are poles of $f$ in $D$
appearing often according to their multiplicities. Then for $z$ with
$|z|=r^{\prime}$, we have

$m(D,z,f)=\frac{1}{2\pi}\int_{0}^{2\pi}\log^{+}|f(R^{\prime}e^{i\theta})|\frac{R^{\prime
2}-|z|^{2}}{|R^{\prime 2}-z|^{2}}{\rm d}\theta$
$\leq\frac{R^{\prime}+|z|}{R^{\prime}-|z|}m(R^{\prime},f)\leq\frac{R^{\prime}+|z|}{R^{\prime}-|z|}T(R^{\prime},f)$

and, by denoting the number of poles of $f$ in $D$ by $n(D,f)$, we have

$\displaystyle N(D,z,f)$ $\displaystyle\leq$
$\displaystyle\sum\limits_{b_{n}\in D}\log\frac{R^{\prime}}{|b_{n}-z|}$
$\displaystyle\leq$ $\displaystyle n(D,f)\log\frac{R^{\prime}}{r^{\prime}-r}$
$\displaystyle=$ $\displaystyle n(r,f)\log\frac{R^{\prime}}{r^{\prime}-r}$
$\displaystyle=$
$\displaystyle\left(\log\frac{R}{r}\right)^{-1}\int_{r}^{R}\frac{n(t,f)}{t}{\rm
d}t\log\frac{R^{\prime}}{r^{\prime}-r}$ $\displaystyle\leq$
$\displaystyle\left(\log\frac{R}{r}\right)^{-1}\left(\log\frac{R^{\prime}}{r^{\prime}-r}\right)T(R,f).$

Therefore we immediately have (3.10). ∎

The following is extracted from the proof of Theorem 1.1 in [45], but the
condition (1.4) in that paper is replaced by the more general condition (1.4)
in this paper.

######  Lemma 6.

Let $f$ be a meromorphic function with (1.4). Let $U$ be a Fatou component of
$f$. Then there exist constants $D>1$, $d>1$ and $R_{0}>0$ such that if for
some $m>0$ and $r\geq R_{0}$,

(3.6) $A(D^{-d}r,D^{d}r)\subseteq f^{m}(U),$

then for all sufficiently large $n$, $f^{n}(U)\supset A_{n}$ and
$A_{n+1}\subset f(A_{n})$ with $A_{n}=A(r_{n},R_{n})$, $R_{n}\geq
r_{n}^{\sigma}$, $r_{n+1}>R_{n}$, $r_{n}\to\infty\ (n\to\infty)$ and
$\sigma>1$.

We remark that if the conclusion of Lemma 6 holds, that is to say, for large
$n$, $f^{n}(U)$ contains large annulus stated in Lemma 6, then the condition
(1.4) can be replaced by the following inequality

(3.7) $\log M(Cr,f)\geq\left(1-\frac{\log\log r}{\log r}\right)T(r,f),\
\forall\ r>r_{0}.$

We will complete the proof of Theorem 2 under the condition (3.7) instead of
(1.4) after we obtain large annulus sequence stated in Lemma 6. Now we are in
the position to prove Theorem 2.

Under the conditions of Theorem 2, in view of Lemma 6, for all sufficiently
large $n$, we have

(3.8) $A(r_{n},R_{n})\subset f^{n}(U)$

with $R_{n}\geq r_{n}^{\sigma}$, $r_{n+1}>R_{n}$, $r_{n}\to\infty\
(n\to\infty)$ and $\sigma>1$. Therefore, for a sufficiently large $m$ and
$n\geq m$ we have $r_{n+1}>R_{n}\geq R_{m}^{\sigma^{n-m}}$. Without loss of
generality for a sufficiently large $m$, we rewrite $r_{m},\ R_{m}$ and
$\sigma>1$ such that

$f^{m}(U)\supset A(r^{\alpha}_{m},R^{\beta}_{m})\supset A(r_{m},R_{m})$

with $R_{m}=r^{\sigma}_{m}$, $(\sigma-1)\log r_{m}>2m^{2}$, $0<\alpha<1$ and
$\beta>1$ (For example in (3.8) for $n=m$ we replace $r_{m}$ by
$r_{m}^{1/\alpha}$ and $R_{m}$ by $R_{m}^{1/\beta}$ with
$\beta=\alpha\sigma$). Take $R^{\prime}_{m}=e^{-m^{2}}R_{m}$ and
$r^{\prime}_{m}=er_{m}$. Applying Lemma 3 to $f$ on $A(r_{m},R_{m})$, we have

(3.9) $\hat{m}(R^{\prime}_{m},f)\geq M(R^{\prime}_{m},f)^{s_{m}},\
s_{m}=\exp\left(-\frac{\pi^{2}}{2}\max\left\\{\frac{1}{\log\frac{R_{m}}{R^{\prime}_{m}}},\frac{1}{\log\frac{R^{\prime}_{m}}{r_{m}}}\right\\}\right).$

Let us estimate $s_{m}$. Since

$\log\frac{R_{m}}{R^{\prime}_{m}}=m^{2}\ \text{and}\
\log\frac{R^{\prime}_{m}}{r_{m}}=(\sigma-1)\log r_{m}-m^{2}>m^{2},$

we have

$s_{m}=\exp\left(-\frac{\pi^{2}}{2m^{2}}\right).$

Take $\tau$ with $1<\tau<\sigma$ such that $s_{m}\sigma>\tau$ and take two
points $a$ and $b$ with $|a|=r^{\prime}_{m}$ and $|b|=R^{\prime}_{m}$. Set
$\hat{R}_{m}=r^{\prime}_{m}e^{m^{3}}$. Finding $\sigma_{m}$ such that
$R^{\prime}_{m}/C={\hat{R}_{m}}^{\sigma_{m}},$ we have

$\sigma_{m}=\frac{\log R^{\prime}_{m}/C}{\log\hat{R}_{m}}=\sigma-\frac{\sigma
m^{3}+m^{2}+\sigma+\log C}{\log r_{m}+m^{3}+1}.$

Therefore, we have $|f(a)|\leq M(r^{\prime}_{m},f)$ and in view of (3.7) and
(3.9), we have

$|f(b)|\geq
M(R^{\prime}_{m},f)^{s_{m}}\geq\exp(s_{m}\tau_{m}T(R^{\prime}_{m}/C,f))$
(3.10)
$=\exp(s_{m}\tau_{m}T({\hat{R}_{m}}^{\sigma_{m}},f))\geq\exp(s_{m}\sigma_{m}\tau_{m}T(\hat{R}_{m},f)),$

where $\tau_{m}=1-\frac{\log\log(R^{\prime}_{m}/C)}{\log(R^{\prime}_{m}/C)}$.
Since $r^{\prime}_{m}>r_{m}>r_{0}^{\sigma^{m}}$ with $r_{0}>e$, we have

$\sigma_{m}>\sigma-\frac{(C+\sigma)m^{3}}{\log
r_{m}}>\sigma-\frac{(C+\sigma)m^{3}}{\sigma^{m}\log r_{0}}$

and $s_{m}\sigma_{m}\tau_{m}\to\sigma>1(m\to\infty)$.

In view of Lemma 5 with $R^{\prime}=r_{m}^{\prime}(m^{2}+1)$,
$r^{\prime}=r^{\prime}_{m}$, $r=r_{m}$ and $R=\hat{R}_{m}$, we have

$\displaystyle\log M(r^{\prime}_{m},f)$ $\displaystyle\leq$
$\displaystyle\left(\frac{m^{2}+2}{m^{2}}+\frac{\log\frac{r^{\prime}_{m}(m^{2}+1)}{r^{\prime}_{m}-r_{m}}}{\log\frac{\hat{R}_{m}}{r_{m}}}\right)T(\hat{R}_{m},f)$
$\displaystyle=$
$\displaystyle\left(\frac{m^{2}+2}{m^{2}}+\frac{\log\frac{e}{e-1}(m^{2}+1)}{m^{3}+1}\right)T(\hat{R}_{m},f)$
$\displaystyle\leq$
$\displaystyle\left(1+\frac{t}{m^{2}}\right)T(\hat{R}_{m},f),$

where $t$ is an absolute constant. Set
$t_{m}=\left(1+\frac{t}{m^{2}}\right)^{-1}$. Thus combining (3.10) together
with the above inequality yields

(3.11) $|f(b)|\geq\exp(s_{m}\sigma_{m}\tau_{m}T(\hat{R}_{m},f))\geq
M(r^{\prime}_{m},f)^{t_{m}s_{m}\sigma_{m}\tau_{m}}\geq|f(a)|^{t_{m}s_{m}\sigma_{m}\tau_{m}}.$

Set $r^{\prime}_{m+1}=|f(a)|$ and $R^{\prime}_{m+1}=|f(b)|$. Then we have
$R^{\prime}_{m+1}\geq(r^{\prime}_{m+1})^{t_{m}s_{m}\sigma_{m}\tau_{m}}$. Set
$R_{m+1}=e^{(m+1)^{2}}R^{\prime}_{m+1}$ and $r_{m+1}=r^{\prime}_{m+1}/e.$

Now let us estimate $d_{U_{m}}(a,b)$ with $U_{m}=f^{m}(U)$. Set
$A^{\prime}_{m}=A(r_{m}^{\alpha},R_{m}^{\beta})$. Since

$\sin\left(\pi\frac{\log\frac{R_{m}^{\beta}}{r^{\prime}_{m}}}{{\rm
mod}(A^{\prime}_{m})}\right)=\sin\left(\pi\frac{(\beta-1/\sigma)\log
R_{m}-1}{(\beta-\alpha/\sigma)\log R_{m}}\right)$
$\to\sin\left(\pi\frac{\sigma\beta-1}{\sigma\beta-\alpha}\right)(m\to\infty)$

and

$\sin\left(\pi\frac{\log\frac{R_{m}^{\beta}}{R^{\prime}_{m}}}{{\rm
mod}(A^{\prime}_{m})}\right)=\sin\left(\pi\frac{(\beta-1)\log
R_{m}+m^{2}}{(\beta-\alpha/\sigma)\log R_{m}}\right)$
$\to\sin\left(\pi\frac{\sigma(\beta-1)}{\sigma\beta-\alpha}\right)(m\to\infty),$

in view of (2.1) we have

$\lambda_{A^{\prime}_{m}}(z)\leq\frac{K\pi}{{\rm
mod}(A^{\prime}_{m})}\frac{1}{|z|},\ r^{\prime}_{m}\leq|z|\leq
R^{\prime}_{m},$

where $K$ is a constant with
$1/K<\min\left\\{\sin\left(\pi\frac{\sigma\beta-1}{\sigma\beta-\alpha}\right),\sin\left(\pi\frac{\sigma(\beta-1)}{\sigma\beta-\alpha}\right)\right\\}$
so that

$\displaystyle\int_{r^{\prime}_{m}}^{R^{\prime}_{m}}\lambda_{A^{\prime}_{m}}(z)|{\rm
d}z|$ $\displaystyle\leq$ $\displaystyle\frac{K\pi}{{\rm
mod}(A^{\prime}_{m})}\int_{r^{\prime}_{m}}^{R^{\prime}_{m}}\frac{{\rm d}t}{t}$
$\displaystyle=$ $\displaystyle\frac{K\pi}{{\rm
mod}(A^{\prime}_{m})}\log\frac{R^{\prime}_{m}}{r^{\prime}_{m}}$
$\displaystyle=$ $\displaystyle\frac{K\pi}{(\beta-\alpha/\sigma)\log
R_{m}}((1-1/\sigma)\log R_{m}-m^{2}-1)$ $\displaystyle\to$
$\displaystyle\frac{K\pi(\sigma-1)}{\sigma\beta-\alpha}(m\to\infty)$

and

$\int_{0}^{\pi}\lambda_{A^{\prime}_{m}}(R^{\prime}_{m}e^{i\theta})R^{\prime}_{m}{\rm
d}\theta\leq\frac{K\pi^{2}}{{\rm mod}(A^{\prime}_{m})}\to 0(m\to\infty).$

Therefore, for sufficiently large $m$ and $\sigma>1$ close to $1$, we have

$\delta=d_{U_{m}}(a,b)\leq d_{A^{\prime}_{m}}(a,b)<\frac{1}{6}.$

In view of the above inequality, (3.11) and (3.10), we have

$\displaystyle e^{-\kappa}\left(\frac{|f(b)|}{|f(a)|}\right)^{1/\delta-1}$
$\displaystyle\geq$ $\displaystyle
e^{-\kappa}\left(|f(b)|^{1-1/(t_{m}s_{m}\sigma_{m}\tau_{m})}\right)^{5}$
$\displaystyle\geq$
$\displaystyle\exp\left[5\left(1-\frac{1}{t_{m}s_{m}\sigma_{m}\tau_{m}}\right)s_{m}\sigma_{m}\tau_{m}T(\hat{R}_{m},f)-\kappa\right]$
$\displaystyle>$
$\displaystyle\exp[3(\sigma-1)T(\hat{R}_{m},f)]>e^{(m+1)^{2}}.$

Thus, for $|f(a)|\leq|f(\hat{z})|\leq|f(b)|$ with $\hat{z}\in U_{m}$ we have

$e^{\kappa}\left(\frac{|f(a)|}{|f(b)|}\right)^{1/\delta}|f(\hat{z})|\leq
e^{\kappa}\left(\frac{|f(a)|}{|f(b)|}\right)^{1/\delta-1}r^{\prime}_{m+1}<r^{\prime}_{m+1}/e=r_{m+1};$

and

$e^{-\kappa}\left(\frac{|f(b)|}{|f(a)|}\right)^{1/\delta}|f(\hat{z})|\geq
e^{-\kappa}\left(\frac{|f(b)|}{|f(a)|}\right)^{1/\delta-1}R^{\prime}_{m+1}>e^{(m+1)^{2}}R^{\prime}_{m+1}=R_{m+1}.$

Since $d_{U_{m}}(a,b)<\frac{1}{6}$, in view of Lemma 1 we have

$U_{m+1}=f(U_{m})\supset
A(r_{m+1},R_{m+1})\supset\overline{A}(r^{\prime}_{m+1},R^{\prime}_{m+1}).$

Thus by noting that $d_{U_{m+1}}(f(a),f(b))\leq d_{U_{m}}(a,b)<\frac{1}{6}$
($\beta$ and $\alpha$ are just used to imply the inequality),
$|f(a)|=r^{\prime}_{m+1}$ and $|f(b)|=R^{\prime}_{m+1}$ , we can repeat the
above step and obtain that

(3.12)
$|f^{n}(b)|\geq|f^{n}(a)|^{\prod_{k=0}^{n-1}s_{k+m}t_{k+m}\sigma_{k+m}\tau_{k+m}}.$

For sufficiently large $m$, we can require, for $n\geq 1$,

$\prod_{k=0}^{n-1}s_{k+m}t_{k+m}\sigma_{k+m}\tau_{k+m}>\frac{\sigma+1}{2}>1.$

Take a point $c\in U$. Define

$h_{n}(z)=\frac{\log|f^{n}(z)|}{\log|f^{n}(c)|},\ \forall\ z\in U.$

Since $U$ is wandering and escaping, we can assume that $|f^{n}(z)|>1$ on $U$
and with (1.1) we have that $h_{n}(z)$ is harmonic and positive on $U$. In
view of (1.1) or by Harnack’s inequality, the family $\\{h_{n}\\}$ of harmonic
functions is locally normal on $U$ and hence

$\overline{h}(z):=\limsup_{n\to\infty}h_{n}(z),\ \forall\ z\in U$

exists and $\overline{h}$ is harmonic and positive on $U$. In view of (3.12),
we have

$h_{n}(b)\geq\frac{\sigma+1}{2}h_{n}(a),\
\overline{h}(b)\geq\frac{\sigma+1}{2}\overline{h}(a)>\overline{h}(a).$

Therefore, $\overline{h}(z)$ is not a constant on $U$. The same argument
implies that

$\underline{h}(z):=\liminf_{n\to\infty}h_{n}(z),\ \forall\ z\in U$

exists and $\underline{h}$ is a non-constant, harmonic and positive function
on $U$.

Suppose that $\overline{h}(z_{0})>\underline{h}(z_{0})$ for some $z_{0}\in U$.
Without any loss of generalities suppose that
$\overline{h}(z_{0})>\overline{h}(c)=1$. Take a real number $\eta$ with
$\max\\{1,\underline{h}(z_{0})\\}<\eta<\overline{h}(z_{0}).$ Since
$\overline{h}(z)$ is not a constant, we can find $z_{1}$ and $z_{2}$ such that
$\overline{h}(z_{1})<1=\overline{h}(c)<\overline{h}(z_{0})<\overline{h}(z_{2})$.
In view of Lemma 1, for some sufficiently large $m$, we have

$f^{m}(U)\supset A\left(|f^{m}(c)|^{\alpha},|f^{m}(c)|^{\beta}\right),$

where $\overline{h}(z_{1})<\alpha<1$ and
$1<\overline{h}(z_{0})<\beta<\overline{h}(z_{2}).$

From the argument of above paragraph we can take $a=f^{m}(c)$ and
$b=f^{m}(z_{0})$ such that

$|f^{n+m}(z_{0})|>|f^{n+m}(c)|^{\eta},\ \forall\ n\in\mathbb{N}.$

This implies that $\underline{h}(z_{0})\geq\eta$. A contradiction is derived.
We have proved that $\overline{h}(z)=\underline{h}(z)$ on $U$, that is to say,

$h(z)=\lim_{n\to\infty}\frac{\log|f^{n}(z)|}{\log|f^{n}(c)|}$

exists on $U$.

Set

$H(z):=\lim_{n\to\infty}\frac{\log|f^{n}(z)|}{\log|f^{n}(z_{0})}=\frac{h(z)}{h(z_{0})}.$

$H$ is a non-constant, harmonic and positive function on $U$. Thus we can find
two points $z_{1}$ and $z_{2}$ in $V$ and $0<\alpha<1$ such that
$H(z_{1})<1-\alpha<1<1+\alpha<H(z_{2})$. For a sufficiently large $m$ we have
$d_{f^{m}(V)}(f^{m}(z_{1}),f^{m}(z_{2}))<\frac{1}{6}$ and hence, in view of
Lemma 1, for $n\geq m$

$f^{n}(V)\supset A(|f^{n}(z_{1})|,|f^{n}(z_{2})|)\supset
A(|f^{n}(z_{0})|^{1-\alpha},|f^{n}(z_{0})|^{1+\alpha}).$

Thus we have (3.6) for sufficiently large $m$ and $f^{m}(V)$ instead. In view
of Lemma 6, for all sufficiently large $n$ and some $\tau>1$, we obtain
$A_{n}=A(r_{n},R_{n})\subseteq f^{n}(V)$ with $R_{n}\geq
r_{n}^{\tau}\to\infty(n\to\infty)$ such that $A_{n+1}\subset f(A_{n})$.

We complete the proof of Theorem 2. ∎

Proof of Theorem 3. In view of Theorem 2, there exist a sufficiently large $m$
and a sufficiently large $r_{m}$ such that for $n\geq m$ and
$r_{n+1}>R_{n}\geq r_{n}^{\sigma}$ with $\sigma>1$ we have

$f^{n}(U)\supset A(r_{n}^{\alpha},R_{n}^{\beta}),\ 0<\alpha<1,\ 1<\beta.$

From (1.9) and (1.4), in view of Lemma 3, Lemma 4 and Lemma 5 in turn, we can
show

$M(R_{m},g)\geq\hat{m}(R_{m},g)>M(r_{m},g)\geq\hat{m}(r_{m},g)$

and in view of the maximal principle, we have

(3.13) $g(A(r_{m},R_{m}))\subseteq A(\hat{m}(r_{m},g),M(R_{m},g)).$

Set $u(z)=f(z)-g(z)$ and $s_{m}=\exp\left(-\frac{\pi^{2}}{2m^{2}}\right)$. For
sufficiently large $m$, we have $s_{m}>\max\\{1/2,\delta\\}$. In view of Lemma
3 for $r=e^{-m^{2}}r_{m}$, $\rho=r_{m}$ and $R=R_{m}$ and by noting that
$\log(1-x)>-2x$ for $0<x<\frac{1}{2}$, we have

$\hat{m}(r_{m},g)\geq\hat{m}(r_{m},f)-M(r_{m},u)\geq
M(r_{m},f)^{s_{m}}-M(r_{m},u)$
$=\left(1-\frac{M(r_{m},u)}{M(r_{m},f)^{s_{m}}}\right)M(r_{m},f)^{s_{m}}\geq
M(r_{m},f)^{s_{m}\gamma_{m}},$

where $\gamma_{m}=1-\frac{4M(r_{m},f)^{\delta-s_{m}}}{\log
M(r_{m},f)}>1-\frac{1}{m^{2}}$ and by noting that $\log(1+x)<x$ for $x>0$, we
have

$M(R_{m},g)\leq M(R_{m},f)+M(R_{m},u)$
$=M(R_{m},f)\left(1+\frac{M(R_{m},u)}{M(R_{m},f)}\right)\leq
M(R_{m},f)^{\tau_{m}},$

where $\tau_{m}=1+\frac{1}{M(R_{m},f)^{1-\delta}\log
M(R_{m},f)}<1+\frac{1}{m^{2}}$ for sufficiently large $m$. Thus it follows
from (3.13) that

(3.14) $g(A(r_{m},R_{m}))\subseteq
A(M(r_{m},f)^{s_{m}\gamma_{m}},M(R_{m},f)^{\tau_{m}}).$

On the other hand, we have

(3.15) $f^{m+1}(U)\supset f(A(r_{m}^{\alpha},R_{m}^{\beta}))\supset
A(M(r_{m}^{\alpha},f),\hat{m}(R_{m}^{\beta}e^{-m^{2}},f)).$

In view of Lemma 3 and (1.4), we have

$\hat{m}(R_{m}^{\beta}e^{-m^{2}},f)\geq
M(R_{m}^{\beta}e^{-m^{2}},f)^{s_{m}}\geq\exp(\sigma_{m}t_{m}T(R_{m}^{\beta}e^{-m^{2}}/C,f)),$

for
$t_{m}=1-\frac{\log\log(R_{m}^{\beta}e^{-m^{2}}/C)}{\log(R_{m}^{\beta}e^{-m^{2}}/C)}>1-\frac{1}{m^{2}}$
for sufficiently large $m$, and in view of Lemma 5, we can show that

$\tau_{m}\log M(R_{m},f)<s_{m}t_{m}T(R_{m}^{\beta}e^{-m^{2}}/C,f)$

and in view of (1.4) and Lemma 4, we have

$s_{m}\gamma_{m}\log M(r_{m},f)>\log M(r_{m}^{\alpha},f).$

Thus $M(R_{m}^{\beta}e^{-m^{2}},f)^{s_{m}}>M(r_{m}^{\alpha},f)$ and from
(3.15) we can deduce

(3.16) $f^{m+1}(U)\supset
A(M(r_{m}^{\alpha},f),M(R_{m}^{\beta}e^{-m^{2}},f)^{s_{m}}).$

Now we set $\hat{r}_{1}=M(r_{m}^{\alpha},f)^{s_{m}\gamma_{m}/\alpha}$ and
$\hat{R}_{1}=M(R_{m},f)^{\tau_{m}}$ so that we have
$M(r_{m},f)^{s_{m}\gamma_{m}}\geq\hat{r}_{1}$. Combining (3.14) and (3.16)
yields

$f^{m+1}(U)\supset
A(\hat{r}_{1}^{\alpha\eta_{m}},\hat{R}_{1}^{\beta_{m}})\supset
A(\hat{r}_{1},\hat{R}_{1})\supset g(A(r_{m},R_{m}))$

with $\eta_{m}=\frac{1}{s_{m}\gamma_{m}}>1$ and
$\beta_{m}=\frac{s_{m}}{\tau_{m}}\frac{\log M(R_{m}^{\beta}e^{-m^{2}},f)}{\log
M(R_{m},f)}\to\beta\ (m\to\infty)$. For sufficiently large $m$, we can require
that $0<\alpha\eta_{m}<\frac{\alpha+1}{2}<1$ and
$1<\frac{\beta+1}{2}<\beta_{m}$.

By the same step as above, we have

$f^{m+2}(U)\supset
A(\hat{r}_{2}^{\alpha\eta_{m}\eta_{m+1}},\hat{R}_{2}^{\beta_{m+1}})\supset
A(\hat{r}_{2},\hat{R}_{2})\supset g^{2}(A(r_{m},R_{m}))$

with
$\hat{r}_{2}=M(\hat{r}_{1}^{\alpha\eta_{m}},f)^{1/(\eta_{m+1}\eta_{m}\alpha)}$,
$\hat{R}_{2}=M(\hat{R}_{1},f)^{\tau_{m+1}}$,
$0<\alpha\eta_{m}\eta_{m+1}<\frac{\alpha+1}{2}<1$ and
$1<\frac{\beta+1}{2}<\beta_{m+1}$.

Inductively, we have

$f^{m+n}(U)\supset g^{n}(A(r_{m},R_{m})).$

Thus $A(r_{m},R_{m})$ is in an escaping Fatou component $W$ of $g$.

Let us notice that in $A(r_{m},R_{m})$ we have

$\displaystyle\log M(Cr,g)$ $\displaystyle\geq$
$\displaystyle\log(M(Cr,f)-M(Cr,u))$ $\displaystyle=$ $\displaystyle\log
M(Cr,f)+\log\left(1-\frac{M(Cr,u)}{M(Cr,f)}\right)$ $\displaystyle\geq$
$\displaystyle\left(1-\frac{2\log\log r}{\log r}\right)T(r,f)$

and for $r_{m}<r+1<r^{\prime}<R^{\prime}<R\leq R_{m}$ we have

$\displaystyle\log M(r^{\prime},g)$ $\displaystyle\leq$ $\displaystyle\log
M(r^{\prime},f)+\log\left(1+\frac{M(r^{\prime}u)}{M(r^{\prime},f)}\right)$
$\displaystyle\leq$
$\displaystyle\left(\frac{R^{\prime}+r^{\prime}}{R^{\prime}-r^{\prime}}+\left(\log\frac{R}{r}\right)^{-1}\log\frac{R^{\prime}}{r^{\prime}-r}\right)T(R,f)+1.$

Thus we can repeat the steps in the proof of Theorem 2 and show the results of
Theorem 2 hold for $g$ in $W$. ∎

We make a remark on (1.9). From the proof of Theorem 3 we see that in order
that Theorem 3 holds for $g$, in fact, we only need to require (1.9) holds in
a sequence of annuli $A(r_{n},r_{n}^{\sigma_{n}})$ instead of $f^{n}(U)$ as
long as $A(r_{n},r_{n}^{\sigma_{n}})\subset f^{n}(U)$ for $\forall\ n\geq m$
with $m$ being large enough and $f(A(r_{n},r_{n}^{\sigma_{n}}))\subset
A(r_{n+1},r_{n+1}^{\sigma_{n+1}})$ with $1<\sigma\leq\sigma_{n}$.

Proof of Theorem 4. With the help of Theorem 3 we are going to find a
meromorphic function $f$ with $\delta(\infty,f)=0$ which has an escaping Fatou
wandering domain $U$ such that the results of Theorem 2 hold for $f$ in $U$.

In [13], Bergweiler, Rippon and Stallard proved the results of Theorem 2 for
entire functions in their multiply connected Fatou components. The first
entire function with multiply connected Fatou component is due to I. N. Baker
[2]. The multiply connected wandering domains which have uniformly perfect
boundary or non-uniformly perfect boundary were found in [14]. For example, in
Theorem 1.3 of [14], an entire function of zero order is constructed such that

$f(A((1+\varepsilon_{k})r_{k},(1-\varepsilon_{k})r_{k+1}))\subset
A((1+\varepsilon_{k+1})r_{k+1},(1-\varepsilon_{k+1})r_{k+2})$

for a sequence $\\{r_{k}\\}$ of positive numbers with $r_{k+1}>2r_{k}^{2}$ and
a sequence $\\{\varepsilon_{k}\\}$ of positive numbers tending to $0$ as
$k\to\infty$. And the Fatou component $U$ containing
$A((1+\varepsilon_{k})r_{k},(1-\varepsilon_{k})r_{k+1})$ may have uniformly
perfect boundaries by suitably choosing $\\{r_{k}\\}$.

Now we put

$g(z)=f(z)+\sum_{n=1}^{\infty}\frac{1}{2^{n}m_{n}}\sum_{k=1}^{m_{n}}\frac{\varepsilon_{n}}{z-z_{k,n}},$

where $m_{n}=[r_{n+1}]+1$ and $[r_{n+1}]$ is the maximal integer not exceeding
$r_{n+1}$, $z_{k,n}=r_{n}e^{i\frac{2k\pi}{m_{n}}}$. It is clear that $g$ is a
meromorphic function. For $|z|=r$ with
$(1+\varepsilon_{k})r_{k}<r<(1-\varepsilon_{k})r_{k+1}$, we have

$\displaystyle|f(z)-g(z)|$ $\displaystyle\leq$
$\displaystyle\sum_{n=1}^{\infty}\frac{1}{2^{n}m_{n}}\sum_{j=1}^{m_{n}}\frac{\varepsilon_{n}}{|z-z_{j,n}|}$
$\displaystyle\leq$ $\displaystyle\sum_{\begin{array}[]{c}n=1\\\
n\not=k,k+1\end{array}}^{\infty}\frac{1}{2^{n}}\frac{\varepsilon_{n}}{|r-r_{n}|}$
$\displaystyle+\frac{1}{2^{k}}\frac{\varepsilon_{k}}{r-r_{k}}+\frac{1}{2^{k+1}}\frac{\varepsilon_{k+1}}{r_{k+1}-r}$
$\displaystyle<$ $\displaystyle\sum_{n=1}^{\infty}\frac{1}{2^{n}}=1,$

and this implies (1.9) for $|z|=r$ with
$(1+\varepsilon_{k})r_{k}<r<(1-\varepsilon_{k})r_{k+1}$. In view of Theorem 3
and from the remark after the proof of Theorem 3, there exists an escaping
wandering domain $W$ of $g$ such that the results of Theorem 2 hold for $g$
and $W$.

Next we calculate the deficiency $\delta(\infty,g)$ of Nevanlinna of $g$. For
$er_{k}<r<(1-\varepsilon_{k})r_{k+1}$, we have $m(r,g)\leq m(r,f)+\log 2$ and

$N(r,g)=\int_{0}^{r}\frac{n(t,g)}{t}{\rm
d}t\geq\int_{r_{k}}^{r}\frac{m_{k}}{t}{\rm d}t\geq r_{k+1}\geq r.$

Then as
$r\in\cup_{k=m}^{\infty}A(er_{k},(1-\varepsilon_{k})r_{k+1})\to\infty$, we
have

$\frac{m(r,g)}{T(r,g)}\leq\frac{m(r,g)}{N(r,g)}\leq\frac{m(r,f)+\log 2}{r}\to
0,$

by noting that $f$ is of zero order. This implies that

$\delta(\infty,g)=\liminf_{r\to\infty}\frac{m(r,g)}{T(r,g)}=0.$

Then $g$ is the desired meromorphic function of Theorem 4. ∎

But we do not know if $W$ has uniformly perfect boundary. Therefore, we beg
the following

Question $\mathcal{B}$: Are there any meromorphic function with zero
Nevanlinna deficiency at poles which has a multiply connected escaping
wandering domain with uniformly perfect boundary?

Perhaps the following approach would be possible to solve Question
$\mathcal{B}$. We try to control the changes of critical values as an entire
function that has a multiply connected escaping wandering domain with
uniformly perfect boundary is small changed to a meromorphic function with
zero Nevanlinna deficiency at poles and then in view of Theorem 3 we show the
meromorphic function also has a multiply connected escaping wandering domain
with uniformly perfect boundary.

##  4\. proofs of Theorems 5-7

Proof of Theorem 5. Put

$K=\prod_{k=1}^{\infty}\left(1+\frac{4}{(k+1)^{2}}\right)^{2},\
K_{n}=\frac{K}{\prod_{k=1}^{n}\left(1+\frac{4}{(k+1)^{2}}\right)}>\sqrt{K}>1,\
\forall\ n\geq 1.$

Since $f$ is transcendental, we have

$\frac{T(r,f)}{\log r}\to\infty(r\to\infty)$

and so we can take a positive number $R$ such that $\forall\ r\geq
R>\max\\{1,r_{0}\\}$, where $r_{0}$ is the one appeared in Lemma 4,
$\hat{T}(r,f)=\exp T(r,f)\geq 9r^{2}$ and

(4.1) $\log M_{n}(R,f)>2^{n}\log R>\frac{2(n+1)^{2}\log(n+2)}{K_{n}},\forall\
n\geq 2.$

We claim that $\forall\ n\geq 2$, we have

(4.2) $\hat{T}_{n}(R^{K},f)\geq(n+1)^{2}M_{n-1}(R,f)^{K_{n-1}}.$

We show the Claim by induction. It is well-known from Poisson-Jensen formula
that for arbitrary real numbers $R$ and $r$ with $R>r>0$, we have

$T(r,f)\leq\log M(r,f)\leq\frac{R+r}{R-r}T(R,f).$

Note that

$\frac{\hat{T}(R^{K},f)-R^{K}}{\hat{T}(R^{K},f)+R^{K}}>1-\frac{2R^{K}}{\hat{T}(R^{K},f)}>1-\frac{1}{2^{2}}$

and

$\left(1+\frac{4}{2^{2}}\right)\left(1-\frac{1}{2^{2}}\right)>1+\frac{1}{2^{2}},\
M(R,f)^{K_{1}/2^{2}}>3^{2}.$

For $n=2$, we have

$\displaystyle\hat{T}_{2}(R^{K},f)$ $\displaystyle\geq$ $\displaystyle
M(R^{K},f)^{\frac{\hat{T}(R^{K},f)-R^{K}}{\hat{T}(R^{K},f)+R^{K}}}$
$\displaystyle\geq$ $\displaystyle M(R^{K},f)^{1-\frac{1}{2^{2}}}$
$\displaystyle\geq$ $\displaystyle M(R,f)^{K_{1}(1+1/2^{2})}$
$\displaystyle\geq$ $\displaystyle 3^{2}M(R,f)^{K_{1}}.$

We assume inductively that (4.2) holds for $n$. Consider the case of $n+1$ for
(4.2). Note that

$\displaystyle\frac{\hat{T}_{n}(R^{K},f)-M_{n-1}(R,f)^{K_{n-1}}}{\hat{T}_{n}(R^{K},f)+M_{n-1}(R,f)^{K_{n-1}}}$
$\displaystyle\geq$
$\displaystyle\frac{(n+1)^{2}-1}{(n+1)^{2}+1}=1-\frac{2}{(n+1)^{2}+1}$

and

$K_{n-1}\left(1-\frac{2}{(n+1)^{2}+1}\right)=K_{n}\left(1+\frac{4}{(n+1)^{2}}\right)\left(1-\frac{2}{(n+1)^{2}+1}\right)$
$>K_{n}+\frac{K_{n}}{(n+1)^{2}}.$

Then in view of (4.2) and (4.1) we have

$\displaystyle\hat{T}_{n+1}(R^{K},f)$ $\displaystyle\geq$ $\displaystyle
M(M_{n-1}(R,f)^{K_{n-1}},f)^{\frac{\hat{T}_{n}(R^{K},f)-M_{n-1}(R,f)^{K_{n-1}}}{\hat{T}_{n}(R^{K},f)+M_{n-1}(R,f)^{K_{n-1}}}}$
$\displaystyle>$ $\displaystyle
M_{n}(R,f)^{K_{n-1}\left(1-\frac{2}{(n+1)^{2}+1}\right)}$ $\displaystyle>$
$\displaystyle(n+2)^{2}M_{n}(R,f)^{K_{n}}.$

We have shown the Claim (4.2).

For any point $z\in M(f)$ in whose definition we choose $R^{K}$ instead of
$R$, we have for some $L\in\mathbb{N}$,

$|f^{n+1+L}(z)|\geq\hat{T}_{n+1}(R^{K},f),\ \forall\ n\in\mathbb{N}.$

In view of (4.2), we deduce

$|f^{n+1+L}(z)|\geq M_{n}(R,f),\ \forall\ n\in\mathbb{N},$

and this implies $z\in A(f)$. It follows that $M(f)\subseteq A(f)$. On the
other hand, it is clear that $M_{n}(R,f)\geq\hat{T}_{n}(R,f)$ and thus for
$z\in A(f)$, it follows from $|f^{n+L}(z)|\geq M_{n}(R,f)$ for some
$L\in\mathbb{N}$ that $|f^{n+L}(z)|\geq\hat{T}_{n}(R,f)$ so that $z\in M(f)$.
We show that $A(f)\subseteq M(f)$.

So far, we have shown that $M(f)=A(f)$.∎

Let us make remarks. Let $f$ be a transcendental entire function. It was shown
in [32] that

$A(f)=\\{z\in\mathbb{C}:\ {\rm there\ exists}\ L\in\mathbb{N}\ {\rm such\
that}\ |f^{n+L}(z)|\geq M(R,f^{n}),\ {\rm for}\ n\in\mathbb{N}\\},$

which is the definition of $A(f)$ introduced in [12]. Since for large $R$, we
have

$T(2R^{3},f^{n})\geq\frac{1}{3}\log M(R^{3},f^{n})\geq\log M(R,f^{n})\geq
T(R,f^{n}),$

we have

$M(f)=\\{z\in\mathbb{C}:\ {\rm there\ exists}\ L\in\mathbb{N}\ {\rm such\
that}\ |f^{n+L}(z)|\geq\hat{T}(R,f^{n}),\ {\rm for}\ n\in\mathbb{N}\\}.$

It is clear that $T(R,f^{n})$ is not available for a meromorphic function $f$
with poles.

Furthermore, it was shown in [32] that the fast escaping set $A(f)$ is
invariant even if $M(r,f)$ is replaced by $\mu(r)=\varepsilon M(r,f)$ for a
fixed $0<\varepsilon<1$ and in [36] by $\mu(r)=\varepsilon(r)M(r,f)$ with
somewhat more involved condition, $0<\varepsilon(r)<1$ and
$\varepsilon(r)r\to\infty(r\to\infty)$. Actually, we can use
$\mu(r)=M(r,f)^{1-\frac{1}{\log_{k}r}}$ in the place of $M(r,f)$ in $A(f)$ to
ensure the invariance of $A(f)$, where $\log_{k}r$ is the $k$th iterate of
$\log r$ with respect to $r$. If $\frac{\log M(r,f)}{\log r\log_{k}r}>2$, then

$rM(r,f)^{-\frac{1}{\log_{k}r}}\to 0(r\to\infty).$

In [33], Rippon and Stallard investigated the following sets:

$Q_{\varepsilon}(f)=\\{z:\ \text{there\ exists\ }L\in\mathbb{N}\ \text{such\
that\ }|f^{n+L}(z)|\geq\mu_{\varepsilon}^{n}(R),\ \forall\ n\in\mathbb{N}\\}$

for $0<\varepsilon<1$ and $\mu_{\varepsilon}(r)=M(r,f)^{\varepsilon}$. And
they showed that $Q_{\varepsilon}(f)=A(f)$ if and only if $f$ is so-called
$\varepsilon$-regular and for $f\in\mathcal{B}$ (Eremenko-Lyubich class
$\mathcal{B}$ which consists of entire functions with bounded singular value
set), $A(f)=\bigcup_{\varepsilon>0}Q_{\varepsilon}(f)$.

Of course, we may also take into account the fast escaping set $M(f)$ of
meromorphic functions in the way mentioned above.

For the proof of Theorem 6 we need the following lemma; see Lemma 1 of [31].

######  Lemma 7.

Let $E_{n},n\geq 0,$ be a sequence of compact sets in $\mathbb{C}$ and
$f:\mathbb{C}\to\hat{\mathbb{C}}$ be a continuous function such that
$E_{n+1}\subset f(E_{n}),\ \text{for}\ n\geq 0.$ Then there exists $\zeta$
such that $f^{n}(\zeta)\in E_{n}$ for $n\geq 0$.

Proof of Theorem 6. For simplicity, we will here complete our proof for $C=1$
and $c=1$. The proof of general case needs just a little cumbersome
calculation.

Put $S=\sqrt[3]{D}>e$. In view of (1.4), we have

$\displaystyle\log M(Sr,f)$ $\displaystyle\geq$
$\displaystyle\left(1-\frac{1}{\log Sr}\right)T(Sr,f)$ $\displaystyle\geq$
$\displaystyle\left(1-\frac{1}{\log Sr}\right)\left(1+\frac{\log S}{\log
r}\right)T(r,f)$ $\displaystyle\geq$ $\displaystyle T(r,f)$

and

$\frac{1}{2\pi}\int_{0}^{2\pi}\log^{+}|f(Sre^{i\theta})|{\rm
d}\theta=m(Sr,f)\leq T(Sr,f).$

Then there exists a point $z_{2}$ with $|z_{2}|=Sr$ such that

$T(r,f)\leq\log|f(z_{2})|\leq T(Sr,f).$

It follows from (1.4) that there is a pint $z_{1}$ with $|z_{1}|=S^{2}r$ such
that

$\log|f(z_{1})|\geq\left(1-\frac{1}{\log S^{2}r}\right)T(S^{2}r,f).$

In view of (2.4) in Lemma 2 with $\tau=\frac{2}{3}$ and $\sigma=\frac{1}{3}$,
we have

$\frac{\pi}{3}\leq d_{A}(z_{1},z_{2})\leq\frac{\sqrt{3}\pi(\pi+1)}{3}<7.52$

for $A=A(r,Dr)$. Since, in view of (1.10), we have

$\left(1-\frac{1}{\log S^{2}r}\right)T(S^{2}r,f)\geq\left(1-\frac{1}{\log
S^{2}r}\right)\left(1+\frac{\log S}{\log Sr}\right)T(Sr,f)$
$=T(Sr,f)+\frac{\log(S/e)}{\log Sr}T(Sr,f)>T(Sr,f)+\kappa\delta+\delta\log D$

with $\delta=d_{A}(z_{1},z_{2})$, we have

$|f(z_{1})|>e^{\kappa\delta}D^{\delta}|f(z_{2})|,\ {\rm i.e.},\
e^{-\kappa}\left(\frac{|f(z_{1})|}{|f(z_{2})|}\right)^{1/\delta}>D.$

In view of Lemma 1, there exists a point $\hat{z}\in A(r,Dr)$ with
$|f(z_{2})|\leq|f(\hat{z})|\leq|f(z_{1})|$ such that

$f(A(r,Dr))\supset\overline{A}(D^{-1}|f(\hat{z})|,D|f(\hat{z})|).$

Now we take

$r_{1}=|f(\hat{z})|\geq|f(z_{2})|\geq e^{T(r,f)}=\hat{T}(r,f)$

and so

$f(A(r,Dr))\supset\overline{A}(r_{1},Dr_{1}).$

Inductively we get a sequence of positive numbers $\\{r_{n}\\}$ with
$r_{n}\geq\hat{T}_{n}(r,f)$ such that
$f(A(r_{n},Dr_{n}))\supset\overline{A}(r_{n+1},Dr_{n+1})$. In view of Lemma 7,
we can find a point $\zeta\in A(r,Dr)$ such that
$|f^{n}(\zeta)|\geq\hat{T}_{n}(r,f)$. That is to say, $\zeta\in M(f)$ and so
$\overline{U}\subset M(f).$ ∎

For the proof of Theorem 7, we need the following lemma, which is Main Lemma
of [18].

######  Lemma 8.

Let $\\{G_{k}\\}_{k=1}^{\infty}$ be a sequence of compact subsets in
$\mathbb{C}$ such that

(i) $\mathbb{C}\setminus G_{k}$ is connected for every $k$;

(ii) $G_{k}\cap G_{m}=\emptyset$ for $k\not=m$;

(iii) ${\rm dist}(0,G_{k})\to\infty(k\to\infty)$.

Then for any sequence of points $\\{z_{k}\\}$ with $z_{k}\in G_{k}$, any
sequence of positive numbers $\\{\varepsilon_{k}\\}$ and any sequence of
complex functions $\\{\phi_{k}\\}$ with $\phi_{k}$ being analytic in $G_{k}$
for every $k$, there exists an entire function $f$ satisfying, for every $k$,

$|f(z)-\phi_{k}(z)|<\varepsilon_{k},\ \forall\ z\in G_{k};$
$f(z_{k})=\phi_{k}(z_{k}),\ f^{\prime}(z_{k})=\phi^{\prime}_{k}(z_{k}).$

The same method as in the proof of Main Lemma of [18] can deduce that for a
sequence of compact sets $\\{G_{k}\\}_{k=1}^{\infty}$ only satisfying (ii) and
(iii), there exists a meromorphic function $f(z)$ such that the results stated
in Lemma 8 hold. This will apply to construction of desired meromorphic
function of Theorem 7.

Proof of Theorem 7. Set

$G_{n}=\overline{B}\left(a_{n},\frac{1}{4}\alpha_{n}\right),\
H_{n}=\overline{B}\left(-a_{n},\frac{1}{4}\alpha_{n}\right),$
$C_{n}=\left\\{z:\ a_{n}+\frac{3}{8}\alpha_{n}\leq{\rm Re}z\leq
a_{n}+\frac{5}{8}\alpha_{n},|{\rm Im}z|\leq\beta_{n}\right\\}$

with $\alpha_{n}=\min\\{a_{n}-a_{n-1},a_{n+1}-a_{n}\\}$, $1<a_{0}<a_{1}$ and
$0<\beta_{n}$ such that $\beta_{n}/(a_{n}-a_{n-1})\to\infty(n\to\infty)$, and

$\phi_{n}(z)=\frac{\alpha_{n+1}}{\alpha_{n}}(z-a_{n})+a_{n+1}:G_{n}\to
G_{n+1},$ $\varphi_{n}(z)=z+a_{n}+e^{a_{n}^{\rho}};$

where when $\rho=\infty$, $e^{a_{n}^{\rho}}$ is replaced by $e^{e^{(\log
a_{n})^{2}}}$. We see that $G_{n},H_{n}$ and $C_{n}$ don’t intersect each
other. In view of Lemma 8, there exists a transcendental entire function $f$
such that for every $n$,

$|f(z)-\phi_{n}(z)|<\varepsilon_{n},\ \forall\ z\in G_{n};$
$|f(z)-\varphi_{n}(z)|<\varepsilon_{n},\ \forall\ z\in H_{n};$
$f(a_{n})=\phi_{n}(a_{n})=a_{n+1},\
f^{\prime}(a_{n})=\frac{\alpha_{n+1}}{\alpha_{n}};$
$f(-a_{n})=\varphi_{n}(-a_{n})=e^{a_{n}^{\rho}},\ f^{\prime}(-a_{n})=1;$
$|f(z)|<\frac{1}{2},\ \forall\ z\in C_{n}\cup B(0,1),f(0)=0,\
f^{\prime}(0)=0.$

Then we have

$M(a_{n},f)\geq|f(-a_{n})|=e^{a_{n}^{\rho}},\ \
\liminf\limits_{n\to\infty}\frac{\log\log M(a_{n},f)}{\log a_{n}}\geq\rho.$

That is to say, the order of $f$ is at least $\rho$.

According to the construction of $f$ and the behavior of $f$ on $G_{n}$, by
the suitable choice of $\\{\varepsilon_{n}\\}$, we can show that
$B\left(a_{1},\frac{1}{5}\alpha_{1}\right)$ is in a Fatou component $U$ of $f$
and

$f^{n}\left(B\left(a_{1},\frac{1}{5}\alpha_{1}\right)\right)\subset U_{n}.$

Since $f^{n}(a_{1})=a_{n+1}\to\infty(n\to\infty)$, $U$ is escaping. And
$C_{m}\cap\cup_{n=0}^{\infty}U_{n}=\emptyset$ for every $m$.

We want to prove that $U$ is not a Baker domain. Suppose that $U$ is a Baker
domain with period $p$, i.e., $f^{p}(U)\subseteq U$. By the Principle of
Hyperbolic Metric, we have, for $n\geq 0$,

$d_{U}(a_{1},a_{p+1})\geq d_{f^{n}(U)}(f^{n}(a_{1}),f^{n}(a_{p+1}))\geq
d_{U_{n}}(a_{n+1},a_{n+p+1}).$

On the other hand, we want to estimate $d_{U_{n}}(a_{n+1},a_{n+p+1})$ from
below. Since Baker domains of an entire function are simply connected and
unbounded, so is $U_{n}$. By the Koebe $\frac{1}{4}$ Theorem, it follows that

$C_{U_{n}}=\min\\{\lambda_{U_{n}}(z)\delta_{U_{n}}(z):\ \forall\ z\in
U_{n}\\}\geq\frac{1}{4}$

where $\delta_{U_{n}}(z)$ is the Euclidean distance from $z$ to
$\partial{U_{n}}$; see [9]. Hence we have

$\displaystyle d_{U_{n}}(a_{n+1},a_{n+p+1})$ $\displaystyle=$
$\displaystyle\inf_{\gamma_{n}}\int_{\gamma_{n}}\lambda_{U_{n}}(z)|{\rm d}z|$
$\displaystyle\geq$
$\displaystyle\frac{1}{4}\inf_{\gamma_{n}}\int_{\gamma_{n}}\frac{1}{\delta_{U_{n}}(z)}|{\rm
d}z|$ $\displaystyle\geq$
$\displaystyle\frac{1}{4}\int_{0}^{\beta_{n+1}}\frac{1}{a_{n+1}-a_{n}}{\rm
d}t$ $\displaystyle=$
$\displaystyle\frac{\beta_{n+1}}{4(a_{n+1}-a_{n})}\to\infty(n\to\infty),$

where the infimum is taken over all curves $\gamma_{n}$ connecting $a_{n+1}$
and $a_{n+p+1}$ in $U_{n}$.

A contradiction is derived and hence we have shown that $U$ is an escaping
wandering domain of $f$. Take a component $U_{0}$ of $f^{-1}(U)$ and a point
$z_{0}\in U_{0}\cup f^{-1}(a_{1})$. Then $f^{n}(z_{0})=a_{n}$. ∎

We omit the specific construction of a meromorphic function which has
wandering domain $U$ such that $f^{n}(z)\sim a_{n}(n\to\infty)$ in $U$.

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