Iterates of Meromorphic Functions on Escaping Fatou Components
aa r X i v : . [ m a t h . C V ] F e b ITERATES OF MEROMORPHIC FUNCTIONS ON ESCAPINGFATOU COMPONENTS
ZHENG JIAN-HUA AND WU CHENG-FA
Abstract.
In this paper, we prove that the ratio of the modulus of the iteratesof two points in an escaping Fatou component may be bounded even if the orbitof the component contains an infinite modulus annulus sequence and this casecannot happen when the maximal modulus of the meromorphic function islarge enough. Therefore, we extend the related results for entire functions toones for meromorphic functions with infinitely many poles. And we investigatethe 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 thefast escaping set under a growth condition of the maximal modulus. Finallywe give examples of wandering domains escaping at arbitrary fast rate andslow rate. Introduction and Main Results
Let f be a meromorphic function which is not a M ¨ obius transformation and let f n , n ∈ N , denote the n th iterate of f . The Fatou set F ( f ) of f is defined as setof points, z ∈ ˆ C , such that { f n } n ∈ N is well-defined and forms a normal family insome neighborhood of z . Therefore, F ( f ) is an open set. Since F ( f ) is completelyinvariant under f , i.e., f ( z ) ∈ F ( f ) if and only if z ∈ 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 componentof F ( f ) such that f n ( U ) ⊆ U n . The complement J ( f ) = ˆ C \ F ( f ) of F ( f ) is calledJulia set of f . Then S ∞ n =0 f − n ( ∞ ) ⊂ J ( f ) and if S ∞ n =0 f − n ( ∞ ) contains threedistinct points in the extended complex plane ˆ C , then S ∞ n =0 f − n ( ∞ ) = J ( f ). Thusevery f n is analytic in F ( f ) for n ≥ U of f with f n | U → ∞ ( n →∞ ). For convenience of statements, we call such an U an escaping Fatou componentof f . Let us first determine what kinds of Fatou components will be escaping to ∞ under iterates.A periodic Fatou component W of f with period p ≥
1, i.e., f p ( W ) ⊆ W = W p and p is the minimal positive integer such that the inclusion holds, is known as aBaker domain if f np | W → a ( n → ∞ ), but f p is not defined at a . Then ∞ must bethe limit value of f np + j in W for some 0 ≤ j ≤ p − W j is unbounded. Ifan escaping Fatou component U of f is periodic, i.e., for some positive integer p , f p ( U ) ⊆ U = U p , then U is a Baker domain of f . The periodic Fatou componentsare classified into five possible types: Attracting domains, Parabolic domains, Siegeldisks, Herman rings and Baker domains; see [10], Theorem 6. Mathematics Subject Classification.
Key words and phrases.
Escaping Sets, Meromorphic Functions, Fatou set, Julia set.Supported by the grant (No. 12071239) of NSF of China.
A Fatou component U is called a wandering domain of f , if for any pair ofpositive integers m = n , U m = U n and actually U m ∩ U n = ∅ . We know that theescaping Fatou component U is either a wandering domain, a Baker domain or apreimage of Baker domain. Many examples of such Baker domains and wanderingdomains 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, forwandering domains; the existence of wandering domains with some special behaviorof geometry, analysis or dynamics and/or for some special class of meromorphicfunctions continues to be an important topic and keeps to attract many interests.Let U be an escaping Fatou component of a meromorphic function f . Generallyfor any two distinct points a and b in U there exist M > N such that for ∀ n ≥ N ,(1.1) | f n ( b ) | /M ≤ | f n ( a ) | ≤ | f n ( b ) | M and if ˆ C \ U contains an unbounded component, we have a more precise inequality(1.2) M − | f n ( a ) | ≤ | f n ( b ) | ≤ M | f n ( a ) | , ∀ n ≥ 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 Bakerdomain 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) mustbe bounded, multiply-connected and wandering.The first entire function with multiply connected Fatou component was con-structed by I. N. Baker [2, 3, 4] and every multiply-connected Fatou component U of a transcendental entire function was shown in T¨opfer [40] and Baker [5] tobe a bounded escaping wandering domain and in [42] and [13] that for all suffi-ciently large n , f n ( U ) contains a round annulus centered at the origin with thelarge modulus tending to ∞ as n → ∞ . But it is not this case for a meromorphicfunction with poles. A meromorphic function may have the Herman rings, multiplyconnected attracting domains, parabolic domains and Baker domains; see [16] forexamples of meromorphic functions with only one pole that have invariant multiplyconnected Fatou component. A meromorphic function was constructed by Baker,Kotus and L¨u [8] to have a multiply connected wandering domain U of preassignedconnectivity such that the limit set of { f n | U } is an infinite set including ∞ . Aplanar region E is said to have connectivity m if ˆ C \ E has m components. Sucha wandering domain of a transcendental entire function must be simply connectedand such an example of entire function was constructed by Eremenko and Lyubich[18]. Some examples of meromorphic functions with finitely or infinitely many poleswhich have simply, doubly or infinitely connected wandering domains and whichchange connectivity of wandering domains under iterates can be found in [30].There exists a meromorphic function with bounded, multiply-connected, escap-ing and wandering domains separating the origin and ∞ where (1.2) holds; seeExample 5.2 of [45]. As we know, there exist the escaping wandering domains suchthat (1.1) holds but (1.2) does not hold. In fact, an entire function does not sat-isfy (1.2) in its multiply connected Fatou component and, however, (1.1) is bestpossible; see Theorem 1.1 in [13]. Furthermore, if (1.2) does not hold, then the U TERATES ON ESCAPING FATOU COMPONENTS 3 contains two points a and b such that(1.3) | f n k ( a ) | / | f n k ( b ) | → ∞ ( k → ∞ ) . We have shown in [45] that in this case, S ∞ n =0 f n ( U ) contains a sequence of annuli A ( r m , R m ) with R m /r m → ∞ ( m → ∞ ) and r m → ∞ ( m → ∞ ); we call such asequence of annuli an infinite modulus annulus sequence; Here and henceforth, for R > r > { z : r < | z | < R } by A ( r, R ). And for allsufficiently large n k , U n k = f n k ( U ) separates the origin and ∞ . A condition wasgiven 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 originwhich form an infinite modulus annulus sequence. We can construct meromorphicfunctions which have wandering domain U such that f n ( U ) surrounds the originand contains a large round annulus centered at the origin with (1.3), but f n − ( U )does not surround the origin by suitably modifying the construction of Example5.4 in [45]. Here we omit the constructing details which are routine but lengthy.This leads us to beg the following Question A : For an escaping Fatou component U of f , if S ∞ n =0 f n ( U ) containsan 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 A is that if (1.2) is satisfied in an escap-ing Fatou component U of f , then whether may S ∞ n =0 f n ( U ) contain an infinitemodulus annulus sequence? It turns out that it is this case. There exist wanderingdomains 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 hasan escaping wandering domain U such that for all sufficiently large n , f n ( U ) ⊃ A ( r n , R n ) with R n /r n → ∞ ( n → ∞ ) and r n → ∞ ( n → ∞ ) , but for any two points a, b ∈ U , there exists an M > such that for all sufficiently large n , M − | f n ( a ) | ≤ | f n ( b ) | ≤ M | f n ( a ) | . In another preparing paper studying the Baker domains, we construct a mero-morphic function which has a Baker domain containing an infinite modulus annulussequence and not essentially conjugating to a M¨obius transformation.However, the answer to Question A is positive for a transcendental entire func-tion; Indeed, (1.2) does not hold for an entire function in its multiply-connectedFatou 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 notationsappearing often below. Here and below M ( r, f ) ( ˆ m ( r, f ), respectively) denotes themaximal (minimal, respectively) modulus of f on the circle {| z | = r } , i.e., M ( r, f ) = max {| f ( z ) | : | z | = r } , ˆ m ( r, f ) = min {| f ( z ) | : | z | = r } . If f has a pole in {| z | = r } , then M ( r, f ) = + ∞ . T ( r, f ) is the characteristic ofNevanlinna of f ; m ( r, f ) is the proximity function of f ; N ( r, f ) is the integratedcounting function of f in Nevanlinna theory (see [20] and [41]). Please see Section2 for their exact definitions. ZHENG AND WU
For an escaping Fatou component U and c ∈ U , define h n ( z ) := log | f n ( z ) | log | f n ( c ) | , ∀ z ∈ U,h ( z ) := lim sup n →∞ h n ( z ) , h ( z ) := lim inf n →∞ h n ( z ) , ∀ z ∈ U. In view of (1.1) we easily see that h ( z ) and h ( z ) exist on U . In [13], Bergweiler,Rippon and Stallard first introduced and proved that h ( z ) := h ( z ) = h ( z ) ona 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 becominga significant function. Theorem 2.
Assume that for some C ≥ , c > and r > , we have (1.4) log M ( Cr, f ) ≥ (cid:18) − c log r (cid:19) T ( r, f ) , ∀ r > r . Then there exist constants
D > , d > and R > such that for a Fatoucomponent U of f , if for some m > and some r ≥ R , we have (1.5) A ( D − d r, D d r ) ⊆ f m ( U ) , then U is an escaping wandering domain, h ( z ) := h ( z ) = h ( z ) exists on U and h ( z ) is a non-constant positive harmonic function on U . Therefore for any point z ∈ U and any neighborhood V of z in U , we have for some constant < α < and allsufficiently large n f n ( V ) ⊃ A ( | f n ( z ) | − α , | f n ( z ) | α ); moreover, there exist two points a and b in V and τ > such that | f n ( a ) | τ ≤ | f n ( b ) | and A n = A ( r n , R n ) ⊆ f n ( V ) with R n ≥ r τn → ∞ ( n → ∞ ) and A n +1 ⊆ f ( A n ) . We can give effective conditions to determine
D, d and R in Theorem 2. We notethat the condition which D, d and R satisfy is actually for the result of Lemma 6to hold. Therefore from the proof of Theorem 1.1 in [45], a rough calculation showsthat D, d and R satisfying the following inequalities are available for Theorem 2:(1.6) D > Ce c +1 , πd cos π d (cid:18) π D (cid:19) < , R ≥ D d r and(1.7) T ( r, f )log r > d log D, ∀ r ≥ R . It is easy to see that that if S ∞ n =0 f n ( U ) contains an infinite modulus annulussequence for an escaping Fatou component U of f , then we have (1.5) with D, d and R such that (1.6) and (1.7) hold.Let us make remarks on (1.4). Theorem 1 tells us that (1.4) cannot be removedin Theorem 2. In Example 5.3 in [45], we constructed a meromorphic functionwhich has an escaping wandering domain U in which there exist two points a and b such that | f n ( b ) | / | f n ( a ) | → ∞ ( n → ∞ ) but h U ( z ) ≡ U . TERATES ON ESCAPING FATOU COMPONENTS 5 If f only has at most finitely many poles, then (1.4) immediately holds. So The-orem 2 extends the related results in [13] to a meromorphic function with infinitelymany poles. If(1.8) T ( r, f ) ≥ N ( r, f ) log r, ∀ r ≥ r , then we have m ( r, f ) = T ( r, f ) − N ( r, f ) ≥ (cid:18) − r (cid:19) T ( r, f ) , ∀ r ≥ r . Hence, (1.8) implies (1.4) and that the Nevanlinna deficiency δ ( ∞ , f ) of f of polesis 1. It is easy to find a meromorphic function f satisfying (1.4) and δ ( ∞ , f ) < f ( z ) = e z − e − z + 1 . Then m ( r, f ) ∼ rπ , N ( r, f ) ∼ rπ ( r → ∞ )and δ ( ∞ , f ) = . Andlog M ( r, f ) ≥ log e r − e − r + 1 ∼ r ∼ πT ( r, f ) ( r → ∞ ) . Then we obtain (1.4) for C = 1.Finally, we mention that Theorem 2 is still true even if (1.4) is satisfied for allpossibly except a set of r with finite logarithmic measure.The existence of large annuli in the orbit of an escaping Fatou component guar-antees the stability of the annuli in the sense that the suitable large annuli stillexist under relatively small changes to the function in question. Theorem 3.
Let f be a meromorphic function satisfying the conditions in Theorem2. If f has an escaping Fatou component U such that S ∞ n =0 f n ( U ) contains aninfinite modulus annulus sequence, then for any meromorphic function g which isanalytic on S ∞ n = N f n ( U ) and such that for some N and for < δ < M ( r, f − g ) ≤ M ( r, f ) δ , for {| z | = r } ⊂ ∞ [ n = N f n ( U ) ,g has an escaping Fatou component V on which the results in Theorem 2 also holdfor g . Theorem 3 was proved in [13] for the case when f and g are entire. Theorem3 gives us a strong argument to be able to find a meromorphic function f with δ ( ∞ , f ) = 0 which has an escaping wandering domain U with properties stated inTheorem 2 under f . This is stated as follows. Theorem 4.
There exists a meromorphic function f with δ ( ∞ , f ) = 0 which hasa 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 Theorem2, 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 componentunder iterates. A fast escaping set of a transcendental entire function f , denotedby A ( f ), was introduced by Bergweiler and Hinkkanen [12] and can be defined [29]by A ( f ) = { z ∈ C : there exists L ∈ N such that | f n + L ( z ) | ≥ M n ( R, f ) , for n ∈ N } , ZHENG AND WU where
R > min z ∈ 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 isrelated 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 powerfulin the study of iterate theory, but it is inferior for a meromorphic function withinfinitely many poles. Although we can define the maximal modulus of a mero-morphic function, basically f n with n > f , set ˆ T ( r, f ) = exp T ( r, f ) and ˆ T n ( r, f ) is the n thiterate of ˆ T ( r, f ) with respect to r . Define M ( f ) = { z ∈ C : there exists L ∈ N such that | f n + L ( z ) | ≥ ˆ T n ( R, f ) , for n ∈ N } . Other equivalent definitions of A ( f ) (cf.[12] and [29]) can be defined with the Nevan-linna characteristic for an entire function f , but not for a meromorphic functionwith poles. In [41], we showed that(1) M ( f ) is non-empty and completely invariant;(2) J ( f ) = ∂M ( f );(3) if U is a Fatou component of f and U ∩ M ( f ) = ∅ , then U ⊂ 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 bein the fast escaping set A ( f ), proved in [32]. But basically, this conclusion does nothold 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 c and R ≥ r with property that for r ≥ R , (1.10) T ( r, f )log r > . κ + log D )log D − c , if a Fatou component U of f satisfies that for some m > and some r > CR , (1.11) A ( r, Dr ) ⊂ f m ( U ) , we have U ⊂ M ( f ) . As we know, a simply connected wandering domain of an entire function maynot be in A ( f ). Are there any simply connected wandering domain in A ( f ), whichis asked in [28]. The existence of such simply connected wandering domains hasbeen shown by Bergweiler [11] and Sixsmith [36].A point z in an escaping Baker domain has escaping ratelog | f n ( z ) | = O ( n ) ( n → ∞ ); TERATES ON ESCAPING FATOU COMPONENTS 7 see [6, 10, 43, 25]. In another paper we find a Baker domain where | f n ( z ) | = o ( n )( n → ∞ ) and f n ( z ) goes forward to ∞ in a slow spiral. For the case ofescaping wandering domains, we can construct escaping wandering domains withany escaping rate under iterates, that is to say, given a sequence { a n } of positivenumbers tending to ∞ , we can find a wandering domain U such that for any point z ∈ U , we have(1.12) | f n ( z ) | ∼ a n ( n → ∞ ) . Theorem 7.
Firstly we need a covering lemma which comes from the hyperbolic metric. LetΩ be a hyperbolic domain in the extended complex plane ˆ C , that is, ˆ C \ Ω containsat least three points. Then there exists on Ω the hyperbolic metric λ Ω ( z ) | d z | withGaussian curvature − λ Ω ( z ) is the hyperbolic density of Ω at z ∈ Ω . Then λ D ( z ) = 1 | z | log | z | , D = { z : | z | > } and(2.1) λ A ( z ) = π | z | mod( A ) sin( π log( R/ | z | ) / mod( A )) , ∀ z ∈ A = A ( r, R ) , where A ( r, R ) = { z : r < | z | < R } and mod( A ) = log R/r is the modulus of A .The hyperbolic distance d Ω ( u, v ) of two points u, v ∈ Ω is defined by d Ω ( u, v ) = inf α Z α λ Ω ( z ) | dz | , where the infimum is taken over all piecewise smooth paths α in Ω joining u and v . Lemma 1. ( [45] ) Let f be analytic on a hyperbolic domain U with f ( U ) . Ifthere exist two distinct points z and z in U such that | f ( z ) | > e κδ | f ( z ) | , where ZHENG AND WU δ = d U ( z , z ) and κ = Γ(1 / / (4 π ) , then there exists a point ˆ z ∈ U such that | f ( z ) | ≤ | f (ˆ z ) | ≤ | f ( z ) | and (2.2) f ( U ) ⊃ A e κ (cid:18) | f ( z ) || f ( z ) | (cid:19) /δ | f (ˆ z ) | , e − κ (cid:18) | f ( z ) || f ( z ) | (cid:19) /δ | f (ˆ z ) | ! . If | f ( z ) | ≥ exp (cid:16) κδ − δ (cid:17) | f ( z ) | and < δ < , then (2.3) f ( U ) ⊃ A ( | f ( z ) | , | f ( z ) | ) . In particular, for δ ≤ and | f ( z ) | ≥ e | f ( z ) | , we have (2.3). In order to be able to cover an effective annulus, we are forced to calculatecarefully the hyperbolic distance in an annulus.
Lemma 2.
Set A = A ( r, R ) with < r < R . Then for any two points z , z ∈ A with | z | ≤ | z | we have max ( π mod( A ) log | z || z | , log log R | z | log R | z | ) ≤ d A ( z , z ) ≤ π mod( A ) ˆ K + π mod( A ) ˆ K log | z || z | , where ˆ K = max j =1 , sin( π log( R/ | z j | ) / mod( A )) and ˆ K = min j =1 , sin( π log( R/ | z j | ) / mod( A )) .In particular, for z , z with R | z | = (cid:0) Rr (cid:1) σ , R | z | = (cid:0) Rr (cid:1) τ and < σ ≤ τ < , we have (2.4) max n ( τ − σ ) π, log τσ o ≤ d A ( z , z ) ≤ π ˆ K mod( A ) + ( τ − σ ) π ˆ K , with ˆ K = max { sin( σπ ) , sin( τ π ) } and ˆ K = min { sin( σπ ) , sin( τ π ) } .Proof. We prove the first inequality. In view of (2.1), for all z ∈ A we have λ A ( z ) = π | z | mod( A ) sin( π log( R/ | z | ) / mod( A )) ≥ | z | log R | z | and λ A ( z ) ≥ π | z | mod( A ) . Therefore d A ( z , z ) ≥ Z γ | d z || z | log R | z | ≥ log log R | z | log R | z | and d A ( z , z ) ≥ π mod( A ) log | z || z | , where γ is the geodesic curve in A connecting z and z .Next we prove the second inequality. For all z with | z | ≤ | z | ≤ | z | we havesin( π log( R/ | z | ) / mod( A )) ≥ min j =1 , sin( π log( R/ | z j | ) / mod( A )) = ˆ K . Without loss of generality assume that ˆ K attains the maximal value at z . Then d A ( z , z ) ≤ π mod( A ) ˆ K Z γ | d z || z | + π mod( A ) ˆ K Z γ | d z || z |≤ π mod( A ) ˆ K + π mod( A ) ˆ K log | z || z | , TERATES ON ESCAPING FATOU COMPONENTS 9 where γ is the shortest arc from | z | e i arg z to z and γ = { z = te i arg z : | z | ≤ t ≤ | z |} . (cid:3) Please pay attention! The bounds of d A ( z , z ) only depend on mod( A ), butindependent of the size of R and r . Lemma 3. ( [45] ) Let h ( z ) be analytic on the annulus B = A ( r, R ) with < r
0) for n (cid:16) t, f (cid:17) 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 δ ( ∞ , f ) = lim inf r →∞ m ( r, f ) T ( r, f ) = 1 − lim sup r →∞ N ( r, f ) T ( r, f ) . We write the conclusions about the Nevanlinna characteristic we shall use belowas 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) T ( r,f )log r → ∞ ( r → ∞ ) ;(2) There exists r > such that for r ≤ r < R , (2.6) T ( R, f ) ≥ log R log r T ( r, f ); (3) There exists r > such that for r ≤ r < R and any positive integer n , wehave (2.7) ˆ T n ( R, f ) ≥ ˆ T n ( r, f ) log R log r , where ˆ T ( r, f ) = e T ( r,f ) and ˆ T n ( r, f ) = ˆ T ( ˆ T n − ( r, f ) , f ) , ˆ T ( r, f ) = r. Since T ( r, f ) is a logarithmic convex function, the result (2) in Lemma 4 followsfrom the result (1) in Lemma 4 and further (3) in Lemma 4 follows (see [45]). 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 andlet f ( z ) be analytic on W . Assume that E is a set which intersects every componentof C \ W . Then for any ε > , there exists a rational function R ( z ) such that allpoles of R ( z ) lie in E and | f ( z ) − R ( z ) | < ε, ∀ z ∈ W. Now let us go to the proof of Theorem 1.Take four sequences { r n } , { R n } , { r ′ n } and { R ′ n } such that 10 < r n < r ′ n 3, 2 ≤ R n /R ′ n ≤ 3, 9 R n < r n +1 and R n /r n = R ′ n +1 /r ′ n +1 .Thus R n r n = R ′ n +1 r ′ n +1 ≤ R n +1 r n +1 , and R n r n ≥ n − R r → ∞ ( n → ∞ ).Define T n ( z ) = r ′ n +1 r n z : A ( r n , R n ) → A ( r ′ n +1 , R ′ n +1 ) . Take a n and b n such that b n > R n + n , b n R n → n → ∞ ) and a n < r n − n , a n r n → n → ∞ ). Set A n = A ( r n , R n ) , C n = { z : | z | = b n or a n } and B n = B (0 , r n / r > { ε n } such that ε n +1 < ε n and ε < . Inview of the Runge’s Theorem, we have a rational function f ( z ) such that | f ( z ) − T ( z ) | < ε , ∀ z ∈ A ; | f ( z ) | < ε , ∀ z ∈ B | f ( z ) | < ε , ∀ z ∈ C and inductively, we have rational function f n +1 ( z ) such that (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n +1 X k =1 f k ( z ) − T n +1 ( z ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) < ε n +1 , ∀ z ∈ A n +1 ; | f n +1 ( z ) | < ε n +1 , ∀ z ∈ B n +1 and (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n +1 X k =1 f k ( z ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) < ε n +1 , ∀ z ∈ C n +1 . Write f ( z ) = P ∞ n =1 f n ( z ). Since the series is uniformly convergent on any compactsubset of C , f ( z ) is a meromorphic function on C .For z ∈ B , we have | f ( z ) | ≤ ∞ X n =1 | f n ( z ) | < ∞ X n =1 ε n < , that is to say, f ( B ) ⊂ B (0 , 1) and so B is contained in an invariant Fatou com-ponent of f . For z ∈ C n , we have, by noting that C n ⊂ B m for m > n , | f ( z ) | ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X k =1 f k ( z ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + ∞ X k = n +1 | f k ( z ) | < ε n + ∞ X k = n +1 | f k ( z ) | < ∞ X k = n ε k < TERATES ON ESCAPING FATOU COMPONENTS 11 and so f ( C n ) ⊂ B (0 , 1) and C n is contained in a preperiodic Fatou component of f . Since for z ∈ A n = A ( r n , R n ) ⊂ B n +1 ,(3.1) | f ( z ) − T n ( z ) | ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X k =1 f k ( z ) − T n ( z ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∞ X k = n +1 f k ( z ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) < ∞ X k = n ε k = ε ′ n ( say ) , we have(3.2) f ( A ( r n , R n )) ⊂ A (cid:0) r ′ n +1 − ε ′ n , R ′ n +1 + ε ′ n (cid:1) ⊂ A ( r n +1 , R n +1 ) . Therefore, A n is contained in a wandering domain U n of f and f : U n → U n +1 is proper. Since f is univalent in A n by the Rouch´e Theorem, U n is not doublyconnected. 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 mod-ulus of annulus A m tends to infinity as m does, which contracts to the conformalinvariant property of annulus modulus. Therefore, U n must be infinitely connected.For a point a ∈ A n , it follows from (3.1) that r ′ n +1 r n | a | − ε ′ n ≤ | f ( a ) | ≤ r ′ n +1 r n | a | + ε ′ n and r ′ n +1 − ε ′ n r n ≤ | f ( a ) || a | ≤ r ′ n +1 + ε ′ n r n . Inductively, from (3.2) we have m Y k =1 r ′ n + k − ε ′ n + k − r n + k − ≤ | f m ( a ) || a | ≤ m Y k =1 r ′ n + k + ε ′ n + k − r n + k − . We note that ∞ Y k =1 r ′ n + k + ε ′ n + k − r ′ n + k − ε ′ n + k − = ∞ Y k =1 (cid:18) ε ′ n + k − r ′ n + k − ε ′ n + k − (cid:19) < exp ∞ X k =1 ε ′ n + k − r ′ n + k − ε ′ n + k − < e . Thus for two points a and b in A n , we have(3.3) | f m ( a ) || f m ( b ) | ≤ | a || b | m Y k =1 r ′ n + k + ε ′ n + k − r ′ n + k − ε ′ n + k − < | a || b | e , ∀ m ∈ N . Now we need to treat two cases:Case (I): | f m ( b ) | ≤ R ′ m < R m ≤ | f m ( a ) | ;Case (II): | f m ( b ) | ≤ r m < r ′ m ≤ | f m ( a ) | .Below we only prove that Case (I) would be impossible. The same method yieldsthe 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 d U ( a, b ) ≥ d U m ( f m ( a ) , f m ( b )) ≥ d E m ( f m ( b ) , f m ( a )) ≥ d E m ( R ′ m , R m )= Z R m R ′ m λ E m ( z ) | d z | = Z R m R ′ m πt mod( E m ) sin( π log b m t / mod( E m )) d t ≥ Z R m R ′ m d tt log b m t = log log b m R ′ m log b m R m = log R m R ′ m log b m R m ! ≥ log b m R m ! → ∞ ( m → ∞ ) . A contradiction is derived. Then for all sufficiently large m > f m ( a ) and f m ( b )lie in A m or A ( a m , r ′ m ) or A ( R ′ m , b m ) at the same time.By noting that 2 ≤ r ′ n /a n ≤ ≤ b n /R ′ n ≤ a, b ∈ U and all sufficiently large n | f n ( a ) || f n ( b ) | ≤ max (cid:26) , | a || b | e (cid:27) . (cid:3) For the proof of Theorem 2, we first establish two lemmas. Lemma 5. Let f be a meromorphic function on { z : | z | ≤ R } and f be analyticon A ( r, R ) with < r < R/ . Then for r + 1 < r ′ < R ′ ≤ R , on {| z | = r ′ } , wehave (3.4) log + | f ( z ) | ≤ R ′ + r ′ R ′ − r ′ + (cid:18) log Rr (cid:19) − log R ′ r ′ − r ! T ( R, f ) . Proof. In view of the Poisson formula, for z ∈ D = { z : | z | < R ′ } with f ( z ) =0 , ∞ , we have(3.5) log | f ( z ) | ≤ m ( D, z, f ) + N ( D, z, f ) , where m ( D, z, f ) = 12 π Z ∂D log + | f ( ζ ) | ∂G D ( ζ, z ) ∂~n d s,N ( D, z, f ) = X b n ∈ D G D ( b n , z ) , TERATES ON ESCAPING FATOU COMPONENTS 13 G D is the Green function for D and all b n are poles of f in D appearing oftenaccording to their multiplicities. Then for z with | z | = r ′ , we have m ( D, z, f ) = 12 π Z π log + | f ( R ′ e iθ ) | R ′ − | z | | R ′ − z | d θ ≤ R ′ + | z | R ′ − | z | m ( R ′ , f ) ≤ R ′ + | z | R ′ − | z | T ( R ′ , f )and, by denoting the number of poles of f in D by n ( D, f ), we have N ( D, z, f ) ≤ X b n ∈ D log R ′ | b n − z |≤ n ( D, f ) log R ′ r ′ − r = n ( r, f ) log R ′ r ′ − r = (cid:18) log Rr (cid:19) − Z Rr n ( t, f ) t d t log R ′ r ′ − r ≤ (cid:18) log Rr (cid:19) − (cid:18) log R ′ r ′ − r (cid:19) T ( R, f ) . Therefore we immediately have (3.10). (cid:3) 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 com-ponent of f . Then there exist constants D > , d > and R > such that if forsome m > and r ≥ R , (3.6) A ( D − d r, D d r ) ⊆ f m ( U ) , then for all sufficiently large n , f n ( U ) ⊃ A n and A n +1 ⊂ f ( A n ) with A n = A ( r n , R n ) , R n ≥ r σn , r n +1 > R n , r n → ∞ ( n → ∞ ) and σ > . 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 bereplaced by the following inequality(3.7) log M ( Cr, f ) ≥ (cid:18) − log log r log r (cid:19) T ( r, f ) , ∀ r > r . 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 theposition 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 ) ⊂ f n ( U )with R n ≥ r σn , r n +1 > R n , r n → ∞ ( n → ∞ ) and σ > 1. Therefore, for asufficiently large m and n ≥ m we have r n +1 > R n ≥ R σ n − m m . Without loss of generality for a sufficiently large m , we rewrite r m , R m and σ > f m ( U ) ⊃ A ( r αm , R βm ) ⊃ A ( r m , R m )with R m = r σm , ( σ − 1) log r m > m , 0 < α < β > n = m we replace r m by r /αm and R m by R /βm with β = ασ ). Take R ′ m = e − m R m and r ′ m = er m . Applying Lemma 3 to f on A ( r m , R m ), we have(3.9) ˆ m ( R ′ m , f ) ≥ M ( R ′ m , f ) s m , s m = exp − π ( R m R ′ m , R ′ m r m )! . Let us estimate s m . Sincelog R m R ′ m = m and log R ′ m r m = ( σ − 1) log r m − m > m , we have s m = exp (cid:18) − π m (cid:19) . Take τ with 1 < τ < σ such that s m σ > τ and take two points a and b with | a | = r ′ m and | b | = R ′ m . Set ˆ R m = r ′ m e m . Finding σ m such that R ′ m /C = ˆ R σ m m , wehave σ m = log R ′ m /C log ˆ R m = σ − σm + m + σ + log C log r m + m + 1 . Therefore, we have | f ( a ) | ≤ M ( r ′ m , f ) and in view of (3.7) and (3.9), we have | f ( b ) | ≥ M ( R ′ m , f ) s m ≥ exp( s m τ m T ( R ′ m /C, f ))(3.10) = exp( s m τ m T ( ˆ R σ m m , f )) ≥ exp( s m σ m τ m T ( ˆ R m , f )) , where τ m = 1 − log log( R ′ m /C )log( R ′ m /C ) . Since r ′ m > r m > r σ m with r > e , we have σ m > σ − ( C + σ ) m log r m > σ − ( C + σ ) m σ m log r and s m σ m τ m → σ > m → ∞ ).In view of Lemma 5 with R ′ = r ′ m ( m + 1), r ′ = r ′ m , r = r m and R = ˆ R m , wehave log M ( r ′ m , f ) ≤ m + 2 m + log r ′ m ( m +1) r ′ m − r m log ˆ R m r m T ( ˆ R m , f )= m + 2 m + log ee − ( m + 1) m + 1 ! T ( ˆ R m , f ) ≤ (cid:18) tm (cid:19) T ( ˆ R m , f ) , where t is an absolute constant. Set t m = (cid:0) tm (cid:1) − . Thus combining (3.10)together with the above inequality yields(3.11) | f ( b ) | ≥ exp( s m σ m τ m T ( ˆ R m , f )) ≥ M ( r ′ m , f ) t m s m σ m τ m ≥ | f ( a ) | t m s m σ m τ m . Set r ′ m +1 = | f ( a ) | and R ′ m +1 = | f ( b ) | . Then we have R ′ m +1 ≥ ( r ′ m +1 ) t m s m σ m τ m .Set R m +1 = e ( m +1) R ′ m +1 and r m +1 = r ′ m +1 /e. TERATES ON ESCAPING FATOU COMPONENTS 15 Now let us estimate d U m ( a, b ) with U m = f m ( U ). Set A ′ m = A ( r αm , R βm ). Sincesin π log R βm r ′ m mod( A ′ m ) = sin (cid:18) π ( β − /σ ) log R m − β − α/σ ) log R m (cid:19) → sin (cid:18) π σβ − σβ − α (cid:19) ( m → ∞ )and sin π log R βm R ′ m mod( A ′ m ) = sin (cid:18) π ( β − 1) log R m + m ( β − α/σ ) log R m (cid:19) → sin (cid:18) π σ ( β − σβ − α (cid:19) ( m → ∞ ) , in view of (2.1) we have λ A ′ m ( z ) ≤ Kπ mod( A ′ m ) 1 | z | , r ′ m ≤ | z | ≤ R ′ m , where K is a constant with 1 /K < min n sin (cid:16) π σβ − σβ − α (cid:17) , sin (cid:16) π σ ( β − σβ − α (cid:17)o so that Z R ′ m r ′ m λ A ′ m ( z ) | d z | ≤ Kπ mod( A ′ m ) Z R ′ m r ′ m d tt = Kπ mod( A ′ m ) log R ′ m r ′ m = Kπ ( β − α/σ ) log R m ((1 − /σ ) log R m − m − → Kπ ( σ − σβ − α ( m → ∞ )and Z π λ A ′ m ( R ′ m e iθ ) R ′ m d θ ≤ Kπ mod( A ′ m ) → m → ∞ ) . Therefore, for sufficiently large m and σ > δ = d U m ( a, b ) ≤ d A ′ m ( a, b ) < . In view of the above inequality, (3.11) and (3.10), we have e − κ (cid:18) | f ( b ) || f ( a ) | (cid:19) /δ − ≥ e − κ (cid:16) | f ( b ) | − / ( t m s m σ m τ m ) (cid:17) ≥ exp (cid:20) (cid:18) − t m s m σ m τ m (cid:19) s m σ m τ m T ( ˆ R m , f ) − κ (cid:21) > exp[3( σ − T ( ˆ R m , f )] > e ( m +1) . Thus, for | f ( a ) | ≤ | f (ˆ z ) | ≤ | f ( b ) | with ˆ z ∈ U m we have e κ (cid:18) | f ( a ) || f ( b ) | (cid:19) /δ | f (ˆ z ) | ≤ e κ (cid:18) | f ( a ) || f ( b ) | (cid:19) /δ − r ′ m +1 < r ′ m +1 /e = r m +1 ; and e − κ (cid:18) | f ( b ) || f ( a ) | (cid:19) /δ | f (ˆ z ) | ≥ e − κ (cid:18) | f ( b ) || f ( a ) | (cid:19) /δ − R ′ m +1 > e ( m +1) R ′ m +1 = R m +1 . Since d U m ( a, b ) < , in view of Lemma 1 we have U m +1 = f ( U m ) ⊃ A ( r m +1 , R m +1 ) ⊃ A ( r ′ m +1 , R ′ m +1 ) . Thus by noting that d U m +1 ( f ( a ) , f ( b )) ≤ d U m ( a, b ) < ( β and α are just used toimply the inequality), | f ( a ) | = r ′ m +1 and | f ( b ) | = R ′ m +1 , we can repeat the abovestep and obtain that(3.12) | f n ( b ) | ≥ | f n ( a ) | Q n − k =0 s k + m t k + m σ k + m τ k + m . For sufficiently large m , we can require, for n ≥ n − Y k =0 s k + m t k + m σ k + m τ k + m > σ + 12 > . Take a point c ∈ U . Define h n ( z ) = log | f n ( z ) | log | f n ( c ) | , ∀ z ∈ U. Since U is wandering and escaping, we can assume that | f n ( z ) | > U and with(1.1) we have that h n ( z ) is harmonic and positive on U . In view of (1.1) or byHarnack’s inequality, the family { h n } of harmonic functions is locally normal on U and hence h ( z ) := lim sup n →∞ h n ( z ) , ∀ z ∈ U exists and h is harmonic and positive on U . In view of (3.12), we have h n ( b ) ≥ σ + 12 h n ( a ) , h ( b ) ≥ σ + 12 h ( a ) > h ( a ) . Therefore, h ( z ) is not a constant on U . The same argument implies that h ( z ) := lim inf n →∞ h n ( z ) , ∀ z ∈ U exists and h is a non-constant, harmonic and positive function on U .Suppose that h ( z ) > h ( z ) for some z ∈ U . Without any loss of generalitiessuppose that h ( z ) > h ( c ) = 1. Take a real number η with max { , h ( z ) } < η Set H ( z ) := lim n →∞ log | f n ( z ) | log | f n ( z ) = h ( z ) h ( z ) .H is a non-constant, harmonic and positive function on U . Thus we can find twopoints z and z in V and 0 < α < H ( z ) < − α < < α < H ( z ).For a sufficiently large m we have d f m ( V ) ( f m ( z ) , f m ( z )) < and hence, in viewof Lemma 1, for n ≥ mf n ( V ) ⊃ A ( | f n ( z ) | , | f n ( z ) | ) ⊃ A ( | f n ( z ) | − α , | f n ( z ) | α ) . Thus we have (3.6) for sufficiently large m and f m ( V ) instead. In view of Lemma6, for all sufficiently large n and some τ > 1, we obtain A n = A ( r n , R n ) ⊆ f n ( V )with R n ≥ r τn → ∞ ( n → ∞ ) such that A n +1 ⊂ f ( A n ).We complete the proof of Theorem 2. (cid:3) Proof of Theorem 3. In view of Theorem 2, there exist a sufficiently large m anda sufficiently large r m such that for n ≥ m and r n +1 > R n ≥ r σn with σ > f n ( U ) ⊃ A ( r αn , R βn ) , < α < , < β. From (1.9) and (1.4), in view of Lemma 3, Lemma 4 and Lemma 5 in turn, we canshow M ( R m , g ) ≥ ˆ m ( R m , g ) > M ( r m , g ) ≥ ˆ m ( r m , g )and in view of the maximal principle, we have(3.13) g ( A ( r m , R m )) ⊆ A ( ˆ m ( r m , g ) , M ( R m , g )) . Set u ( z ) = f ( z ) − g ( z ) and s m = exp (cid:16) − π m (cid:17) . For sufficiently large m , we have s m > max { / , δ } . In view of Lemma 3 for r = e − m r m , ρ = r m and R = R m andby noting that log(1 − x ) > − x for 0 < x < , we haveˆ m ( r m , g ) ≥ ˆ m ( r m , f ) − M ( r m , u ) ≥ M ( r m , f ) s m − M ( r m , u )= (cid:18) − M ( r m , u ) M ( r m , f ) s m (cid:19) M ( r m , f ) s m ≥ M ( r m , f ) s m γ m , where γ m = 1 − M ( r m ,f ) δ − sm log M ( r m ,f ) > − m and by noting that log(1 + x ) < x for x > M ( R m , g ) ≤ M ( R m , f ) + M ( R m , u )= M ( R m , f ) (cid:18) M ( R m , u ) M ( R m , f ) (cid:19) ≤ M ( R m , f ) τ m , where τ m = 1 + M ( R m ,f ) − δ log M ( R m ,f ) < m for sufficiently large m . Thus itfollows from (3.13) that(3.14) g ( A ( r m , R m )) ⊆ A ( M ( r m , f ) s m γ m , M ( R m , f ) τ m ) . On the other hand, we have(3.15) f m +1 ( U ) ⊃ f ( A ( r αm , R βm )) ⊃ A ( M ( r αm , f ) , ˆ m ( R βm e − m , f )) . In view of Lemma 3 and (1.4), we haveˆ m ( R βm e − m , f ) ≥ M ( R βm e − m , f ) s m ≥ exp( σ m t m T ( R βm e − m /C, f )) , for t m = 1 − log log( R βm e − m /C )log( R βm e − m /C ) > − m for sufficiently large m , and in view ofLemma 5, we can show that τ m log M ( R m , f ) < s m t m T ( R βm e − m /C, f )and in view of (1.4) and Lemma 4, we have s m γ m log M ( r m , f ) > log M ( r αm , f ) . Thus M ( R βm e − m , f ) s m > M ( r αm , f ) and from (3.15) we can deduce(3.16) f m +1 ( U ) ⊃ A ( M ( r αm , f ) , M ( R βm e − m , f ) s m ) . Now we set ˆ r = M ( r αm , f ) s m γ m /α and ˆ R = M ( R m , f ) τ m so that we have M ( r m , f ) s m γ m ≥ ˆ r . Combining (3.14) and (3.16) yields f m +1 ( U ) ⊃ A (ˆ r αη m , ˆ R β m ) ⊃ A (ˆ r , ˆ R ) ⊃ g ( A ( r m , R m ))with η m = s m γ m > β m = s m τ m log M ( R βm e − m ,f )log M ( R m ,f ) → β ( m → ∞ ). For sufficientlylarge m , we can require that 0 < αη m < α +12 < < β +12 < β m .By the same step as above, we have f m +2 ( U ) ⊃ A (ˆ r αη m η m +1 , ˆ R β m +1 ) ⊃ A (ˆ r , ˆ R ) ⊃ g ( A ( r m , R m ))with ˆ r = M (ˆ r αη m , f ) / ( η m +1 η m α ) , ˆ R = M ( ˆ R , f ) τ m +1 , 0 < αη m η m +1 < α +12 < < β +12 < β m +1 .Inductively, we have f m + n ( U ) ⊃ 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 havelog M ( Cr, g ) ≥ log( M ( Cr, f ) − M ( Cr, u ))= log M ( Cr, f ) + log (cid:18) − M ( Cr, u ) M ( Cr, f ) (cid:19) ≥ (cid:18) − r log r (cid:19) T ( r, f )and for r m < r + 1 < r ′ < R ′ < R ≤ R m we havelog M ( r ′ , g ) ≤ log M ( r ′ , f ) + log (cid:18) M ( r ′ u ) M ( r ′ , f ) (cid:19) ≤ R ′ + r ′ R ′ − r ′ + (cid:18) log Rr (cid:19) − log R ′ r ′ − r ! T ( R, f ) + 1 . Thus we can repeat the steps in the proof of Theorem 2 and show the results ofTheorem 2 hold for g in W . (cid:3) We make a remark on (1.9). From the proof of Theorem 3 we see that in orderthat Theorem 3 holds for g , in fact, we only need to require (1.9) holds in a sequenceof annuli A ( r n , r σ n n ) instead of f n ( U ) as long as A ( r n , r σ n n ) ⊂ f n ( U ) for ∀ n ≥ m with m being large enough and f ( A ( r n , r σ n n )) ⊂ A ( r n +1 , r σ n +1 n +1 ) with 1 < σ ≤ σ n . TERATES ON ESCAPING FATOU COMPONENTS 19 Proof of Theorem 4. With the help of Theorem 3 we are going to find a meromor-phic function f with δ ( ∞ , 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 forentire functions in their multiply connected Fatou components. The first entirefunction with multiply connected Fatou component is due to I. N. Baker [2]. Themultiply connected wandering domains which have uniformly perfect boundary ornon-uniformly perfect boundary were found in [14]. For example, in Theorem 1.3of [14], an entire function of zero order is constructed such that f ( A ((1 + ε k ) r k , (1 − ε k ) r k +1 )) ⊂ A ((1 + ε k +1 ) r k +1 , (1 − ε k +1 ) r k +2 )for a sequence { r k } of positive numbers with r k +1 > r k and a sequence { ε k } ofpositive numbers tending to 0 as k → ∞ . And the Fatou component U containing A ((1 + ε k ) r k , (1 − ε k ) r k +1 ) may have uniformly perfect boundaries by suitablychoosing { r k } .Now we put g ( z ) = f ( z ) + ∞ X n =1 n m n m n X k =1 ε 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 kπmn . It is clear that g is a meromorphic function. For | z | = r with(1 + ε k ) r k < r < (1 − ε k ) r k +1 , we have | f ( z ) − g ( z ) | ≤ ∞ X n =1 n m n m n X j =1 ε n | z − z j,n |≤ ∞ X n = 1 n = k, k + 1 12 n ε n | r − r n | + 12 k ε k r − r k + 12 k +1 ε k +1 r k +1 − r< ∞ X n =1 n = 1 , and this implies (1.9) for | z | = r with (1 + ε k ) r k < r < (1 − ε k ) r k +1 . In viewof Theorem 3 and from the remark after the proof of Theorem 3, there exists anescaping wandering domain W of g such that the results of Theorem 2 hold for g and W .Next we calculate the deficiency δ ( ∞ , g ) of Nevanlinna of g . For er k < r < (1 − ε k ) r k +1 , we have m ( r, g ) ≤ m ( r, f ) + log 2 and N ( r, g ) = Z r n ( t, g ) t d t ≥ Z rr k m k t d t ≥ r k +1 ≥ r. Then as r ∈ ∪ ∞ k = m A ( er k , (1 − ε k ) r k +1 ) → ∞ , we have m ( r, g ) T ( r, g ) ≤ m ( r, g ) N ( r, g ) ≤ m ( r, f ) + log 2 r → , by noting that f is of zero order. This implies that δ ( ∞ , g ) = lim inf r →∞ m ( r, g ) T ( r, g ) = 0 . Then g is the desired meromorphic function of Theorem 4. (cid:3) But we do not know if W has uniformly perfect boundary. Therefore, we begthe following Question B : Are there any meromorphic function with zero Nevanlinna defi-ciency at poles which has a multiply connected escaping wandering domain withuniformly perfect boundary? Perhaps the following approach would be possible to solve Question B . We tryto control the changes of critical values as an entire function that has a multiplyconnected escaping wandering domain with uniformly perfect boundary is smallchanged to a meromorphic function with zero Nevanlinna deficiency at poles andthen in view of Theorem 3 we show the meromorphic function also has a multiplyconnected escaping wandering domain with uniformly perfect boundary.4. proofs of Theorems 5-7 Proof of Theorem 5. Put K = ∞ Y k =1 (cid:18) k + 1) (cid:19) , K n = K Q nk =1 (cid:16) k +1) (cid:17) > √ K > , ∀ n ≥ . Since f is transcendental, we have T ( r, f )log r → ∞ ( r → ∞ )and so we can take a positive number R such that ∀ r ≥ R > max { , r } , where r is the one appeared in Lemma 4, ˆ T ( r, f ) = exp T ( r, f ) ≥ r and(4.1) log M n ( R, f ) > n log R > n + 1) log( n + 2) K n , ∀ n ≥ . We claim that ∀ n ≥ 2, we have(4.2) ˆ T n ( R K , f ) ≥ ( n + 1) M n − ( R, f ) K n − . We show the Claim by induction. It is well-known from Poisson-Jensen formulathat for arbitrary real numbers R and r with R > r > 0, we have T ( r, f ) ≤ log M ( r, f ) ≤ R + rR − r T ( R, f ) . Note that ˆ T ( R K , f ) − R K ˆ T ( R K , f ) + R K > − R K ˆ T ( R K , f ) > − and (cid:18) (cid:19) (cid:18) − (cid:19) > , M ( R, f ) K / > . TERATES ON ESCAPING FATOU COMPONENTS 21 For n = 2, we have ˆ T ( R K , f ) ≥ M ( R K , f ) ˆ T ( RK,f ) − RK ˆ T ( RK,f )+ RK ≥ M ( R K , f ) − ≥ M ( R, f ) K (1+1 / ) ≥ M ( R, f ) K . We assume inductively that (4.2) holds for n . Consider the case of n + 1 for(4.2). Note thatˆ T n ( R K , f ) − M n − ( R, f ) K n − ˆ T n ( R K , f ) + M n − ( R, f ) K n − ≥ ( n + 1) − n + 1) + 1 = 1 − n + 1) + 1and K n − (cid:18) − n + 1) + 1 (cid:19) = K n (cid:18) n + 1) (cid:19) (cid:18) − n + 1) + 1 (cid:19) > K n + K n ( n + 1) . Then in view of (4.2) and (4.1) we haveˆ T n +1 ( R K , f ) ≥ M ( M n − ( R, f ) K n − , f ) ˆ Tn ( RK,f ) − Mn − R,f ) Kn − Tn ( RK,f )+ Mn − R,f ) Kn − > M n ( R, f ) K n − (cid:16) − n +1)2+1 (cid:17) > ( n + 2) M n ( R, f ) K n . We have shown the Claim (4.2).For any point z ∈ M ( f ) in whose definition we choose R K instead of R , we havefor some L ∈ N , | f n +1+ L ( z ) | ≥ ˆ T n +1 ( R K , f ) , ∀ n ∈ N . In view of (4.2), we deduce | f n +1+ L ( z ) | ≥ M n ( R, f ) , ∀ n ∈ N , and this implies z ∈ A ( f ). It follows that M ( f ) ⊆ A ( f ). On the other hand, it isclear that M n ( R, f ) ≥ ˆ T n ( R, f ) and thus for z ∈ A ( f ), it follows from | f n + L ( z ) | ≥ M n ( R, f ) for some L ∈ N that | f n + L ( z ) | ≥ ˆ T n ( R, f ) so that z ∈ M ( f ). We showthat A ( f ) ⊆ M ( f ).So far, we have shown that M ( f ) = A ( f ). (cid:3) Let us make remarks. Let f be a transcendental entire function. It was shownin [32] that A ( f ) = { z ∈ C : there exists L ∈ N such that | f n + L ( z ) | ≥ M ( R, f n ) , for n ∈ N } , which is the definition of A ( f ) introduced in [12]. Since for large R , we have T (2 R , f n ) ≥ 13 log M ( R , f n ) ≥ log M ( R, f n ) ≥ T ( R, f n ) , we have M ( f ) = { z ∈ C : there exists L ∈ N such that | f n + L ( z ) | ≥ ˆ T ( R, f n ) , for n ∈ 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 invarianteven if M ( r, f ) is replaced by µ ( r ) = εM ( r, f ) for a fixed 0 < ε < µ ( r ) = ε ( r ) M ( r, f ) with somewhat more involved condition, 0 < ε ( r ) < ε ( r ) r → ∞ ( r → ∞ ). Actually, we can use µ ( r ) = M ( r, f ) − 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 oflog r with respect to r . If log M ( r,f )log r log k r > 2, then rM ( r, f ) − k r → r → ∞ ) . In [33], Rippon and Stallard investigated the following sets: Q ε ( f ) = { z : there exists L ∈ N such that | f n + L ( z ) | ≥ µ nε ( R ) , ∀ n ∈ N } for 0 < ε < µ ε ( r ) = M ( r, f ) ε . And they showed that Q ε ( f ) = A ( f ) if andonly if f is so-called ε -regular and for f ∈ B (Eremenko-Lyubich class B whichconsists of entire functions with bounded singular value set), A ( f ) = S ε> Q ε ( f ).Of course, we may also take into account the fast escaping set M ( f ) of mero-morphic 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 ≥ , be a sequence of compact sets in C and f : C → ˆ C be acontinuous function such that E n +1 ⊂ f ( E n ) , for n ≥ . Then there exists ζ suchthat f n ( ζ ) ∈ E n for n ≥ .Proof of Theorem 6. For simplicity, we will here complete our proof for C = 1and c = 1. The proof of general case needs just a little cumbersome calculation.Put S = √ D > e . In view of (1.4), we havelog M ( Sr, f ) ≥ (cid:18) − Sr (cid:19) T ( Sr, f ) ≥ (cid:18) − Sr (cid:19) (cid:18) S log r (cid:19) T ( r, f ) ≥ T ( r, f )and 12 π Z π log + | f ( Sre iθ ) | d θ = m ( Sr, f ) ≤ T ( Sr, f ) . Then there exists a point z with | z | = Sr such that T ( r, f ) ≤ log | f ( z ) | ≤ T ( Sr, f ) . It follows from (1.4) that there is a pint z with | z | = S r such thatlog | f ( z ) | ≥ (cid:18) − S r (cid:19) T ( S r, f ) . In view of (2.4) in Lemma 2 with τ = and σ = , we have π ≤ d A ( z , z ) ≤ √ π ( π + 1)3 < . A = A ( r, Dr ). Since, in view of (1.10), we have (cid:18) − S r (cid:19) T ( S r, f ) ≥ (cid:18) − S r (cid:19) (cid:18) S log Sr (cid:19) T ( Sr, f ) TERATES ON ESCAPING FATOU COMPONENTS 23 = T ( Sr, f ) + log( S/e )log Sr T ( Sr, f ) > T ( Sr, f ) + κδ + δ log D with δ = d A ( z , z ), we have | f ( z ) | > e κδ D δ | f ( z ) | , i . e ., e − κ (cid:18) | f ( z ) || f ( z ) | (cid:19) /δ > D. In view of Lemma 1, there exists a point ˆ z ∈ A ( r, Dr ) with | f ( z ) | ≤ | f (ˆ z ) | ≤ | f ( z ) | such that f ( A ( r, Dr )) ⊃ A ( D − | f (ˆ z ) | , D | f (ˆ z ) | ) . Now we take r = | f (ˆ z ) | ≥ | f ( z ) | ≥ e T ( r,f ) = ˆ T ( r, f )and so f ( A ( r, Dr )) ⊃ A ( r , Dr ) . Inductively we get a sequence of positive numbers { r n } with r n ≥ ˆ T n ( r, f ) suchthat f ( A ( r n , Dr n )) ⊃ A ( r n +1 , Dr n +1 ). In view of Lemma 7, we can find a point ζ ∈ A ( r, Dr ) such that | f n ( ζ ) | ≥ ˆ T n ( r, f ). That is to say, ζ ∈ M ( f ) and so U ⊂ M ( f ) . (cid:3) For the proof of Theorem 7, we need the following lemma, which is Main Lemmaof [18]. Lemma 8. Let { G k } ∞ k =1 be a sequence of compact subsets in C such that(i) C \ G k is connected for every k ;(ii) G k ∩ G m = ∅ for k = m ;(iii) dist(0 , G k ) → ∞ ( k → ∞ ) .Then for any sequence of points { z k } with z k ∈ G k , any sequence of positive numbers { ε k } and any sequence of complex functions { φ k } with φ k being analytic in G k forevery k , there exists an entire function f satisfying, for every k , | f ( z ) − φ k ( z ) | < ε k , ∀ z ∈ G k ; f ( z k ) = φ k ( z k ) , f ′ ( z k ) = φ ′ k ( z k ) . The same method as in the proof of Main Lemma of [18] can deduce that fora sequence of compact sets { G k } ∞ k =1 only satisfying (ii) and (iii), there exists ameromorphic function f ( z ) such that the results stated in Lemma 8 hold. This willapply to construction of desired meromorphic function of Theorem 7. Proof of Theorem 7. Set G n = B (cid:18) a n , α n (cid:19) , H n = B (cid:18) − a n , α n (cid:19) ,C n = (cid:26) z : a n + 38 α n ≤ Re z ≤ a n + 58 α n , | Im z | ≤ β n (cid:27) with α n = min { a n − a n − , a n +1 − a n } , 1 < a < a and 0 < β n such that β n / ( a n − a n − ) → ∞ ( n → ∞ ), and φ n ( z ) = α n +1 α n ( z − a n ) + a n +1 : G n → G n +1 ,ϕ n ( z ) = z + a n + e a ρn ; where when ρ = ∞ , e a ρn is replaced by e e (log an )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 entirefunction f such that for every n , | f ( z ) − φ n ( z ) | < ε n , ∀ z ∈ G n ; | f ( z ) − ϕ n ( z ) | < ε n , ∀ z ∈ H n ; f ( a n ) = φ n ( a n ) = a n +1 , f ′ ( a n ) = α n +1 α n ; f ( − a n ) = ϕ n ( − a n ) = e a ρn , f ′ ( − a n ) = 1; | f ( z ) | < , ∀ z ∈ C n ∪ B (0 , , f (0) = 0 , f ′ (0) = 0 . Then we have M ( a n , f ) ≥ | f ( − a n ) | = e a ρn , lim inf n →∞ log log M ( a n , f )log a n ≥ ρ. That is to say, the order of f is at least ρ .According to the construction of f and the behavior of f on G n , by the suitablechoice of { ε n } , we can show that B (cid:0) a , α (cid:1) is in a Fatou component U of f and f n (cid:18) B (cid:18) a , α (cid:19)(cid:19) ⊂ U n . Since f n ( a ) = a n +1 → ∞ ( n → ∞ ), U is escaping. And C m ∩ ∪ ∞ n =0 U n = ∅ forevery m .We want to prove that U is not a Baker domain. Suppose that U is a Bakerdomain with period p , i.e., f p ( U ) ⊆ U . By the Principle of Hyperbolic Metric, wehave, for n ≥ d U ( a , a p +1 ) ≥ d f n ( U ) ( f n ( a ) , f n ( a p +1 )) ≥ 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. SinceBaker domains of an entire function are simply connected and unbounded, so is U n .By the Koebe Theorem, it follows that C U n = min { λ U n ( z ) δ U n ( z ) : ∀ z ∈ U n } ≥ δ U n ( z ) is the Euclidean distance from z to ∂U n ; see [9]. Hence we have d U n ( a n +1 , a n + p +1 ) = inf γ n Z γ n λ U n ( z ) | d z |≥ 14 inf γ n Z γ n δ U n ( z ) | d z |≥ Z β n +1 a n +1 − a n d t = β n +1 a n +1 − a n ) → ∞ ( n → ∞ ) , where the infimum is taken over all curves γ 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 wan-dering domain of f . Take a component U of f − ( U ) and a point z ∈ U ∪ f − ( a ).Then f n ( z ) = a n . (cid:3) TERATES ON ESCAPING FATOU COMPONENTS 25 We omit the specific construction of a meromorphic function which has wander-ing domain U such that f n ( z ) ∼ a n ( n → ∞ ) in U . References [1] I. N. Baker, The iteration of polynomials and transcendental entire functions, J. Aust. Math.Soc. (Ser. A) (1981), 483-495[2] I. N. Baker, Multiply connected domains of normality in ieteration theory, Math. Z., (1963),206-214[3] I. N. Baker, The domains of normality of an entire functions, Ann. Acad. Sci. Fenn. Ser. A IMath. 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