A class of Newton maps with Julia sets of Lebesgue measure zero
AA class of Newton maps with Julia sets of Lebesguemeasure zero
Mareike Wolff
Abstract
Let g ( z ) = (cid:82) z p ( t ) exp( q ( t )) dt + c where p, q are polynomials and c ∈ C , and let f be the function from Newton’s method for g . We show that under suitable assump-tions the Julia set of f has Lebesgue measure zero. Together with a theorem byBergweiler, our result implies that f n ( z ) converges to zeros of g almost everywherein C if this is the case for each zero of g (cid:48)(cid:48) . In order to prove our result, we establishgeneral conditions ensuring that Julia sets have Lebesgue measure zero.2020 Mathematical Subject Classification: 37F10 (Primary), 30D05 (Secondary) For a rational function f : ˆ C → ˆ C that is not constant and not a M¨obius transformation,or a transcendental meromorphic function f : C → ˆ C , let f n denote the n th iterateof f . The Fatou set , F ( f ), is the set of all z such that all iterates f n are defined andform a normal family in a neighbourhood of z . Its complement, J ( f ), is called the Julia set . For an introduction to the iteration theory of meromorphic functions, see,for example, [23] for rational functions and [2] for transcendental functions.Let g be a non-constant meromorphic function. Newton’s root finding method for g consists of iterating the function f ( z ) = z − g ( z ) g (cid:48) ( z ) . (1.1)We also call f the Newton map corresponding to g . The zeros of g are preciselythe attracting fixed points of f , and the simple zeros of g are even superattractingfixed points. Recall that, more generally, a periodic point z of period p of a mero-morphic function f is called attracting , indifferent or repelling depending on whether | ( f p ) (cid:48) ( z ) | < | ( f p ) (cid:48) ( z ) | = 1 or | ( f p ) (cid:48) ( z ) | >
1. The periodic point z is called superat-tracting if ( f p ) (cid:48) ( z ) = 0. An indifferent periodic point z is called rationally indifferent if ( f p ) (cid:48) ( z ) is a q th root of unity for some q ∈ N , otherwise it is called irrationallyindifferent .We will investigate the Lebesgue measure of Julia sets of Newton maps correspond-ing to functions of the form g ( z ) = (cid:90) z p ( t ) e q ( t ) dt + c (1.2)where p is a polynomial with p (cid:54)≡ q is a non-constant polynomial and c ∈ C . In the following, we will assume that g is not of the form g ( z ) = ˜ p ( z ) e ˜ q ( z ) (1.3) a r X i v : . [ m a t h . D S ] J a n Mareike Wolff with polynomials ˜ p and ˜ q . Then g has infinitely many zeros and f is transcendental.Newton’s method for functions of the form (1.3) has been studied by Haruta [9].Let dist( · , · ) denote the Euclidean distance in C . We will prove the following result. Theorem 1.1.
Let g be of the form (1.2) but not of the form (1.3) , and let f be thecorresponding Newton map. Denote the zeros of g (cid:48)(cid:48) which are not zeros of g or g (cid:48) by z , ..., z N . Suppose that for all j ∈ { , ..., N } , the point z j is attracted by a periodiccycle, that is, there exists a periodic cycle C of f such that lim n →∞ dist( f n ( z j ) , C ) = 0 .Then the Lebesgue measure of J ( f ) is zero. Jankowski [11, §
3] proved that if f is the Newton map corresponding to a function g of the form g ( z ) = r ( z ) e az + b where r is a rational function and a, b ∈ C \ { } , andif for each of the zeros, z , ..., z N , of g (cid:48)(cid:48) that are not zeros of g or g (cid:48) , the iterates f n ( z j )converge to a finite limit as n → ∞ , then the Julia set of f has Lebesgue measure zero.Note that if r is a polynomial, then g can be written in the form (1.2) with q ( t ) = at and p ( t ) = ar ( t ) + r (cid:48) ( t ). Also, under the assumptions of Jankowski’s result, f n ( z j )is attracted by a cycle of period 1 for all j ∈ { , ..., N } . So Jankowski’s theorem forpolynomial r is a special case of Theorem 1.1. The essential new difficulties we have todeal with in our proof come from the fact that we allow q to have degree greater thanone.Bergweiler [3, Theorem 3] also investigated Newton’s method for functions of theform (1.2). He proved the following result. Theorem (Bergweiler) . Let g be of the form (1.2) but not of the form e az + b with a, b ∈ C , and let f be the corresponding Newton map. Denote the zeros of g (cid:48)(cid:48) which are notzeros of g or g (cid:48) by z , ..., z N . If f n ( z j ) converges to a finite limit for all j ∈ { , ..., N } ,then f n ( z ) converges to zeros of g on an open dense subset of C . It is not difficult to see that under the assumptions of Bergweiler’s theorem, f n ( z j )converges to an attracting fixed point of f and hence a zero of g for all j ∈ { , ..., N } .So the Theorem says that f n ( z ) converges to zeros of g on an open dense subset of C ,provided this is the case for each zero z of g (cid:48)(cid:48) that is not a zero of g or g (cid:48) .A component U of the Fatou set F ( f ) is called periodic if there is p ∈ N with f p ( U ) ⊂ U , the component U is called preperiodic if there is l ∈ N such that f l ( U )is contained in a periodic Fatou component, and U is called a wandering domain ifit is not (pre)periodic. It is known (see, e.g., [2, § U is a periodic Fatoucomponent of period p of f , then either f np | U converges to an attracting periodic pointin U ( immediate basin of attraction ), f np | U converges to a rationally indifferent periodicpoint in ∂ U ( parabolic domain ), f np | U converges to some z ∈ ∂ U and f p ( z ) is notdefined ( Baker domain ), or f np | U is conjugate to a rotation of a disk ( Siegel disk ) oran annulus (
Herman ring ).Bergweiler’s theorem is proved by showing that under the given assumptions, f has neither wandering domains nor parabolic domains, Baker domains, Siegel disks orHerman rings.The following corollary is a direct consequence of Theorem 1.1 and Bergweiler’stheorem. Corollary 1.2.
Let g be of the form (1.2) but not of the form (1.3) , and let f bethe corresponding Newton map. Denote the zeros of g (cid:48)(cid:48) which are not zeros of g or g (cid:48) by z , ..., z N . If f n ( z j ) converges to a finite limit for all j ∈ { , ..., N } , then f n ( z ) converges to zeros of g for almost all z ∈ C . ulia sets of measure zero g ( z ) = (cid:82) z e − t dt + c with −√ π/ < c < √ π/
2, see [3, § F ( f ).In order to prove Theorem 1.1, we will first prove a general Theorem giving con-ditions ensuring that the Julia set of a meromorphic function has Lebesgue measurezero. This may be of independent interest. For a meromorphic function f , we denoteby sing( f − ) the set of singular values of f , that is, the set of critical and asymptoticvalues of f and limit points of those. For n ≥
0, let N n = { z : f n ( z ) is not defined } .Let P ( f ) := ∞ (cid:91) n =0 f n (sing( f − ) \ N n )denote the postsingular set of f . For z ∈ C and r >
0, let D ( z , r ) denote the opendisk centred at z with radius r . Also, let meas( · ) denote Lebesgue measure, and formeasurable A , B ⊂ C with 0 < meas( B ) < ∞ , letdens( A , B ) = meas( A ∩ B )meas( B )denote the density of A in B .Following [15], we call a measurable set A ⊂ C thin at ∞ if there exist R , ε > z ∈ C , we havedens( A , D ( z, R )) < − ε . Additionally, we introduce the concept that A is thin at z ∈ C if there exist δ , ε > z ∈ D ( z , δ ), we havedens( A , D ( z, | z − z | )) < − ε . (1.4)We call A uniformly thin at B ⊂ C if there are δ , ε > z ∈ B . Theorem 1.3.
Let f be a meromorphic function that is not constant and not a M¨obiustransformation. Suppose that there exists R > such that(i) P ( f ) ∩ J ( f ) ∩ D (0 , R ) is a finite set;(ii) J ( f ) is thin at ∞ ;(iii) J ( f ) is uniformly thin at ( P ( f ) ∩ C ) \ D (0 , R ) .Then the Lebesgue measure of J ( f ) is zero. McMullen [15, Proposition 7.3] proved that if f is entire, P ( f ) is compact, P ( f ) ∩J ( f ) = ∅ and J ( f ) is thin at ∞ , then meas( J ( f )) = 0. A meromorphic function f for which P ( f ) is compact and does not intersect J ( f ) is called hyperbolic . Thereare various results on iteration of hyperbolic meromorphic functions; see, for exam-ple, [4, 19, 20, 22, 25]. Stallard [21] extended McMullen’s result to entire functions f with possibly unbounded postsingular set such that dist( P ( f ) , J ( f )) > J ( f ) isthin at ∞ . Meromorphic functions f with dist( P ( f ) , J ( f )) > topologically hyperbolic , and have also been considered in [1, 14].Jankowski [11, 12] extended Stallard’s result by allowing that f is meromorphicand that there are certain exceptions to the condition dist( P ( f ) , J ( f )) >
0. A more
Mareike Wolff general result was later obtained by Zheng [24, Theorem 5] who proved that if f is ameromorphic function such that the set P ( f ) ∩ J ( f ) is finite, there exists R > J ( f ) ∩ C ) \ D (0 , R ) , P ( f )) > J ( f ) is thin at ∞ , then meas( J ( f )) = 0.The results by McMullen, Stallard, Jankowki and Zheng mentioned above are spe-cial cases of Theorem 1.3 since the condition that dist(( J ( f ) ∩ C ) \ D (0 , R ) , P ( f )) > J ( f ) is uniformly thin at ( P ( f ) ∩ C ) \ D (0 , R (cid:48) ) for R (cid:48) > R . Our theoremis the first of this kind to allow infinitely many postsingular values in the Julia set oran unbounded sequence of postsingular values whose distance to the Julia set tends tozero.In general, the condition that J ( f ) is thin at ∞ cannot be dropped. If | α | issmall, then the postsingular set of f ( z ) = sin( αz ) is a compact subset of F ( f ), andMcMullen [15] showed that J ( f ) has positive measure. However, there are resultswhere instead of assuming that J ( f ) is thin at ∞ other conditions are imposed, see [5,Theorem 8], [24, Theorems 3 and 4].This article is structured as follows. In Section 2, we prove Theorem 1.3. In theremaining part of this paper, we prove Theorem 1.1. First, in Section 3, we introducethe change of variables w = q ( z ). In Section 4, we give asymptotic representations of g and f . In Section 5, we introduce a class of subsets of C whose preimages under q are connected to the asymptotic behaviour of f . In Section 6, we investigate thepostsingular set of f . In Section 7-10, we investigate the location and size of the set q ( F ( f )). Finally, in Section 11, we complete the proof of Theorem 1.1. In this section, we prove Theorem 1.3. The following Lemma is an easy consequence ofthe well-known Koebe 1 / Lemma 2.1.
Let z ∈ C and r > , and let f : D ( z , r ) → C be holomorphic andinjective. Then f ( D ( z , r )) ⊃ D (cid:18) f ( z ) , | f (cid:48) ( z ) | r (cid:19) . Moreover, for ρ ∈ (0 , , f ( D ( z , ρr )) ⊂ D (cid:18) f ( z ) , ρ (1 − ρ ) | f (cid:48) ( z ) | r (cid:19) and min z ∈D ( z ,ρr ) | f (cid:48) ( z ) | max z ∈D ( z ,ρr ) | f (cid:48) ( z ) | ≥ (cid:18) − ρ ρ (cid:19) . For
A ⊂ C , denote the forward orbit of A by O + ( A ) := ∞ (cid:91) n =0 f n ( A \ N n )where N n = { z : f n ( z ) is not defined } , and for B ⊂ ˆ C , let O − ( B ) := ∞ (cid:91) n =1 f − n ( B )be the backward orbit of B . For z ∈ ˆ C , write O ± ( z ) := O ± ( { z } ) . ulia sets of measure zero z ∈ ˆ C an exceptional point of the meromorphic function f if O − ( z ) is finite.It is not difficult to see that any meromorphic function that is not constant and not aM¨obius transformation has at most two exceptional points. Lemma 2.2.
Let f be a meromorphic function that is not constant and not a M¨obiustransformation. If f is transcendental, in addition suppose that O − ( ∞ ) is finite. Let K be a compact subset of C that contains no exceptional point of f , and let U ⊂ C beopen with U ∩ J ( f ) (cid:54) = ∅ . Then there is n ∈ N such that K ⊂ f n ( U ) for all n ≥ n . This is due to Fatou for rational [7, p.39] and entire functions [8, p.356]. His prooffor entire functions also works for transcendental meromorphic functions where O − ( ∞ )is finite. We also require the following result. Lemma 2.3.
Let f be a meromorphic function that is not constant and not a M¨obiustransformation, and let C = { z , f ( z ) , ..., f p ( z ) = z } be a periodic cycle of f . Supposethat z ∈ J ( f ) is attracted by C . Then there exists n ∈ N such that f n ( z ) ∈ C . Clearly, the hypotheses imply that
C ⊂ J ( f ). It is not difficult to see that theconclusion of Lemma 2.3 is true for repelling cycles. For rationally indifferent cycles,the result follows from the Leau flower theorem [16, § z ∈ J ( f ) is not a point ofdensity of J ( f ). We will then use Lemma 2.4 and the Lebesgue density theorem toprove Theorem 1.3. Lemma 2.4.
Let f be a meromorphic function that is not constant and not a M¨obiustransformation, and let z ∈ J ( f ) \ O − ( P ( f ) ∪ {∞} ) . Suppose that there exist sequences ( n k ) ∈ N N with lim k →∞ n k = ∞ and ( r k ) ∈ (0 , ∞ ) N satisfying the following conditions:(i) D ( f n k ( z ) , r k ) ∩ P ( f ) = ∅ for all k ∈ N ;(ii) there is ε > such that dens( F ( f ) , D ( f n k ( z ) , r k )) ≥ ε for all k ∈ N .Then z is not a point of density of J ( f ) .Proof. Let ω := (cid:114) − ε . For k ∈ N , let z k := f n k ( z ) , D k := D ( z k , r k ) and D (cid:48) k := D ( z k , ωr k ) . Since D k ∩ P ( f ) = ∅ , there is a branch ϕ k of f − n k defined in D k with ϕ k ( z k ) = z . ByKoebe’s theorems (see Lemma 2.1), D (cid:16) z, ω r k | ϕ (cid:48) k ( z k ) | (cid:17) ⊂ ϕ k ( D (cid:48) k ) ⊂ D (cid:18) z, ω (1 − ω ) r k | ϕ (cid:48) k ( z k ) | (cid:19) . (2.1)We claim that lim k →∞ (cid:12)(cid:12) ϕ (cid:48) k ( z k ) (cid:12)(cid:12) r k = 0 . (2.2)If this was not true, there would be δ > D ( z, δ ) ⊂ ϕ k ( D (cid:48) k ) for infinitelymany k , and hence f n k ( D ( z, δ )) ⊂ D (cid:48) k for infinitely many k . If f is transcendentaland O − ( ∞ ) is infinite, this is impossible because O − ( ∞ ) is dense in J ( f ). Supposethat f is rational or O − ( ∞ ) is finite. Fix v ∈ P ( f ) ∩ C and let K be of the form Mareike Wolff K = { z : | z − v | = ρ } where ρ is chosen such that K does not contain any exceptionalpoint of f . Then by Lemma 2.2, K ⊂ f n k ( D ( z, δ )) ⊂ D (cid:48) k ⊂ D k for all large k . But thisimplies v ∈ D k , contradicting (i). This proves (2.2).We will now show that lim sup r → dens( F ( f ) , D ( z, r )) > , that is, z is not a point of density of J ( f ). We havedens( F ( f ) , ϕ k ( D (cid:48) k )) ≥ (cid:32) min ζ ∈D (cid:48) k | ϕ (cid:48) k ( ζ ) | max ζ ∈D (cid:48) k | ϕ (cid:48) k ( ζ ) | (cid:33) dens( F ( f ) , D (cid:48) k )= (cid:32) min ζ ∈D (cid:48) k | ϕ (cid:48) k ( ζ ) | max ζ ∈D (cid:48) k | ϕ (cid:48) k ( ζ ) | (cid:33) meas( D (cid:48) k ∩ F ( f ))meas D (cid:48) k ≥ (cid:32) min ζ ∈D (cid:48) k | ϕ (cid:48) k ( ζ ) | max ζ ∈D (cid:48) k | ϕ (cid:48) k ( ζ ) | (cid:33) · meas( D k ∩ F ( f )) − meas( D k \ D (cid:48) k )meas D k . Hence, by the Koebe distortion theorem and (ii),dens( F ( f ) , ϕ k ( D (cid:48) k )) ≥ (cid:18) − ω ω (cid:19) · (cid:18) ε − πr k − πr k ω πr k (cid:19) = (cid:18) − ω ω (cid:19) · ε . (2.3)By (2.3) and (2.1),dens (cid:18) F ( f ) , D (cid:18) z, ω (1 − ω ) | ϕ (cid:48) k ( z k ) | r k (cid:19)(cid:19) ≥ dens( F ( f ) , ϕ k ( D (cid:48) k )) · dens (cid:18) ϕ k ( D (cid:48) k ) , D (cid:18) z, ω (1 − ω ) | ϕ (cid:48) k ( z k ) | r k (cid:19)(cid:19) ≥ (cid:18) − ω ω (cid:19) ε ·
116 (1 − ω ) . Proof of Theorem 1.3.
We first show that meas( P ( f ) ∩ J ( f )) is zero. In order todo so, write P ( f ) ∩ J ( f ) ∩ C = P ∪ P with P := P ( f ) ∩ J ( f ) ∩ D (0 , R ) and P := ( P ( f ) ∩ J ( f ) ∩ C ) \ D (0 , R ). Since P is a finite set, we only have to show thatmeas( P ) = 0.Since J ( f ) is uniformly thin at ( P ( f ) ∩ C ) \ D (0 , R ), there are δ , ε > v ∈ P ( f ) ∩ C with | v | > R and all ζ ∈ D ( v, δ ), we havedens( F ( f ) , D ( ζ, | ζ − v | )) > ε . (2.4)Let z ∈ P and r ∈ (0 , δ ). Then D ( z + r/ , r/ ⊂ D ( z, r ) and dens( F ( f ) , D ( z + r/ , r/ > ε . Thus,dens( F ( f ) , D ( z, r )) ≥ dens (cid:16) F ( f ) , D (cid:16) z + r , r (cid:17)(cid:17) · dens (cid:16) D (cid:16) z + r , r (cid:17) , D ( z, r ) (cid:17) > ε . Hence, z is not a point of density of J ( f ). By the Lebesgue density theorem (see, e.g.,[13, Corollary 2.14]), the Lebesgue measure of P is zero. So meas( P ( f ) ∩J ( f ) ∩ C ) = 0and hence also meas( O − ( P ( f ) ∩ J ( f ) ∩ C )) = 0. Since O − ( ∞ ) is countable, we obtainthat meas( O − (( P ( f ) ∩ J ( f )) ∪ {∞} )) = 0.Next, we show that each z ∈ J ( f ) \ O − ( P ( f ) ∪ {∞} ) satisfieslim sup n →∞ dist( f n ( z ) , P ) > . (2.5) ulia sets of measure zero n →∞ dist( f n ( z ) , P ) = 0. We show that then z ∈O − ( P ), contradicting our assumption.Because P is finite, there is a subsequence, ( f n k ( z )), that converges to some w ∈ P .For all j ∈ N , we have f j ( w ) = lim k →∞ f n k + j ( z ) ∈ P . Thus, w is preperiodic, that is, f l ( w ) is periodic for some l ∈ N . Assume without loss of generality that l = 0, that is,there is p ∈ N with f p ( w ) = w. Let α > D ( ζ, α ) with ζ ∈ P are pairwise disjoint, and let β ∈ (0 , α ) such that f ( D ( f j ( w ) , β )) ⊂ D ( f j +1 ( w ) , α ) for all j ∈ { , ..., p − } . Then byperiodicity, this is true for all j ≥
0. For large k , we have dist( f n k + j ( z ) , P ) < β forall j ≥ f n k ( z ) ∈ D ( w, β ). Then f n k +1 ( z ) ∈ D ( f ( w ) , α ). Since the disks D ( ζ, α )with ζ ∈ P are disjoint, we have | f n k +1 ( z ) − ζ | > α > β for all ζ ∈ P \ { f ( w ) } .Thus, | f n k +1 ( z ) − f ( w ) | < β . Inductively, we obtain that f n k + j ( z ) ∈ D ( f j ( w ) , β ) forall j ∈ N . Thus, f n ( z ) is attracted by the cycle { w, f ( w ) , ..., f p − ( w ) , f p ( w ) = w } . ByLemma 2.3, z is eventually mapped to this cycle, so z ∈ O − ( P ).Now let z ∈ J ( f ) \ O − ( P ( f ) ∪ {∞} ). By (2.5), there exist a subsequence ( f n k ( z ))and η > f n k ( z ) , P ) > η (2.6)for all k ∈ N . We will show that z satisfies the assumptions of Lemma 2.4 and hence isnot a point of density of J ( f ). Let d k := dist( f n k ( z ) , P ( f )) , and let z k ∈ P ( f ) with | f n k ( z ) − z k | = d k . First suppose that the sequence ( d k ) is bounded, say d k ≤ γ for all k . We distinguishthree cases. | z k | ≤ R for infinitely many k . By passing to a subsequence if necessary,we can assume that | z k | ≤ R for all k . Then ( f n k ( z )) is bounded, and by again passingto a subsequence, we can assume that f n k ( z ) converges to some w ∈ J ( f ). By (2.6),we have w / ∈ P . If | w | > R , then for large k , we have d k = | f n k ( z ) − z k | ≥ | f n k ( z ) | − | z k | ≥ | f n k ( z ) | − R > | f n k ( z ) − w | . Thus, w / ∈ P ( f ), so ν := dist( w, P ( f )) >
0. For large k , D (cid:16) w, ν (cid:17) ⊂ D (cid:16) f n k ( z ) , ν (cid:17) ⊂ D ( w, ν ) . Thus, D ( f n k ( z ) , ν/ ∩ P ( f ) = ∅ , anddens (cid:16) F ( f ) , D (cid:16) f n k ( z ) , ν (cid:17)(cid:17) ≥ dens (cid:16) D (cid:16) w, ν (cid:17) , D (cid:16) f n k ( z ) , ν (cid:17)(cid:17) · dens (cid:16) F ( f ) , D (cid:16) w, ν (cid:17)(cid:17) = 14 dens (cid:16) F ( f ) , D (cid:16) w, ν (cid:17)(cid:17) > . By Lemma 2.4, z is not a point of density of J ( f ). | z k | > R and d k ≤ δ for infinitely many k , without loss of generality forall k . Then by (2.4), dens( F ( f ) , D ( f n k ( z ) , d k )) > ε . By Lemma 2.4, z is not a point of density of J ( f ). Mareike Wolff | z k | > R and d k > δ for infinitely many k , without loss of generality forall k . Let w k := z k + δ d k ( f n k ( z ) − z k ) . Then | f n k ( z ) − w k | = (cid:18) − δ d k (cid:19) | f n k ( z ) − z k | = d k − δ and hence D ( w k , δ ) ⊂ D ( f n k ( z ) , d k ) . Also, | w k − z k | = δ . By the hypotheses,dens( F ( f ) , D ( f n k ( z ) , d k )) ≥ dens( D ( w k , δ ) , D ( f n k ( z ) , d k )) · dens( F ( f ) , D ( w k , δ )) ≥ δ d k ε ≥ δ γ ε . By Lemma 2.4, z is not a point of density of J ( f ).Now suppose that the sequence ( d k ) is unbounded. Since J ( f ) is thin at ∞ , thereare R , ε > v ∈ C , we havedens( F ( f ) , D ( v, R )) > ε . By passing to a subsequence if necessary, we can assume that d k ≥ R for all k . Then D ( f n k ( z ) , R ) ∩ P ( f ) = ∅ for all k . Also,dens( F ( f ) , D ( f n k ( z ) , R )) > ε . By Lemma 2.4, z is not a point of density of J ( f ).Altogether, it follows that the set of density points of J ( f ) has Lebesgue measurezero. By the Lebesgue density theorem, the Lebesgue measure of J ( f ) is zero. Throughout the remaining part of the paper, let g be defined by (1.2), that is, g ( z ) = (cid:90) z p ( t ) e q ( t ) dt + c. Remark 3.1.
Suppose that q ( t ) = at d + O ( t d − ) as t → ∞ , where a ∈ C \ { } and d ≥
1. Let α ∈ C with α d = a . Then q ( t/α ) = t d + O ( t d − ) as t → ∞ , g ( z/α ) = (cid:90) z/α p ( t ) e q ( t ) dt + c = (cid:90) z α p (cid:18) tα (cid:19) e q ( t/α ) dt + c, and Newton’s method for g ( z/α ) is conjugate to Newton’s method for g via z (cid:55)→ αz .Thus, we can and will assume without loss of generality that a = 1, that is, q ( t ) = t d + O ( t d − )as t → ∞ . Also, since the functions g and b · g for b ∈ C \ { } have the same zeros and Newton’smethods for g and b · g coincide, we can and will assume without loss of generality that p has the form p ( t ) = dt m + O ( t m − )as t → ∞ , where d = deg( q ). ulia sets of measure zero f be defined by (1.1), that is, f is the Newton map corresponding to g . In orderto prove Theorem 1.1, it will be useful to consider the change of variables w = q ( z ).Let R > q are contained in D (0 , R ) and such that for | z | ≥ (1 / R /d , we have 12 d | z | d ≤ | q ( z ) | ≤ d | z | d . (3.1)Define G := C \ ( D (0 , R ) ∪ [0 , ∞ )) . Lemma 3.2.
There exists c > such that the set q − ( G ) consists of d components, S , ..., S d , satisfying S j ⊂ (cid:26) z : | z | > R /d , j − πd − c | z | < arg( z ) < jπd + c | z | (cid:27) and S j ⊃ (cid:26) z : | z | > R /d , j − πd + c | z | < arg( z ) < jπd − c | z | (cid:27) for all j ∈ { , ..., d } . Moreover, q maps each S j conformally onto G .Proof. Since G is simply connected and contains no critical values of q , its preimage q − ( G ) consists of d components, and q maps each of them conformally onto G . By(3.1), q (cid:18) D (cid:18) , R /d (cid:19)(cid:19) ⊂ D (0 , R )and q (cid:16) C \ D (cid:16) , R /d (cid:17)(cid:17) ⊂ C \ D (0 , R ) . Also, for z ∈ C , we havearg( q ( z )) = arg (cid:20) z d (cid:18) O (cid:18) z (cid:19)(cid:19)(cid:21) ≡ d arg( z )+arg (cid:18) O (cid:18) z (cid:19)(cid:19) ≡ d arg( z )+ O (cid:18) z (cid:19) mod 2 π as z → ∞ . Thus,arg( z ) ≡ arg( q ( z )) d + O (cid:18) z (cid:19) mod 2 πd as z → ∞ . Using that q is surjective, we obtain the desired conclusion.For j ∈ { , ..., d } , let ϕ j be the branch of q − defined in G with ϕ j ( G ) = S j . g and f In this section, we give asymptotic representations for g ( ϕ j ( w )) , g ( z ) , f ( ϕ j ( w )) , f ( z ).Let λ := d − − md . Then p ( z ) q (cid:48) ( z ) = z − λd (cid:18) O (cid:18) z (cid:19)(cid:19) (4.1)as z → ∞ and, for j ∈ { , ..., d } , (cid:12)(cid:12)(cid:12)(cid:12) p ( ϕ j ( w )) q (cid:48) ( ϕ j ( w )) (cid:12)(cid:12)(cid:12)(cid:12) = | w | − λ (cid:18) O (cid:18) | w | /d (cid:19)(cid:19) (4.2)as w → ∞ in G .0 Mareike Wolff
Lemma 4.1.
Let j ∈ { , ..., d } . Then there exists c j ∈ C such that g ( ϕ j ( w )) = c j + p ( ϕ j ( w )) q (cid:48) ( ϕ j ( w )) (cid:18) λw + O (cid:18) | w | /d (cid:19)(cid:19) e w as w → ∞ in G . In terms of z = ϕ j ( w ), Lemma 4.1 says the following. Corollary 4.2.
For j ∈ { , ..., d } , we have g ( z ) = c j + p ( z ) q (cid:48) ( z ) (cid:18) λz d + O (cid:18) z d +1 (cid:19)(cid:19) e q ( z ) as z → ∞ in S j .Proof of Lemma 4.1. Let x ∈ ( −∞ , − R ) = G ∩ ( −∞ ,
0] and w ∈ G . Then g ( ϕ j ( w )) = (cid:90) ϕ j ( w )0 p ( t ) e q ( t ) dt + c = (cid:90) ϕ j ( w ) ϕ j ( x ) p ( t ) e q ( t ) dt + (cid:90) ϕ j ( x )0 p ( t ) e q ( t ) dt + c = (cid:90) ϕ j ( w ) ϕ j ( x ) p ( t ) e q ( t ) dt + g ( ϕ j ( x ))= (cid:90) wx ϕ (cid:48) j ( s ) p ( ϕ j ( s )) e s ds + g ( ϕ j ( x )) . Let r ( s ) := ϕ (cid:48) j ( s ) p ( ϕ j ( s )) = p ( ϕ j ( s )) q (cid:48) ( ϕ j ( s )) . Repeated integration by parts yields (cid:90) wx r ( s ) e s ds = (cid:0) r ( s ) − r (cid:48) ( s ) + r (cid:48)(cid:48) ( s ) (cid:1) e s (cid:12)(cid:12)(cid:12) wx − (cid:90) wx r (cid:48)(cid:48)(cid:48) ( s ) e s ds. We have r (cid:48) ( s ) = ϕ (cid:48) j ( s ) q (cid:48) ( ϕ j ( s )) p (cid:48) ( ϕ j ( s )) − q (cid:48)(cid:48) ( ϕ j ( s )) p ( ϕ j ( s )) q (cid:48) ( ϕ j ( s )) = (cid:18) q (cid:48) ( ϕ j ( s )) · p ( ϕ j ( s )) q ( ϕ j ( s )) (cid:19) · (cid:18) q ( ϕ j ( s )) p ( ϕ j ( s )) · q (cid:48) ( ϕ j ( s )) p (cid:48) ( ϕ j ( s )) − q (cid:48)(cid:48) ( ϕ j ( s )) p ( ϕ j ( s )) q (cid:48) ( ϕ j ( s )) (cid:19) = p ( ϕ j ( s )) q (cid:48) ( ϕ j ( s )) s · q ( ϕ j ( s )) q (cid:48) ( ϕ j ( s )) p (cid:48) ( ϕ j ( s )) /p ( ϕ j ( s )) − q ( ϕ j ( s )) q (cid:48)(cid:48) ( ϕ j ( s )) q (cid:48) ( ϕ j ( s )) = r ( s ) s · m − ( d − d (cid:18) O (cid:18) | s | /d (cid:19)(cid:19) = − λs r ( s ) (cid:18) O (cid:18) | s | /d (cid:19)(cid:19) . Also, a computation shows that r (cid:48)(cid:48) ( s ) = r ( s ) O (cid:18) s (cid:19) and r (cid:48)(cid:48)(cid:48) ( s ) = r ( s ) O (cid:18) s (cid:19) ulia sets of measure zero s → ∞ . With h ( x ) := ( r ( x ) − r (cid:48) ( x ) + r (cid:48)(cid:48) ( x )) e x , we obtain (cid:90) wx r ( s ) e s ds = r ( w ) e w (cid:18) λw + O (cid:18) | w | /d (cid:19)(cid:19) − h ( x ) − (cid:90) wx r (cid:48)(cid:48)(cid:48) ( s ) e s ds. We have (cid:90) wx r (cid:48)(cid:48)(cid:48) ( s ) e s ds = (cid:90) w −| w | r (cid:48)(cid:48)(cid:48) ( s ) e s ds + (cid:90) −| w |−∞ r (cid:48)(cid:48)(cid:48) ( s ) e s ds − (cid:90) x −∞ r (cid:48)(cid:48)(cid:48) ( s ) e s ds. To estimate (cid:82) w −| w | r (cid:48)(cid:48)(cid:48) ( s ) e s ds , let γ be the part of the circle with centre 0 and radius | w | that connects −| w | and w in G . Then Re s ≤ Re w for s ∈ γ . We obtain (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:90) w −| w | r (cid:48)(cid:48)(cid:48) ( s ) e s ds (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ length( γ ) · max s ∈ γ (cid:12)(cid:12) r (cid:48)(cid:48)(cid:48) ( s ) e s (cid:12)(cid:12) ≤ O ( | w | ) | r ( w ) | O (cid:18) | w | (cid:19) e Re w = | r ( w ) | O (cid:18) | w | (cid:19) | e w | . Let us now estimate (cid:82) −| w |−∞ r (cid:48)(cid:48)(cid:48) ( s ) e s ds . By (4.2), we have | r ( s ) | ∼ | s | − λ as | s | → ∞ .First suppose that λ ≥
0. Using that r (cid:48)(cid:48)(cid:48) ( s ) = r ( s ) O (1 /s ), we obtain (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:90) −| w |−∞ r (cid:48)(cid:48)(cid:48) ( s ) e s ds (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ | r ( w ) | e −| w | O (cid:18) | w | (cid:19) (cid:90) −| w |−∞ e s + | w | ds ≤ | r ( w ) e w | O (cid:18) | w | (cid:19) (cid:90) −∞ e s ds = | r ( w ) e w | O (cid:18) | w | (cid:19) . Now suppose that λ <
0. Then (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:90) −| w |−∞ r (cid:48)(cid:48)(cid:48) ( s ) e s ds (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ O (cid:18) | w | (cid:19) (cid:90) −| w |−∞ | s | − λ e s ds. Integration by parts yields (cid:90) −| w |−∞ | s | − λ e s ds = O ( | w | − λ e −| w | ) ≤ O ( | r ( w ) e w | )and hence (cid:90) −| w |−∞ r (cid:48)(cid:48)(cid:48) ( s ) e s ds = r ( w ) e w O (cid:18) | w | (cid:19) . Altogether, we obtain the desired conclusion with c j = g ( ϕ j ( x )) − h ( x ) + (cid:90) x −∞ r (cid:48)(cid:48)(cid:48) ( s ) e s ds. For the function f from Newton’s method for g , Lemma 4.1 yields the following. Corollary 4.3.
For j ∈ { , ..., d } , we have f ( ϕ j ( w )) = ϕ j ( w ) − q (cid:48) ( ϕ j ( w )) (cid:18) λw + O (cid:18) | w | /d (cid:19)(cid:19) − c j e − w p ( ϕ j ( w )) as w → ∞ in G . In terms of z , Corollary 4.3 says the following. Corollary 4.4.
For j ∈ { , ..., d } , we have f ( z ) = z − q (cid:48) ( z ) (cid:18) λz d + O (cid:18) | z | d +1 (cid:19)(cid:19) − c j e − q ( z ) p ( z ) as z → ∞ in S j . Mareike Wolff
For a more detailed study of the behaviour of f ◦ ϕ j , we will divide the complex planeinto several sets depending on how large | e − w | is compared to some power of | w | . Moreprecisely, we consider sets whose boundary points satisfyRe w = µ log | w | − log α (5.1)for certain µ ∈ R and α >
0. Such sets were also considered by Jankowski [11]. In thissection, we will show that given µ ∈ R , α > y ∈ R of sufficiently large modulus,there is a unique x y ∈ R such that w = x y + iy satisfies (5.1). We also give a proof ofseveral properties of the mapping y (cid:55)→ x y which in part are also shown in [11, § Lemma 5.1.
Let µ ∈ R , α > and y ∈ R with | y | ≥ | µ | . Then there exists a unique x y ∈ R with x y = µ log | x y + iy | − log α. (5.2) If x > x y , then x > µ log | x + iy | − log α. (5.3) If x < x y , then x < µ log | x + iy | − log α. (5.4) Proof.
Let ϕ : R → R , ϕ ( x ) = x − µ log | x + iy | = x − µ x + y ) . Then ϕ ( x ) → ∞ as x → ∞ , and ϕ ( x ) → −∞ as x → −∞ . Thus, ϕ is surjective, sothere exists x y satisfying (5.2).Also, ϕ (cid:48) ( x ) = 1 − µxx + y . Since | µx | x + y ≤ | µ | max {| x | , | y |} max { x , y } = | µ | max {| x | , | y |} ≤ , we have ϕ (cid:48) ( x ) ≥ . Thus, ϕ is strictly increasing, which implies (5.3) and (5.4). In particular, ϕ is injective,so x y is unique.For µ ∈ R and α >
0, let γ µ,α : ( −∞ , − | µ | ] ∪ [2 | µ | , ∞ ) → R , γ µ,α ( y ) = x y . Lemma 5.2.
Let µ ∈ R and α > .(i) The function γ µ,α is continuously differentiable.(ii) If µ > , then lim | y |→∞ γ µ,α ( y ) = ∞ . If µ < , then lim | y |→∞ γ µ,α ( y ) = −∞ . For µ = 0 , γ µ,α ≡ − log α .(iii) | γ (cid:48) µ,α ( y ) | ≤ | µ | / | y | . In particular, lim | y |→∞ γ (cid:48) µ,α ( y ) = 0 uniformly in α . ulia sets of measure zero (iv) For α > β > , we have
23 log αβ ≤ γ µ,β ( y ) − γ µ,α ( y ) ≤ αβ and lim | y |→∞ ( γ µ,β ( y ) − γ µ,α ( y )) = log αβ . Proof.
For µ = 0, the results are obvious. We will prove the Lemma for µ >
0, theproof for µ < x = µ log | x + iy | − log α is equivalent to y = α /µ e (2 /µ ) x − x . The function ψ ( x ) = α /µ e (2 /µ ) x − x satisfies lim x →−∞ ψ ( x ) = −∞ and lim x →∞ ψ ( x ) = ∞ . (5.5)Let x := max { x : ψ ( x ) = 4 µ } . Then ψ ( x ) > µ for x > x . Also, ψ (cid:48) ( x ) = 2 µ α /µ e (2 /µ ) x − x = 2 µ ( ψ ( x ) + x − µx ) . It is not difficult to see that x − µx ≥ − µ / x ∈ R . Thus, ψ (cid:48) ( x ) ≥ µ (cid:18) ψ ( x ) − µ (cid:19) > x > x . In particular, ψ : [ x , ∞ ) → [4 µ , ∞ ) is bijective. This implies that γ µ,α ( y ) = ψ − ( y )is a continuously differentiable function. By (5.5), (ii) is satisfied. Also, by (5.6) andsince y ≥ µ , we have | γ (cid:48) µ,α ( y ) | = (cid:12)(cid:12)(cid:12)(cid:12) yψ (cid:48) ( ψ − ( y )) (cid:12)(cid:12)(cid:12)(cid:12) ≤ | y | (2 /µ )( ψ ( ψ − ( y )) − µ /
4) = µ | y | y − µ / ≤ µ | y | , that is, (iii) is satisfied. To prove (iv), let y ∈ R with | y | ≥ µ be fixed, and let ϕ beas in the proof of Lemma 5.1. Let x := γ µ,α ( y ) and x := γ µ,β ( y ) . Then by the meanvalue theorem, log αβ = ϕ ( x ) − ϕ ( x ) = ϕ (cid:48) ( ξ )( x − x )for some ξ ∈ [ x , x ]. In the proof of Lemma 5.1, we have seen that ϕ (cid:48) ( ξ ) ≥ /
2, andthe same arguments show that ϕ (cid:48) ( ξ ) ≤ /
2. Also, ϕ (cid:48) ( ξ ) → | y | → ∞ .For µ ∈ R , α > ν ≥ | µ | , define H ( µ, α, ν ) := { w : Re w ≥ µ log | w | − log( α ) , | Im w | ≥ ν } = { x + iy : | y | ≥ ν, x ≥ γ µ,α ( y ) } . Also, let Γ( µ, α ) := { w : | Im w | ≥ | µ | , Re w = µ log | w | − log α } = { γ µ,α ( y ) + iy : | y | ≥ | µ |} . Mareike Wolff
Remark 5.3.
Note that if w ∈ Γ( µ, α ), then | e − w | = e − Re w = α | w | − µ ;if w ∈ H ( µ, α, ν ), then | e − w | ≤ α | w | − µ ;and if w ∈ C \ H ( µ, α, ν ) with | Im w | ≥ ν , then | e − w | > α | w | − µ . f Recall that g ( z ) = (cid:90) z p ( t ) e q ( t ) dt + c where p ( t ) = t d + O ( t d − ) and q ( t ) = dt m + O ( t m − ) as t → ∞ , and f is the Newtonmap corresponding to g . Let us assume throughout the rest of the paper that g and f satisfy the assumptions of Theorem 1.1.In this section, we determine the location of the singular values of f . Lemma 6.1. [3, §
7, p.238]
The function f does not have finite asymptotic values. So each singular value of f in C must be a critical value or a limit point of criticalvalues. We have f (cid:48) ( z ) = g ( z ) g (cid:48)(cid:48) ( z ) g (cid:48) ( z ) . Thus, the critical points of f are:1. the zeros of g that are not zeros of g (cid:48) . These are superattracting fixed points of f and form a discrete subset of C .2. the zeros of g (cid:48)(cid:48) that are not zeros of g or g (cid:48) . There are only finitely many of these, z , ..., z N , and by assumption, each z j is attracted by a periodic cycle.In particular, the set of critical values of f does not have limit points in C . So everysingular value of f in C is a critical value, and all but finitely many of them aresuperattracting fixed points. Lemma 6.2.
The set P ( f ) ∩ J ( f ) is finite.Proof. Since the superattracting fixed points of f form a discrete subset of C , the set P ( f ) ∩ J ( f ) is contained in the closure of O + ( { z , ..., z N } ). Each z j is attracted bya periodic cycle C . In particular, O + ( z j ) is bounded and has only finitely many limitpoints. If z j ∈ J ( f ), then Lemma 2.3 yields that z j is eventually mapped to C , so theforward orbit of z j is finite.By Corollary 4.2, g ( z ) = c j + p ( z ) q (cid:48) ( z ) (cid:18) λz d + O (cid:18) z d +1 (cid:19)(cid:19) e q ( z ) as z → ∞ in S j , for j ∈ { , ..., d } . We will see later that if c j (cid:54) = 0, then g has infinitelymany zeros in S j . It is easy to see that this cannot be the case for c j = 0. However,we will show now that under the assumptions of Theorem 1.1, the case c j = 0 does notoccur. ulia sets of measure zero Lemma 6.3. If c j = 0 for some j ∈ { , ..., d } , then f has a Baker domain.Proof. If c j = 0, then Corollary 4.4 yields that f ( z ) = z − dz d − + O (cid:18) z d (cid:19) as z → ∞ in S j . The claim now follows from [6, § §
11] (see also [10, Theorem 2]).
Corollary 6.4.
If the assumptions of Theorem 1.1 are satisfied, then c j (cid:54) = 0 for all j ∈ { , ..., d } .Proof. A theorem by Bergweiler [3, Theorem 2] says that if g and f are defined by (1.2)and (1.1), then every cycle of Baker domains of f contains a singularity of f − . Thiscannot be true under the assumptions of Theorem 1.1.We now investigate the location of the zeros of g . It turns out that the imagesunder q of all but finitely many of them are close to the curves Γ( λ, / | c j | ) defined inSection 5. More precisely, we have the following. Lemma 6.5.
For j ∈ { , ..., d } and k ∈ Z , let v j,k ∈ Γ( λ, / | c j | ) such that Im v j,k = (cid:40) arg( − c j ) + λ ( π/ π ( j − kπ if k ≥ − c j ) + λ ( − π/ πj ) + 2 kπ if k < . If z ∈ S j is a zero of g and | z | is large, then there exists k ∈ Z such that q ( z ) = v j,k + o (1) (6.1) as | z | → ∞ . Vice versa, if j ∈ { , ..., d } and | k | is large, then g has a zero z ∈ S j satisfying (6.1) .Proof. First suppose that z ∈ S j is a zero of g . By Corollary 4.2 and (4.1), g ( z ) = c j + z − dλ (1 + o (1)) e q ( z ) as z → ∞ , and hence e q ( z ) = − c j z dλ (1 + o (1)) . (6.2)Thus, Re q ( z ) = log (cid:12)(cid:12)(cid:12) e q ( z ) (cid:12)(cid:12)(cid:12) = log | c j | + dλ log | z | + o (1)= log | c j | + λ log | q ( z ) | + o (1)= λ log | q ( z ) | − log 1 | c j | + o (1) . In particular, Re q ( z ) = o ( | q ( z ) | ) as z → ∞ and hencearg q ( z ) = ± π o (1) . (6.3)Let us now assume that Im q ( z ) > q ( z ) = π/ o (1) . The proof in thecase where Im q ( z ) < q ( z ) ≡ arg( − c j ) + dλ arg( z ) + o (1) mod 2 π. (6.4)6 Mareike Wolff
We have arg q ( z ) = arg( z d (1 + o (1))) ≡ d arg z + o (1) mod 2 π and hence arg z ≡ d arg q ( z ) + o (1) ≡ π d + o (1) mod 2 πd . Since z ∈ S j , this impliesarg z ≡ π d + 2 π ( j − d + o (1) mod 2 π. (6.5)Inserting (6.5) into (6.4) yieldsIm q ( z ) ≡ arg( − c j ) + λ (cid:16) π π ( j − (cid:17) + o (1) mod 2 π. This completes the proof of the first part of Lemma 6.5.Let us now prove the second part. As before, we will give the proof only for k > k < ϕ j is the branch of q − that maps C \ ( D (0 , R ) ∪ [0 , ∞ )) onto S j . For small ε >
0, let G j,k be the interior of the set of all v ∈ H (cid:18) λ, ε | c j | , | λ | (cid:19) \ H (cid:18) λ, − ε | c j | , | λ | (cid:19) satisfying | Im v − Im v j,k | < π. We will use the minimum principle to show that g ◦ ϕ j has a zero in G j,k . For v ∈ G j,k ,we have Re( v ) = o ( | v | ), and hencearg( v ) = π o (1)as | v | → ∞ . Similar arguments as above and the definition of v j,k yieldarg( ϕ j ( v )) ≡ π d + 2 π ( j − d + o (1) ≡ arg( − c j ) − Im( v j,k ) − dλ + o (1) mod 2 πdλ . (6.6)In particular, this is true for v = v j,k . Also, since v j,k ∈ Γ( λ, / | c j | ), we have e − Re v j,k = 1 | c j | | v j,k | − λ = 1 | c j | | ϕ j ( v j,k ) | − dλ (1 + o (1)) , that is, | ϕ j ( v j,k ) | − dλ = | c j | e − Re v j,k (1 + o (1)) . Using Lemma 4.1 and (4.1), we obtain( g ◦ ϕ j )( v j,k ) = c j + ϕ j ( v j,k ) − dλ (1 + o (1)) exp( v j,k )= c j + | ϕ j ( v j,k ) | − dλ exp( − idλ arg( ϕ j ( v j,k )))(1 + o (1)) exp( v j,k )= c j + | c j | exp( − Re v j,k ) exp( i (arg( − c j ) − Im( v j,k )))(1 + o (1)) exp( v j,k )= c j − c j (1 + o (1)) = o (1) . Next, we will show that g ◦ ϕ j is bounded below on ∂ G j,k . ulia sets of measure zero v ∈ Γ( λ, (1 + ε ) / | c j | ), then | ( g ◦ ϕ j )( v ) | = | c j + ϕ j ( v ) − dλ (1 + o (1)) e v | ≥ || c j | − | v | − λ e Re v (1 + o (1)) | = (cid:12)(cid:12)(cid:12)(cid:12) | c j | − | c j | ε (1 + o (1)) (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) ε | c j | ε − o (1) (cid:12)(cid:12)(cid:12)(cid:12) ≥ ε | c j | , provided ε is sufficiently small and k is sufficiently large. An analogous estimate yieldsthat if v ∈ Γ( λ, (1 − ε ) / | c j | ), then | ( g ◦ ϕ j )( v ) | ≥ (cid:12)(cid:12)(cid:12)(cid:12) | c j | − ε (1 + o (1)) − | c j | (cid:12)(cid:12)(cid:12)(cid:12) ≥ ε | c j | , provided ε is sufficiently small and k is sufficiently large. If Im( v ) = Im( v j,k ) ± π , thenby (6.6), arg( ϕ j ( v ) − dλ e v ) ≡ arg( − c j ) ± π + o (1) ≡ arg( c j ) + o (1) mod 2 π. Thus, for v ∈ G j,k with Im v = Im v j,k ± π , we have | ( g ◦ ϕ j )( v ) | = (cid:12)(cid:12)(cid:12) c j + ϕ j ( v ) − dλ (1 + o (1)) e v (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) | c j | exp( i arg( c j )) + | v | − λ | e v | exp( i arg c j + o (1))(1 + o (1)) (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) | c j | + | v | − λ | e v | (1 + o (1)) (cid:12)(cid:12)(cid:12) ≥ | c j | . We obtain that if k is sufficiently large, then | ( g ◦ ϕ j )( v j,k ) | = o (1) < min v ∈ ∂ G j,k | v | . By the minimum principle, g ◦ ϕ j has a zero w ∈ G j,k . The first part of the Lemmayields that z := ϕ j ( w ) satisfies (6.1). Corollary 6.6.
Let j ∈ { , ..., d } and let z ∈ S j be a zero of g . Then arg( z ) = (cid:40) π/ (2 d ) + 2 π ( j − /d + o (1) if Im q ( z ) > − π/ (2 d ) + 2 πj/d + o (1) if Im q ( z ) < as | z | → ∞ . In particular, dist( z, ∂ S j ) ≥ (cid:18) d + o (1) (cid:19) | z | as | z | → ∞ .Proof. The first part is stated in (6.5) in the case where Im q ( z ) >
0, and follows from(6.3) with similar arguments in the case where Im q ( z ) <
0. We obtaindist( z, ∂ S j ) = sin (cid:16) π d + o (1) (cid:17) | z | ≥ (cid:18) d + o (1) (cid:19) | z | . Mareike Wolff { w : | Im w | ≥ ν }\H (cid:18) λ − , β | c j | , ν (cid:19) H (cid:18) λ − , α | c j | , ν (cid:19) \H (cid:18) λ, β | c j | , ν (cid:19) H (cid:18) λ, | c j | , ν (cid:19) { w : | Im w | ≥ ν }\H (cid:18) λ − , β | c j | , ν (cid:19) H (cid:18) λ − , α | c j | , ν (cid:19) \H (cid:18) λ, β | c j | , ν (cid:19) H (cid:18) λ, | c j | , ν (cid:19) Figure 1: An illustration of the sets H ( λ, / | c j | , ν ), H ( λ − , α / | c j | , ν ) \ H ( λ, β / | c j | , ν )and { w : | Im w | ≥ ν } \ H ( λ − , β / | c j | , ν ) in the case where λ > q ( F ( f )) : first part For j ∈ { , ..., d } , let F j := F ( f ) ∩ S j . In Section 7-9, we will investigate the location and size of q ( F j ) in three differentsubsets of C , using the sets H ( µ, α, ν ) introduced in Section 5. The first subset is H ( λ, / | c j | , ν ), the second one is H ( λ − , α / | c j | , ν ) \ H ( λ, β / | c j | , ν ) for small α > β >
0, and the third set is { w : | Im w | ≥ ν } \ H ( λ − , β / | c j | , ν ) for large β >
0. See Figure 1 for an illustration of these sets.In this section, we investigate the location and size of q ( F j ) in H ( λ, / | c j | , ν ) for j ∈ { , ..., d } and large ν >
0. Recall that the branch ϕ j of q − maps H ( λ, / | c j | , ν ) toa subset of S j . Lemma 7.1.
Let j ∈ { , ..., d } . There exists ν > such that ( f ◦ ϕ j )( H ( λ, / | c j | , ν )) ⊂ S j . In particular, if ( q ◦ f ◦ ϕ j ) k ( w ) ∈ H ( λ, / | c j | , ν ) for all k ∈ { , ..., n − } , then ( f n ◦ ϕ j )( w ) ∈ S j and ( q ◦ f ◦ ϕ j ) n ( w ) = ( q ◦ f n ◦ ϕ j )( w ) . ulia sets of measure zero Proof.
Let w ∈ H ( λ, / | c j | , ν ) . By Corollary 4.3, (4.2) and Remark 5.3, | ( f ◦ ϕ j )( w ) − ϕ j ( w ) |≤ | q (cid:48) ( ϕ j ( w )) | (cid:18) O (cid:18) | w | (cid:19) + | q (cid:48) ( ϕ j ( w )) c j e − w || p ( ϕ j ( w )) | (cid:19) = 1 | q (cid:48) ( ϕ j ( w )) | (cid:18) O (cid:18) | w | (cid:19) + | w | λ | c j e − w | (cid:18) O (cid:18) | w | /d (cid:19)(cid:19)(cid:19) ≤ | q (cid:48) ( ϕ j ( w )) | = 3 | ϕ (cid:48) j ( w ) | if | w | is sufficiently large. For ν ≥
12 + R , with R as in Section 3, we obtain f ( ϕ j ( w )) ∈ D ( ϕ j ( w ) , | ϕ (cid:48) j ( w ) | ) ⊂ D (cid:18) ϕ j ( w ) , ν − R | ϕ (cid:48) j ( w ) | (cid:19) . On the other hand, by Koebe’s 1 / S j ⊃ ϕ j ( D ( w, ν − R )) ⊃ D (cid:18) ϕ j ( w ) , ν − R | ϕ (cid:48) j ( w ) | (cid:19) , whence the claim follows.Next, we derive an asymptotic expression for h j ( w ) = ( q ◦ f ◦ ϕ j )( w )in H ( λ, / | c j | , ν ) for large ν > Lemma 7.2.
Let j ∈ { , ..., d } . There exists ν > such that h j ( w ) = w − m + 1 − d d · w + O (cid:18) | w | /d (cid:19) − c j e − w ϕ j ( w ) dλ (cid:18) O (cid:18) | w | /d (cid:19)(cid:19) (7.1) as w → ∞ in H ( λ, / | c j | , ν ) . Remark 7.3.
In fact, for any α >
0, there exists ν > h j has an asymp-totic expression of the form (7.1) in H ( λ, α/ | c j | , ν ). We will need that α > H ( λ, α/ | c j | , ν ) ⊃ H ( λ, / | c j | , ν ). Proof of Lemma 7.2.
By Corollary 4.3, we have f ( ϕ j ( w )) = ϕ j ( w ) − η ( w ) q (cid:48) ( ϕ j ( w ))where η ( w ) = 1 + λw + O (cid:18) | w | /d (cid:19) + c j e − w q (cid:48) ( ϕ j ( w )) p ( ϕ j ( w ))as | w | → ∞ . Note that η is bounded in H ( λ, / | c j | , ν ). Taylor expansion of q around ϕ j ( w ) yields h j ( w ) = q ( f ( ϕ j ( w ))) = d (cid:88) k =0 k ! q ( k ) ( ϕ j ( w ))( f ( ϕ j ( w )) − ϕ j ( w )) k = d (cid:88) k =0 ( − k k ! q ( k ) ( ϕ j ( w )) q (cid:48) ( ϕ j ( w )) k η ( w ) k = w − η ( w ) + 12 q (cid:48)(cid:48) ( ϕ j ( w )) q (cid:48) ( ϕ j ( w )) η ( w ) + d (cid:88) k =3 ( − k k ! O (cid:18) w k − (cid:19) η ( w ) k = w − η ( w ) + 12 q (cid:48)(cid:48) ( ϕ j ( w )) q (cid:48) ( ϕ j ( w )) η ( w ) + O (cid:18) w (cid:19) (7.2)0 Mareike Wolff as w → ∞ in H ( λ, / | c j | , ν ). Using that λ = ( d − − m ) /d , we have − η ( w ) = − m + 1 − dd · w + O (cid:18) | w | /d (cid:19) − c j e − w ϕ j ( w ) d − − m (cid:18) O (cid:18) | w | /d (cid:19)(cid:19) . (7.3)Moreover, q (cid:48)(cid:48) ( ϕ j ( w )) q (cid:48) ( ϕ j ( w )) = d − d · w (cid:18) O (cid:18) | w | /d (cid:19)(cid:19) and η ( w ) = (cid:18) O (cid:18) w (cid:19) + c j e − w ϕ j ( w ) dλ (cid:18) O (cid:18) | w | /d (cid:19)(cid:19)(cid:19) = 1 + O (cid:18) w (cid:19) + c j e − w ϕ j ( w ) dλ · O (1) . Hence,12 q (cid:48)(cid:48) ( ϕ j ( w )) q (cid:48) ( ϕ j ( w )) η ( w ) = d − d w + O (cid:18) | w | /d (cid:19) + c j e − w ϕ j ( w ) dλ O (cid:18) w (cid:19) . (7.4)Combining (7.2), (7.3) and (7.4) yields the desired conclusion.For the derivative of h j , we obtain the following. Lemma 7.4.
Let j ∈ { , ..., d } . There exists ν > such that h (cid:48) j ( w ) = 1 + O (cid:18) | w | /d (cid:19) + c j e − w ϕ j ( w ) dλ (cid:18) O (cid:18) | w | /d (cid:19)(cid:19) as w → ∞ in H ( λ, / | c j | , ν ) .Proof. Suppose ν ≥ ν + 1. By Lemma 7.2, there are holomorphic functions, a , a ,satisfying a ( w ) = O (1 / | w | /d ) and a ( w ) = O (1 / | w | /d ) as w → ∞ such that h j ( w ) = w − m + 1 − d d · w + a ( w ) − c j e − w ϕ j ( w ) dλ (1 + a ( w ))for w ∈ H ( λ, / | c j | , ν − a (cid:48) ( w ) = O (1 / | w | /d ) and a (cid:48) ( w ) = O (1 / | w | /d ) as w → ∞ in H ( λ, / | c j | , ν ). Also,dd w e − w ϕ j ( w ) dλ = − e − w ϕ j ( w ) dλ (cid:18) − dλϕ j ( w ) ϕ (cid:48) j ( w ) (cid:19) = − e − w ϕ j ( w ) dλ (cid:18) O (cid:18) w (cid:19)(cid:19) . Thus, dd w (cid:16) c j e − w ϕ j ( w ) dλ (1 + a ( w )) (cid:17) = − c j e − w ϕ j ( w ) dλ (cid:18) O (cid:18) | w | /d (cid:19)(cid:19) . We obtain h (cid:48) j ( w ) = 1 − m + 1 − d d w + a (cid:48) ( w ) + c j e − w ϕ j ( w ) dλ (cid:18) O (cid:18) | w | /d (cid:19)(cid:19) = 1 + O (cid:18) | w | /d (cid:19) + c j e − w ϕ j ( w ) dλ (cid:18) O (cid:18) | w | /d (cid:19)(cid:19) as w → ∞ in H ( λ, / | c j | , ν ). ulia sets of measure zero z ∈ S j is a superattracting fixedpoint of f , then q ( z ) lies close to the curve Γ( λ, / | c j | ). Also, every horizontal stripof width 2 π + ε with ε > z is a superattracting fixedpoint of f and A ∗ ( z ) is its immediate basin of attraction, then q ( A ∗ ( z )) contains adisk of a fixed radius around q ( z ). We then consider preimages of this disk underiterates of h j = q ◦ f ◦ ϕ j . The function h j is not locally invertible at q ( z ), butif α is slightly smaller than 1, then h j has a local inverse function, ψ j , defined in H ( λ, α/ | c j | , ν ). If α is sufficiently close to 1, then H ( λ, α/ | c j | , ν ) intersects the diskscontained in q ( A ∗ ( z )). We then show that the images of this intersection under ψ j have a certain size and are more or less evenly distributed in H ( λ, / | c j | , ν ). The ideahere is that if w ∈ H ( λ, / | c j | , ν ) is sufficiently far from the boundary, then Lemma 7.2yields h j ( w ) ≈ w −
1, and hence ψ j ( w ) ≈ w + 1. See Figure 2 for an illustration of theabovementioned approach. ψ j ψ j ψ j ψ j ψ j ψ j Figure 2: The images of superattracting fixed points of f under q lie close to thedashed line. The white disks around them are contained in the images of the basins ofattraction of the superattracting fixed points. To the right of the solid line, the inverse ψ j of h j is defined. The grey disks lie in the intersection of the images of the basins ofattraction under q and the domain of definition of ψ j , and their images under iterationof ψ j are contained in q ( F j ). Lemma 7.5. If z is a zero of g but not a zero of g (cid:48) and | z | is sufficiently large, then A ∗ ( z ) ⊃ D (cid:18) z , d | z | d − (cid:19) . For the proof, we require the following theorem which essentially says that undersuitable assumptions the solution of B¨ottcher’s functional equation in a neighbour-hood of a superattracting fixed point extends to a conformal map defined in the entireimmediate basin of attraction.
Theorem 7.6.
Let h be a meromorphic function, and let z be a superattracting fixedpoint of multiplicity k of h . Suppose that A ∗ ( z ) contains no critical point other than Mareike Wolff z and no asymptotic value of h . Then there is a conformal map Φ : D (0 , → A ∗ ( z ) satisfying Φ(0) = z and h (Φ( z )) = Φ( z k ) for all z ∈ D (0 , . A proof of this theorem can be found, for example, in [23, p. 65, Theorem 4]. There,the result is stated for rational functions, but the proof also works for meromorphicfunctions without asymptotic values in A ∗ ( z ). Proof of Lemma 7.5.
Let z be a zero of g that is not a zero of g (cid:48) , and assume thatnone of the finitely many zeros of g (cid:48)(cid:48) lies in A ∗ ( z ). Then z is a superattracting fixedpoint of f , and there are no other critical points of f in A ∗ ( z ). Also, f (cid:48)(cid:48) ( z ) = g (cid:48) ( z ) g (cid:48)(cid:48) ( z ) + g ( z ) g (cid:48) ( z ) g (cid:48)(cid:48)(cid:48) ( z ) − g ( z ) g (cid:48)(cid:48) ( z ) g (cid:48) ( z ) , and hence f (cid:48)(cid:48) ( z ) = g (cid:48)(cid:48) ( z ) g (cid:48) ( z ) (cid:54) = 0 . By Theorem 7.6, there is a conformal map Φ : D (0 , → A ∗ ( z ) satisfying f (Φ( z )) =Φ( z ) and Φ(0) = z . Differentiating the equation f (Φ( z )) = Φ( z ) twice yields f (cid:48)(cid:48) (Φ( z ))Φ (cid:48) ( z ) + f (cid:48) (Φ( z ))Φ (cid:48)(cid:48) ( z ) = 2Φ (cid:48) ( z ) + 4 z Φ (cid:48)(cid:48) ( z ) . For z = 0, we obtain f (cid:48)(cid:48) ( z )Φ (cid:48) (0) = 2Φ (cid:48) (0)and hence | Φ (cid:48) (0) | = 2 | f (cid:48)(cid:48) ( z ) | . We have f (cid:48)(cid:48) ( z ) = g (cid:48)(cid:48) ( z ) g (cid:48) ( z ) = ( p ( z ) q (cid:48) ( z ) + p (cid:48) ( z )) e q ( z ) p ( z ) e q ( z ) = q (cid:48) ( z ) + p (cid:48) ( z ) p ( z ) = dz d − (cid:18) O (cid:18) z (cid:19)(cid:19) . Hence, by Koebe’s 1 / A ∗ ( z ) = Φ( D (0 , ⊃ D (cid:18) z , | Φ (cid:48) (0) | (cid:19) = D (cid:18) z , | f (cid:48)(cid:48) ( z ) | (cid:19) ⊃ D (cid:18) z , d | z | d − (cid:19) if | z | is sufficiently large. Corollary 7.7.
Let z ∈ C be a zero of g that is not a zero of g (cid:48) . If | z | is sufficientlylarge, then q ( A ∗ ( z )) ⊃ D (cid:18) q ( z ) , (cid:19) . Proof. If | z | is sufficiently large, then by Lemma 7.5, A ∗ ( z ) ⊃ D (cid:18) z , d | z | d − (cid:19) , and q is injective in this disk. Koebe’s 1 / q ( A ∗ ( z )) ⊃ q (cid:18) D (cid:18) z , d | z | d − (cid:19)(cid:19) ⊃ D (cid:18) q ( z ) , | q (cid:48) ( z ) | d | z | d − (cid:19) . Since q (cid:48) ( z ) = dz d − (1 + O (1 /z )) as z → ∞ , the claim follows. ulia sets of measure zero h j . Lemma 7.8.
Let α ∈ (0 , , ε ∈ (0 , − α ) and j ∈ { , ..., d } . There exists ν > such that for each w ∈ H ( λ, α/ | c j | , ν ) , there is a unique w ∈ H ( λ, α/ | c j | , ν − with h j ( w ) = w . More precisely, w ∈ D ( w + 1 , α + ε ) .Proof. Let w ∈ H ( λ, α/ | c j | , ν − . By Lemma 7.2 and Remark 5.3, | h j ( w ) − ( w − | ≤ O (cid:18) | w | (cid:19) + | c j e − w || w | λ (cid:18) O (cid:18) | w | /d (cid:19)(cid:19) ≤ O (cid:18) | w | (cid:19) + α (cid:18) O (cid:18) | w | /d (cid:19)(cid:19) < α + ε, provided ν and hence | w | is sufficiently large. If h j ( w ) = w , we obtain | w − ( w + 1) | = | w − ( w − | = | h j ( w ) − ( w − | < α + ε, that is, w ∈ D ( w + 1 , α + ε ). On the other hand, Lemma 5.2 yields that D ( w + 1 , α + ε ) ⊂ H ( λ, α/ | c j | , ν − ν is sufficiently large. Thus, for w ∈ ∂ D ( w + 1 , α + ε ), | ( h j ( w ) − w ) − ( w − − w ) | = | h j ( w ) − ( w − | < α + ε = | w − − w | . By Rouch´e’s theorem, there is a unique w ∈ D ( w + 1 , α + ε ) satisfying h j ( w ) = w .By Lemma 7.8, there is a subset H j ⊂ H ( λ, α/ | c j | , ν −
1) such that h j maps H j conformally onto H ( λ, α/ | c j | , ν ). Let ψ j : H ( λ, α/ | c j | , ν ) → H j be the correspondinginverse function. The next Lemma yields that if | Im w | is sufficiently large, then alliterates ψ nj ( w ) are defined and tend to ∞ as n → ∞ in a horizontal strip whose widthis bounded independent of w . Lemma 7.9.
Let α ∈ (0 , , ε ∈ (0 , − α ) and j ∈ { , ..., d } . Then there exist ν > and C > such that ψ nj ( w ) is defined for all w ∈ H ( λ, α/ | c j | , ν ) and all n ∈ N , andsatisfies(i) Re ψ nj ( w ) ≥ Re w + n (1 − α − ε ); (ii) | ψ nj ( w ) | ≥ max { n, | w |} · − α − ε ;(iii) | Im ψ nj ( w ) − Im w | ≤ C ;(iv) e − Re ψ nj ( w ) | ψ nj ( w ) | λ = O ( e − n (1 − α − ε ) / ) . For the proof, we require the following Lemma.
Lemma 7.10.
For all n ∈ N , we have ∞ (cid:88) n = n k ≤ n . Proof.
We have ∞ (cid:88) n = n k ≤ n + ∞ (cid:88) k = n +1 (cid:90) kk − t dt = 1 n + (cid:90) ∞ n t dt = 1 n + 1 n ≤ n . Mareike Wolff
Proof of Lemma 7.9.
Let δ := 1 − α − ε. First note that if ψ nj ( w ) is defined, then Lemma 7.8 yields that ψ kj ( w ) ∈ D ( ψ k − j ( w ) +1 , α + ε ) for all k ∈ { , ..., n } , and henceRe ψ nj ( w ) ≥ Re w + nδ. So ψ nj ( w ) satisfies (i). Also, if n ≤ | w | /
2, then | ψ nj ( w ) | ≥ | w | − n (1 + α + ε ) ≥ | w | − | w | α + ε ) = | w | δ ≥ nδ. If n > | w | /
2, then | ψ nj ( w ) | ≥ Re ψ nj ( w ) ≥ Re w + nδ ≥ λ log | w | − log α | c j | + nδ ≥ nδ ≥ | w | δ , provided | w | and hence also n is sufficiently large. In particular, ψ nj ( w ) satisfies (ii).Let n w := (cid:98)| w |(cid:99) . We will show by induction that if w ∈ H ( λ, α/ | c j | , ν ) for sufficiently large ν >
0, then ψ nj ( w ) is defined for all n ∈ N and | Im ψ nj ( w ) − Im w | ≤ C (cid:48) min (cid:26) n | w | , (cid:27) + n w n (cid:88) k = n w k + n (cid:88) k =1 k /d + n (cid:88) k =1 e − kδ/ , (7.5)where C (cid:48) does not depend on w or n . Note that by Lemma 7.10, C (cid:48) min (cid:26) n | w | , (cid:27) + n w n (cid:88) k = n w k + n (cid:88) k =1 k /d + n (cid:88) k =1 e − kδ/ ≤ C (cid:48) (cid:32) ∞ (cid:88) k =1 k /d + ∞ (cid:88) k =1 e − kδ/ (cid:33) =: C < ∞ . So (7.5) implies (iii). Clearly, (7.5) is true for n = 0. Now suppose that (7.5) holdswith n replaced by n −
1. By Lemma 7.8, ψ nj ( w ) is defined if and only if | Im ψ n − j ( w ) | >ν . This is satisfied if | Im w | > ν + C . By Lemma 7.2, | Im ψ nj ( w ) − Im ψ n − j ( w ) | = | Im ψ nj ( w ) − Im h j ( ψ nj ( w )) |≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) m + 1 − d d Im (cid:32) ψ nj ( w ) (cid:33)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + O (cid:32) | ψ nj ( w ) | /d (cid:33) + 2 | c j | e − Re ψ nj ( w ) | ψ nj ( w ) | λ , provided | w | is sufficiently large. By (ii),1 | ψ nj ( w ) | /d = O (cid:18) n /d (cid:19) . If n ≤ | w | , then we estimate the first summand by (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Im (cid:32) ψ nj ( w ) (cid:33)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ | ψ nj ( w ) | = O (cid:18) | w | (cid:19) . ulia sets of measure zero n > | w | , then by Lemma 7.8, (ii) and the induction hypothesis, (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Im (cid:32) ψ nj ( w ) (cid:33)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = | Im ψ nj ( w ) || ψ nj ( w ) | ≤ | Im ψ n − j ( w ) | + α + ε | ψ nj ( w ) | ≤ | Im w | + C + α + ε ) δ n ≤ | w | δ n = n w · O (cid:18) n (cid:19) , provided | w | is sufficiently large.Moreover, if λ ≥
0, then by (i), Lemma 7.8 and Remark 5.3, | c j | e − Re ψ nj ( w ) | ψ nj ( w ) | λ ≤ | c j | e − Re w ( | w | + n (1 + α + ε )) λ e − nδ ≤ α | w | − λ ( | w | + n (1 + α + ε )) λ e − nδ = α (cid:18) n | w | (1 + α + ε ) (cid:19) λ e − nδ ≤ α (1 + n (1 + α + ε )) λ e − nδ = O ( e − nδ/ ) , provided | w | ≥
1. If λ <
0, then by (i), (ii) and Remark 5.3, | c j | e − Re ψ nj ( w ) | ψ nj ( w ) | λ ≤ (cid:18) δ (cid:19) λ | w | λ | c j | e − Re w e − nδ ≤ (cid:18) δ (cid:19) λ αe − nδ . In particular, (iv) is satisfied. Also, if n ≤ | w | , thenIm ψ nj ( w ) − Im ψ n − j ( w ) = O (cid:18) | w | (cid:19) + O (cid:18) n /d (cid:19) + O (cid:16) e − nδ/ (cid:17) , and if n > | w | , thenIm ψ nj ( w ) − Im ψ n − j ( w ) = n w O (cid:18) n (cid:19) + O (cid:18) n /d (cid:19) + O ( e − nδ/ ) . Thus, ψ nj ( w ) satisfies (7.5), and hence also (iii).We now estimate the derivative of ψ nj . Lemma 7.11.
Let α ∈ (0 , and j ∈ { , ..., d } . There are ν > and B > such that | ( ψ nj ) (cid:48) ( w ) | ≥ B for all w ∈ H ( λ, α/ | c j | , ν ) and all n ∈ N .Proof. We have( ψ nj ) (cid:48) = n − (cid:89) k =0 ψ (cid:48) j ◦ ψ kj = 1 (cid:81) n − k =0 h (cid:48) j ◦ ψ k +1 j = 1 (cid:81) nk =1 h (cid:48) j ◦ ψ kj . By Lemma 7.4 and Lemma 7.9, | h (cid:48) j ( ψ kj ( w )) | ≤ O (cid:32) | ψ kj ( w ) | /d (cid:33) + | c j | e − Re ψ kj ( w ) | ψ kj ( w ) | λ (cid:32) O (cid:32) | ψ kj ( w ) | /d (cid:33)(cid:33) ≤ O (cid:18) k /d (cid:19) + O (cid:16) e − k (1 − α − ε ) / (cid:17) . Since the infinite product (cid:81) ∞ k =1 (1+ O (1 /k /d )+ O ( e − k (1 − α − ε ) / )) converges, we obtainthe desired conclusion.6 Mareike Wolff
Recall that F j = F ( f ) ∩ S j . Lemma 7.12.
For j ∈ { , ..., d } and k ∈ Z , let v j,k be as in Lemma 6.5 and w j,k := v j,k + 1 / . There is ϑ > such that if | k | is sufficiently large, then D ( ψ nj ( w j,k ) , ϑ ) ⊂ q ( F j ) for all n ∈ N . Remark 7.13.
For sufficiently large | k | , the point v j,k is close to q ( z ) for some at-tracting fixed point z of f . The function ψ j is not defined in q ( z ) and v j,k . Therefore,we introduce the point w j,k which is in the domain of definition of ψ j for large | k | andalso lies in q ( A ∗ ( z )). Proof of Lemma 7.12.
By Lemma 6.5, there is a zero z of g satisfying q ( z ) = v j,k + o (1). Thus, w j,k = q ( z ) + 1 /
26 + o (1). If | k | is sufficiently large, we obtain D (cid:18) w j,k , (cid:19) ⊂ D (cid:18) q ( z ) + 126 , (cid:19) ⊂ D (cid:18) q ( z ) , (cid:19) . By Corollary 7.7, this yields D (cid:18) w j,k , (cid:19) ⊂ q ( A ∗ ( z )) . Let ν > − / / − / < α < . Then2 log 1 α < − . (7.6)Since v j,k ∈ Γ( λ, / | c j | ), by (7.6) and Lemma 5.2 (iv) and (iii), we get D (cid:18) w j,k , (cid:19) ⊂ H (cid:18) λ, α | c j | , ν (cid:19) if | k | is sufficiently large. By Koebe’s 1 / ψ nj (cid:18) D (cid:18) w j,k , (cid:19)(cid:19) ⊃ D (cid:18) ψ nj ( w j,k ) , | ( ψ nj ) (cid:48) ( w j,k ) | · (cid:19) ⊃ D (cid:18) ψ nj ( w j,k ) , B · (cid:19) . With ϑ := B/ (4 ·
27) we thus have h nj ( D ( ψ nj ( w j,k ) , ϑ )) ⊂ D (cid:18) w j,k , (cid:19) ⊂ q ( F j ) . Since by Lemma 7.1, h nj ( w ) = ( q ◦ f ◦ ϕ j ) n ( w ) = ( q ◦ f n ◦ ϕ j )( w )for w ∈ H ( λ, α/ | c j | , ν ), this implies D ( ψ nj ( w j,k ) , ϑ ) ⊂ q ( F j ) , provided | k | is sufficiently large.The final result of this section says that q ( F j ) has positive density in rectangles ofsufficiently large side lengths that are contained in H ( λ, / | c j | , ν ). Lemma 7.14.
There are D , ν, η > such that for all j ∈ { , ..., d } and any rectangle R ⊂ H ( λ, / | c j | , ν ) with sides parallel to the real and imaginary axis whose vertical andhorizontal side lengths are both at least D , we have dens( q ( F j ) , R ) ≥ η . ulia sets of measure zero Proof.
First suppose that R = { w : x ≤ Re w ≤ x , y ≤ Im ≤ y } (7.7)where 2 π + 2( C + ϑ ) ≤ x − x , y − y ≤ π + 2( C + ϑ )) , (7.8)with C as in Lemma 7.9 and ϑ as in Lemma 7.12. Let v j,k be as in Lemma 6.5 and w j,k = v j,k + 1 /
26. There is k ∈ Z such that y + C + ϑ ≤ Im w j,k = Im v j,k ≤ y − C − ϑ .Also, by Lemma 7.8, there is n ∈ N such that x + ϑ < Re ψ nj ( w j,k ) < x − ϑ . By Lemma7.9, we have y + ϑ ≤ Im ψ nj ( w j,k ) ≤ y − ϑ . Thus, D ( ψ nj ( w j,k ) , ϑ ) ⊂ R . Also, by Lemma 7.12, D ( ψ nj ( w j,k ) , ϑ ) ⊂ q ( F j ) . We obtaindens( q ( F j ) , R ) ≥ meas( D ( ψ nj ( w j,k ) , ϑ ))meas R ≥ πϑ π + 2( C + ϑ )) =: η . Now, if
R ⊂ H ( λ, / | c j | , ν ) is any rectangle whose horizontal and vertical side lengthboth exceed D := 2 π + 2( C + ϑ ), then R can be written as the union of rectangles ofthe form (7.7) that satisfy (7.8) and have pairwise disjoint interior, whence the claimfollows. q ( F ( f )) : second part In this section, we investigate the density of q ( F ( f )) in subsets of H ( λ − , α / | c j | , ν ) \H ( λ, β / | c j | , ν ) for small α > β > h j in H ( λ − , α / | c j | , ν ) \H ( λ, β / | c j | , ν ). Lemma 8.1.
Let ε > and j ∈ { , ..., d } . Then there are α , β , ν > such that forall w ∈ H ( λ − , α / | c j | , ν ) \ H ( λ, β / | c j | , ν ) , we have (cid:12)(cid:12)(cid:12)(cid:12) h j ( w ) − w − c j e − w ϕ j ( w ) dλ − (cid:12)(cid:12)(cid:12)(cid:12) < ε. Proof.
Taylor expansion of q around ϕ j ( w ) yields h j ( w ) = q ( f ( ϕ j ( w ))) = w + d (cid:88) k =1 q ( k ) ( ϕ j ( w )) k ! ( f ( ϕ j ( w )) − ϕ j ( w )) k . Thus, h j ( w ) − w − c j e − w ϕ j ( w ) dλ − q (cid:48) ( ϕ j ( w ))( f ( ϕ j ( w )) − ϕ j ( w )) − c j e − w ϕ j ( w ) dλ − d (cid:88) k =2 q ( k ) ( ϕ j ( w )) k ! ( f ( ϕ j ( w )) − ϕ j ( w )) k − c j e − w ϕ j ( w ) dλ . (8.1)By Corollary 4.3, f ( ϕ j ( w )) = ϕ j ( w ) − q (cid:48) ( ϕ j ( w )) (cid:18) O (cid:18) | w | (cid:19) + c j e − w ϕ j ( w ) dλ (cid:18) O (cid:18) | w | /d (cid:19)(cid:19)(cid:19) (8.2)8 Mareike Wolff as w → ∞ . Hence, q (cid:48) ( ϕ j ( w ))( f ( ϕ j ( w )) − ϕ j ( w )) − c j e − w ϕ j ( w ) dλ − O (1 / | w | ) c j e − w ϕ j ( w ) dλ + O (cid:18) | w | /d (cid:19) . (8.3)For w ∈ C \ H ( λ, β / | c j | , ν ) with | Im w | ≥ ν , we have (cid:12)(cid:12)(cid:12) c j e − w ϕ j ( w ) dλ (cid:12)(cid:12)(cid:12) ≥ β , provided ν is sufficiently large. Inserting this into (8.3) yields (cid:12)(cid:12)(cid:12)(cid:12) q (cid:48) ( ϕ j ( w ))( f ( ϕ j ( w )) − ϕ j ( w )) − c j e − w ϕ j ( w ) dλ − (cid:12)(cid:12)(cid:12)(cid:12) ≤ β + O (cid:18) | w | /d (cid:19) < εd (8.4)if β and | w | are sufficiently large.Also, for w ∈ H ( λ − , α / | c j | , ν ), we have (cid:12)(cid:12)(cid:12) c j e − w ϕ j ( w ) d ( λ − (cid:12)(cid:12)(cid:12) ≤ α , provided ν is sufficiently large. By (8.2), this yields | f ( ϕ j ( w )) − ϕ j ( w ) | ≤ | q (cid:48) ( ϕ j ( w )) | (cid:18) O (cid:18) | w | (cid:19) + 3 α ϕ j ( w ) d (cid:19) ≤ d α | ϕ j ( w ) | , (8.5)provided | w | is sufficiently large. If k ≥
2, then by (8.4) and (8.5), we have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q ( k ) ( ϕ j ( w )) k ! · ( f ( ϕ j ( w )) − ϕ j ( w )) k − c j e − w ϕ j ( w ) dλ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) q (cid:48) ( ϕ j ( w ))( f ( ϕ j ( w )) − ϕ j ( w )) − c j e − w ϕ j ( w ) dλ (cid:12)(cid:12)(cid:12)(cid:12) · (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q ( k ) ( ϕ j ( w )) k ! q (cid:48) ( ϕ j ( w )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) · | f ( ϕ j ( w )) − ϕ j ( w ) | k − ≤ (cid:16) εd (cid:17) · (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) q ( k ) ( ϕ j ( w )) k ! q (cid:48) ( ϕ j ( w )) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) · (cid:18) d α | ϕ j ( w ) | (cid:19) k − ≤ (cid:16) εd (cid:17) (cid:18) dk (cid:19) d | ϕ j ( w ) | − k +1 (cid:18) d α | ϕ j ( w ) | (cid:19) k − = (cid:16) εd (cid:17) (cid:18) dk (cid:19) · k − d k α k − < εd (8.6)if | w | is sufficiently large and α is sufficiently small. Inserting (8.4) and (8.6) into (8.1)yields the desired conclusion.We will now proceed as follows. First, we show that h j maps the intersectionof certain horizontal strips with H ( λ − , α / | c j | , ν ) \ H ( λ, β / | c j | , ν ) into H ( λ, /c ∗ , ν )where c ∗ = max l | c l | . The idea is that if Im w lies in certain intervals, then the argumentof − c j e − w ϕ j ( w ) dλ is small, and using that h j ( w ) ≈ w − c j e − w ϕ j ( w ) dλ by Lemma 8.1,one can deduce that Re h j ( w ) is large. By Section 7, the set q ( F ( f )) has positivedensity in large bounded subsets of H ( λ, /c ∗ , ν ). Together with the invariance of F ( f )under f , we deduce that q ( F ( f )) has positive density in large bounded subsets of H ( λ − , α / | c j | , ν ) \ H ( λ, β / | c j | , ν ).The next Lemma deals with the mapping behaviour of f in certain horizontal stripsin H ( λ − , α / | c j | , ν ) \ H ( λ, β / | c j | , ν ). For j ∈ { , ..., d } and n ∈ Z , let y jn := (cid:40) arg( − c j ) + λ ( π/ π ( j − nπ if n ≥ − c j ) + λ ( − π/ πj ) + 2 nπ if n < . ulia sets of measure zero Lemma 8.2.
Let ε ∈ (0 , π/ . Then there are α , β , ν > such that the followingholds. Let j ∈ { , ..., d } . Suppose that w lies in the closure of H ( λ − , α / | c j | , ν ) \H ( λ, β / | c j | , ν ) and there exists n ∈ Z with | Im w − y jn | ≤ π/ . Let β ≥ β such that w ∈ Γ( λ, β/ | c j | ) , and let θ := Im w − y jn . Then, | h j ( w ) − w | ≤ (1 + ε ) β, (1 − ε ) β cos( | θ | + ε ) ≤ Re( h j ( w ) − w ) ≤ (1 + ε ) β and (1 − ε ) β sin( | θ | − ε ) ≤ | Im( h j ( w ) − w ) | ≤ (1 + ε ) β sin( | θ | + ε ) . Proof.
By Lemma 8.1, (cid:12)(cid:12)(cid:12)(cid:12) h j ( w ) − w − c j e − w ϕ j ( w ) dλ − (cid:12)(cid:12)(cid:12)(cid:12) ≤ ε , (8.7)provided α is sufficiently small and β and ν are sufficiently large. Thus, (cid:16) − ε (cid:17) (cid:12)(cid:12)(cid:12) c j e − w ϕ j ( w ) dλ (cid:12)(cid:12)(cid:12) ≤ | h j ( w ) − w | ≤ (cid:16) ε (cid:17) (cid:12)(cid:12)(cid:12) c j e − w ϕ j ( w ) dλ (cid:12)(cid:12)(cid:12) . Since w ∈ Γ( λ, β/ | c j | ), this yields(1 − ε ) β ≤ | h j ( w ) − w | ≤ (1 + ε ) β if ν is sufficiently large. Also, by (8.7), (cid:12)(cid:12)(cid:12)(cid:12) arg (cid:18) h j ( w ) − w − c j e − w ϕ j ( w ) dλ (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤ arcsin (cid:16) ε (cid:17) ≤ π ε. (8.8)We havearg w = arg q ( ϕ j ( w )) ≡ arg( ϕ j ( w ) d (1 + o (1))) ≡ d arg ϕ j ( w ) + o (1) mod 2 π as w → ∞ . Since w lies in the closure of H ( λ − , α / | c j | , ν ) \ H ( λ, β / | c j | , ν ), we havearg w = sgn(Im( w )) π o (1) = sgn( n ) π o (1)if | n | is sufficiently large. Hence,arg ϕ j ( w ) ≡ sgn( n ) π d + o (1) mod 2 πd . Since ϕ j ( w ) ∈ S j , we obtainarg ϕ j ( w ) ≡ (cid:40) π/ (2 d ) + 2 π ( j − /d + o (1) if n > − π/ (2 d ) + 2 πj/d + o (1) if n < π. Thus, arg (cid:16) − c j e − w ϕ j ( w ) dλ (cid:17) ≡ arg( − c j ) − Im( w ) + dλ arg ϕ j ( w ) ≡ − θ − nπ + o (1) ≡ − θ + o (1) mod 2 π. By (8.8), this implies | θ | − ε ≤ | arg( h j ( w ) − w ) | ≤ | θ | + ε if | w | is sufficiently large. We obtainRe( h j ( w ) − w ) ≤ | h j ( w ) − w | ≤ (1 + ε ) β, Re( h j ( w ) − w ) = | h j ( w ) − w | cos(arg( h j ( w ) − w )) ≥ (1 − ε ) β cos( | θ | + ε ) , | Im( h j ( w ) − w ) | = | h j ( w ) − w | · | sin(arg( h j ( w ) − w )) | ≤ (1 + ε ) β sin( | θ | + ε ) , | Im( h j ( w ) − w ) | = | h j ( w ) − w | · | sin(arg( h j ( w ) − w )) | ≥ (1 − ε ) β sin( | θ | − ε ) . Mareike Wolff
Let c ∗ := max ≤ l ≤ d | c l | . (8.9)The following Lemma says that h j maps the intersection of H ( λ − , α / | c j | , ν ) \H ( λ, β / | c j | , ν ) with certain horizontal strips into H ( λ, /c ∗ , ν ). Lemma 8.3.
There are α , β , ν > such that for all j ∈ { , ..., d } , n ∈ Z and all w in the closure of H ( λ − , α / | c j | , ν ) \ H ( λ, β / | c j | , ν ) with | Im w − y jn | ≤ π/ , we have h j ( w ) ∈ H ( λ, /c ∗ , ν ) .Proof. We have w ∈ Γ( λ, β/ | c j | ) for some β ≥ β . Since also w ∈ H ( λ − , α / | c j | , ν ),we have λ log | w | − log β = Re w ≥ ( λ −
1) log | w | − log α and hence β ≤ α | w | . (8.10)Let ε ∈ (0 , π/ θ := Im w − y jn , and suppose that α , β , ν are chosen such that theconclusion of Lemma 8.2 holds. Then | Im h j ( w ) − Im w | ≤ (1 + ε ) β sin( | θ | + ε ) ≤ β. By (8.10) and since | Im w | = (1 + o (1)) | w | for w in the closure of H ( λ − , α / | c j | , ν ) \H ( λ, β / | c j | , ν ), this yields | Im h j ( w ) | ≥ | Im w | − β = (1 + o (1)) | w | − β ≥ (1 + o (1)) | w | − α | w | > ν, provided | w | is sufficiently large and α is sufficiently small.Also, by Lemma 8.2 and (8.10), | h j ( w ) − w | ≤ (1 + ε ) β ≤ (1 + ε ) α | w | . If α is sufficiently small, we obtain12 | w | ≤ | h j ( w ) | ≤ | w | and hence 12 | h j ( w ) | ≤ | w | ≤ | h j ( w ) | . (8.11)By Lemma 8.2 and (8.11),Re h j ( w ) ≥ Re w + (1 − ε ) β cos (cid:16) π ε (cid:17) = λ log | w | − log β + (1 − ε ) β cos (cid:16) π ε (cid:17) ≥ λ log | h j ( w ) | − | λ | log 2 − log β + (1 − ε ) β cos (cid:16) π ε (cid:17) ≥ λ log | h j ( w ) | − log 1 c ∗ if β and hence β is sufficiently large. Thus, h j ( w ) ∈ H ( λ, /c ∗ , ν ) . Let us now define several sets. We start with subsets Q jn,k , ˜ Q jn,k ⊂ C \H ( λ, β / | c j | , ν )for j ∈ { , ..., d } , k ∈ N and n ∈ Z . ulia sets of measure zero < θ < / (6 π ) arccos(5 / j ∈ { , ..., d } , k ∈ N and n ∈ Z , let Q jn,k bethe set of all w ∈ H (cid:18) λ, k +1 β | c j | , ν (cid:19) \ H (cid:18) λ, k β | c j | , ν (cid:19) such that | Im w − y jn | ≤ θ . Also, let ˜ Q jn,k be the set of all w ∈ H (cid:18) λ, k +2 β | c j | , ν (cid:19) \ H (cid:18) λ, k − β | c j | , ν (cid:19) such that | Im w − y jn | ≤ πθ . See Figure 3 for an illustration of these sets. Note that Q jn,k ⊂ ˜ Q jn,k . If ˜ Q jn,k ⊂H ( λ − , α / | c j | , ν ), then by Lemma 8.3, we have h j ( ˜ Q jn,k ) ⊂ H ( λ, /c ∗ , ν ).Γ (cid:18) λ, k +2 β | c j | (cid:19) Γ (cid:18) λ, k +1 β | c j | (cid:19) Γ (cid:18) λ, k β | c j | (cid:19) Γ (cid:18) λ, k − β | c j | (cid:19) Im w = y jn ˜ Q jn,k Q jn,k Figure 3: An illustration of the sets Q jn,k and ˜ Q jn,k .Moreover, let R jn,k be the rectangle containing all v ∈ C satisfying34 2 k β < Re v − λ log | n | <
52 2 k β and | Im v − y jn | < · k β θ . Also, let ˜ R jn,k be the rectangle containing all v ∈ C satisfying58 2 k β < Re v − λ log | n | < · k β and | Im v − y jn | < · k β θ . Note that R jn,k ⊂ ˜ R jn,k . Lemma 8.4.
There are α , β , ν, n > such that the following holds. If j ∈ { , ..., d } , n ∈ Z with | n | ≥ n and k ∈ N are such that ˜ Q jn,k ⊂ H ( λ − , α / | c j | , ν ) , then h j ( Q jn,k ) ⊂ R jn,k and h j ( ˜ Q jn,k ) ⊃ ˜ R jn,k . Mareike Wolff
Proof.
For w ∈ H ( λ − , α / | c j | , ν ) \ H ( λ, β / | c j | , ν ), we have Re w = o ( | w | ) and hence | w | = (1 + o (1)) | Im w | as w → ∞ . If | Im w − y jn | ≤ πθ and | n | is sufficiently large,we obtain | n | ≤ | w | ≤ e | n | . For w ∈ ˜ Q jn,k , this implies thatRe w ≥ λ log | w | − log 2 k +2 β | c j | ≥ λ log | n | − | λ | − log 2 k +2 β | c j | (8.12)and Re w ≤ λ log | w | − log 2 k − β | c j | ≤ λ log | n | + 2 | λ | − log 2 k − β | c j | . (8.13)Let ε > w ∈ Q jn,k ⊂ ˜ Q jn,k . By Lemma 8.2 and (8.12), and since0 < θ < / (6 π ) arccos(5 / < (1 /
2) arccos(5 / h j ( w ) ≥ Re w + (1 − ε )2 k β cos( θ + ε ) ≥ Re w + (1 − ε )2 k β cos(2 θ ) > λ log | n | − | λ | − log 2 k +2 β | c j | + (1 − ε )2 k β · > λ log | n | + 34 2 k β if ε is sufficiently small and β is sufficiently large. Analogously,Re h j ( w ) ≤ Re w + (1 + ε )2 k +1 β ≤ λ log | n | + 2 | λ | − log 2 k − β | c j | + (1 + ε )2 k +1 β < λ log | n | + 52 2 k β if ε is sufficiently small and β is sufficiently large. Moreover, by Lemma 8.2, | Im h j ( w ) − y jn | ≤ | Im h j ( w ) − Im w | + | Im w − y jn |≤ (1 + ε )2 k +1 β sin( θ + ε ) + | Im w − y jn |≤ (1 + ε )2 k +1 β ( θ + ε ) + θ < · k β θ if ε is sufficiently small and β is sufficiently large. Thus, h j ( Q jn,k ) ⊂ R jn,k .In the following, we show that h j ( ∂ ˜ Q jn,k ) ∩ ˜ R jn,k = ∅ . Since we have already shownthat h j ( ˜ Q jn,k ) ∩ ˜ R jn,k (cid:54) = ∅ , this implies that h j ( ˜ Q jn,k ) ⊃ ˜ R jn,k . If w ∈ Γ( λ, k − β / | c j | ) and β is large, then by Lemma 8.2 and (8.13),Re h j ( w ) ≤ Re w + (1 + ε )2 k − β ≤ λ log | n | + 2 | λ | − log 2 k − β | c j | + (1 + ε )2 k − β < λ log | n | + 58 2 k β . ulia sets of measure zero w ∈ Γ( λ, k +2 β / | c j | ) and | Im w − y jn | ≤ πθ , then by Lemma 8.2 and (8.12), andsince 0 < θ < / (6 π ) arccos(5 / h j ( w ) ≥ Re w + (1 − ε )2 k +2 β cos(5 πθ + ε ) ≥ λ log | n | − | λ | − log 2 k +2 β | c j | + (1 − ε )2 k +2 β cos(6 πθ ) > λ log | n | − | λ | − log 2 k +2 β | c j | + (1 − ε )2 k +2 β · > λ log | n | + 3 · k β , provided ε is sufficiently small and β is sufficiently large.If | Im w − y jn | = 5 πθ and w ∈ H ( λ, k +2 β / | c j | , ν ) \ H ( λ, k − β / | c j | , ν ), then byLemma 8.2, | Im h j ( w ) − y jn | ≥ | Im h j ( w ) − Im( w ) | − | Im w − y jn |≥ (1 − ε )2 k − β sin(5 πθ − ε ) − πθ ≥ (1 − ε )2 k − β π (5 πθ − ε ) − πθ > · k β θ , provivded ε is sufficiently small and β is sufficiently large. Thus, h j ( ∂ ˜ Q jn,k ) ⊂ C \ ˜ R jn,k .Next, we prove that the density of q ( F j ) in ˜ Q jn,k is bounded below by a positiveconstant. Lemma 8.5.
There are α , β , ν, δ, n > such that for all j ∈ { , ..., d } , n ∈ Z with | n | ≥ n and k ∈ N with ˜ Q jn,k ⊂ H ( λ − , α / | c j | , ν ) , we have dens( q ( F j ) , ˜ Q jn,k ) ≥ δ. Proof.
By Section 6, in particular Lemma 6.5, the function h j = q ◦ f ◦ ϕ j has nocritical points in ˜ Q jn,k if ν and β are sufficiently large. By Lemma 8.4, h j ( ˜ Q jn,k ) ⊃ ˜ R jn,k and h j ( Q jn,k ) ⊂ R jn,k . Let U be the component of h − j ( ˜ R jn,k ) containing Q jn,k . Then Q jn,k ⊂ U ⊂ ˜ Q jn,k . Since ˜ R jn,k is simply connected, h j maps U conformally onto ˜ R jn,k .Let ψ : ˜ R jn,k → U be the corresponding inverse function. By Lemma 8.3, h j ( ˜ Q jn,k ) ⊂ H (cid:18) λ, c ∗ , ν (cid:19) ⊂ C \ ( D (0 , R ) ∪ [0 , ∞ )) = q ( S l )for all l ∈ { , ..., d } . Hence, there exists l ∈ { , ..., d } such that ( f ◦ ϕ j )( ˜ Q jn,k ) ⊂ S l . Wehave ψ ( q ( F l ) ∩ ˜ R jn,k ) = q ( F j ) ∩ U . By the Koebe distortion theorem, ψ has boundeddistortion in R jn,k independent of n , k and j . We obtaindens( q ( F j ) , ˜ Q jn,k ) ≥ dens( q ( F j ) , ψ ( R jn,k )) · dens( ψ ( R jn,k ) , ˜ Q jn,k )= dens( ψ ( q ( F l ) ∩ R jn,k ) , ψ ( R jn,k )) · dens( ψ ( R jn,k ) , ˜ Q jn,k ) ≥ c dens( q ( F l ) , R jn,k ) · dens( Q jn,k , ˜ Q jn,k )4 Mareike Wolff for some c > n, k and j . If β is sufficiently large, then by Lemma7.14, dens( q ( F l ) , R jn,k ) ≥ η . Moreover, by Lemma 5.2,meas Q jn,k ≥
23 log 2 · θ and meas ˜ Q jn,k ≤ · πθ . Hence, dens( q ( F j ) , ˜ Q jn,k ) ≥ cη log 215 π log 8 =: δ. The last Lemma of this Section says that there is a positive lower bound for thedensity of q ( F j ) in any sufficiently large rectangle contained in H ( λ − , α / | c j | , ν ) \H ( λ, β / | c j | , ν ) . Lemma 8.6.
There are α , β , ν, D , η > such that for all j ∈ { , ..., d } and anyrectangle R ⊂ H ( λ − , α / | c j | , ν ) \ H ( λ, β / | c j | , ν ) with sides parallel to the real andimaginary axis and side lengths at least D , we have dens( q ( F j ) , R ) ≥ η . Proof.
Suppose that D ≥ ,D ≥ y ln + 2 π + 10 πθ and D ≥ | y l − n | + 2 π + 10 πθ for all l ∈ { , ..., d } , with n as in Lemma 8.5. Let R ⊂ H ( λ − , α / | c j | , ν ) \ H ( λ, β / | c j | , ν ) be a rectanglewith sides parallel to the real and imaginary axis and side lengths at least D . By thedefinition of ˜ Q jn,k and Lemma 5.2, there are k ∈ N and n ∈ Z with | n | ≥ n such that˜ Q jn,k ⊂ R . If, in addition, the side lengths of R do not exceed 2 D , then by Lemma 8.5 and Lemma5.2, dens( q ( F j ) , R ) ≥ dens( q ( F j ) , ˜ Q jn,k ) · dens( ˜ Q jn,k , R ) ≥ δ / · πθ D . Since any general rectangle with side lengths at least D can be written as the union ofrectangles with side lengths between D and 2 D which are disjoint up to the boundary,the claim follows. q ( F ( f )) : third part For ν >
0, let G ν := { w : | Im w | ≥ ν } . In this section, we investigate the density of q ( F ( f )) in subsets of G ν \H ( λ − , β / | c j | , ν )for large β >
0. First, we give an approximation for h j in G ν \ H ( λ − , β / | c j | , ν ). Lemma 9.1.
Let ε > and j ∈ { , ..., d } . Then there are β , ν > such that for all w ∈ G ν \ H ( λ − , β / | c j | , ν ) , we have (cid:12)(cid:12)(cid:12)(cid:12) h j ( w )( − c j /d ) d e − dw w − m − (cid:12)(cid:12)(cid:12)(cid:12) < ε. ulia sets of measure zero Proof.
By Corollary 4.3, f ( ϕ j ( w )) = ϕ j ( w ) − q (cid:48) ( ϕ j ( w )) (cid:18) O (cid:18) | w | (cid:19)(cid:19) − c j e − w p ( ϕ j ( w ))= O ( | w | /d ) − c j d e − w ϕ j ( w ) − m (cid:18) O (cid:18) | w | /d (cid:19)(cid:19) (9.1)as w → ∞ . Note that the O ( · )-terms do not depend on β . For w ∈ G ν \ H ( λ − , β / | c j | , ν ), we have (cid:12)(cid:12)(cid:12) c j d e − w ϕ j ( w ) − m (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) c j d e − w (cid:12)(cid:12)(cid:12) · | w | λ − /d (1 + o (1)) ≥ β | w | /d d (9.2)if | w | is sufficiently large. In particular, | f ( ϕ j ( w )) | ≥ β d | w | /d if β and | w | are sufficiently large, and hence h j ( w ) = q ( f ( ϕ j ( w ))) = f ( ϕ j ( w )) d (cid:18) O (cid:18) | w | /d (cid:19)(cid:19) as w → ∞ in G ν \ H ( λ − , β / | c j | , ν ). Also, by (9.1) and (9.2), (cid:12)(cid:12)(cid:12)(cid:12) f ( ϕ j ( w ))( − c j /d ) e − w ϕ j ( w ) − m − (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) O ( | w | /d )( − c j /d ) e − w ϕ j ( w ) − m + O (cid:18) | w | /d (cid:19)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ dβ O (1) + O (cid:18) | w | /d (cid:19) , where the O ( · )-terms do not depend on β . Hence, we can achieve that (cid:12)(cid:12)(cid:12)(cid:12) f ( ϕ j ( w )) d (( − c j /d ) e − w ϕ j ( w ) − m ) d − (cid:12)(cid:12)(cid:12)(cid:12) ≤ ε β and ν sufficiently large. Also, (cid:16) − c j d e − w ϕ j ( w ) − m (cid:17) d = (cid:16) − c j d (cid:17) d e − dw w − m (cid:18) O (cid:18) | w | /d (cid:19)(cid:19) as w → ∞ , whence the claim follows.We proceed similarly as in Section 8, that is, we show that h j maps certain subsets of G ν \H ( λ − , β / | c j | , ν ) into H ( λ, /c ∗ , ν ). We then apply the results of Section 7 to showthat q ( F ( f )) has positive density in large bounded subsets of G ν \ H ( λ − , β / | c j | , ν ) . For n ∈ Z , k ∈ N and j ∈ { , ..., d } , let P jn,k be the set of all w ∈ H (cid:18) λ − , k +2 β | c j | , ν (cid:19) \ H (cid:18) λ − , k − β | c j | , ν (cid:19) satisfying (2 n − πd ≤ Im w ≤ n + 1) πd . There are θ jn,k ∈ [ − π, π ) and r jn,k > w ∈ P jn,k , we have | w | = r jn,k (1 + o (1)) and arg( w ) = θ jn,k + o (1)as | n | → ∞ . Let t jn,k ∈ [2 nπ/d, n + 1) π/d ) with t jn,k ≡ arg( − c j ) − md θ jn,k mod 2 πd . Mareike Wolff
Lemma 9.2.
Let θ ∗ ∈ (0 , π/ (4 d )) . Then there are β , ν > such that the fol-lowing holds. Let j ∈ { , ..., d } , k ∈ N and w ∈ H ( λ − , k +2 β / | c j | , ν ) \ H ( λ − , k − β / | c j | , ν ) such that there exists n ∈ Z with t jn,k − π/ (4 d ) ≤ Im w ≤ t jn,k − θ ∗ .Let β ∈ [2 k − β , k +2 β ] such that w ∈ Γ( λ − , β/ | c j | ) and let θ := t jn,k − Im w . Then (cid:18) βd (cid:19) d r jn,k cos(2 dθ ) < Re h j ( w ) < (cid:18) βd (cid:19) d r jn,k and π (cid:18) βd (cid:19) d r jn,k dθ < Im h j ( w ) < (cid:18) βd (cid:19) d r jn,k dθ. Proof.
Let ε > (cid:12)(cid:12)(cid:12)(cid:12) h j ( w )( − c j /d ) d e − dw w − m − (cid:12)(cid:12)(cid:12)(cid:12) < ε if β and ν are sufficiently large. Thus,(1 − ε ) (cid:18) | c j | d (cid:19) d e − d Re w | w | − m ≤ | h j ( w ) | ≤ (1 + ε ) (cid:18) | c j | d (cid:19) d e − d Re w | w | − m . (9.3)Since w ∈ Γ( λ − , β/ | c j | ), we have | w | − − m e − d Re w = | w | d ( λ − e − d Re w = (cid:18) β | c j | (cid:19) d . Thus, (cid:18) | c j | d (cid:19) d e − d Re w | w | − m = (cid:18) βd (cid:19) d | w | = (cid:18) βd (cid:19) d r jn,k (1 + o (1))as | n | → ∞ . Inserting the last equation into (9.3) yields34 (cid:18) βd (cid:19) d r jn,k < | h j ( w ) | < (cid:18) βd (cid:19) d r jn,k (9.4)if ε is sufficiently small and | n | is sufficiently large. Also, by Lemma 9.1, (cid:12)(cid:12)(cid:12)(cid:12) arg( h j ( w )) − arg (cid:18)(cid:16) − c j d (cid:17) d e − dw w − m (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) < arcsin( ε ) ≤ π ε. (9.5)We have arg (cid:18)(cid:16) − c j d (cid:17) d e − dw w − m (cid:19) ≡ d arg( − c j ) − d Im w − m arg w ≡ d arg( − c j ) − dt jn,k + dθ − mθ jn,k + o (1) ≡ dθ + o (1) mod 2 π as | n | → ∞ . By (9.5), this yields dθ < arg h j ( w ) < dθ (9.6) ulia sets of measure zero ε is sufficiently small compared to θ ∗ . By (9.4), (9.6) and the fact that (2 /π ) x ≤ sin x ≤ x for 0 ≤ x ≤ π/
2, we obtainRe h j ( w ) ≤ | h j ( w ) | < (cid:18) βd (cid:19) d r jn,k , Re h j ( w ) = | h j ( w ) | cos(arg h j ( w )) > (cid:18) βd (cid:19) d r jn,k cos(2 dθ ) , Im h j ( w ) = | h j ( w ) | sin(arg h j ( w )) < (cid:18) βd (cid:19) d r jn,k sin(2 dθ ) ≤ (cid:18) βd (cid:19) d r jn,k dθ, Im h j ( w ) = | h j ( w ) | sin(arg h j ( w )) > (cid:18) βd (cid:19) d r jn,k sin (cid:18) dθ (cid:19) ≥ π (cid:18) βd (cid:19) d r jn,k dθ. Let us now define several sets. We start with subsets T jn,k , ˜ T jn,k ⊂ G ν \ H ( λ − , β / | c j | , ν ). Let 0 < θ < · d +1 dπ arccos (cid:18) (cid:19) . For n ∈ Z , k ∈ N and j ∈ { , ..., d } , let T jn,k be the set of all w ∈ H (cid:18) λ − , k +1 β | c j | , ν (cid:19) \ H (cid:18) λ − , k β | c j | , ν (cid:19) satisfing t jn,k − θ ≤ Im w ≤ t jn,k − θ . Also, let ˜ T jn,k be the set of all w ∈ H (cid:18) λ − , k +2 β | c j | , ν (cid:19) \ H (cid:18) λ − , k − β | c j | , ν (cid:19) satisfying t jn,k − d +1 πθ ≤ Im w ≤ t jn,k − · d π θ . Note that T jn,k ⊂ ˜ T jn,k . See Figure 4 for an illustration of T jn,k and ˜ T jn,k .Moreover, let U jn,k be the rectangle containing all v ∈ C satisfying1116 (cid:18) k β d (cid:19) d r jn,k < Re v < (cid:18) k +1 β d (cid:19) d r jn,k and 38 π (cid:18) k β d (cid:19) d r jn,k dθ < Im v < (cid:18) k +1 β d (cid:19) d r jn,k dθ . Also, let ˜ U jn,k be the rectangle containing all v ∈ C satisfying58 (cid:18) k β d (cid:19) d r jn,k < Re v < (cid:18) k +1 β d (cid:19) d r jn,k and 14 π (cid:18) k β d (cid:19) d r jn,k dθ < Im v < (cid:18) k +1 β d (cid:19) d r jn,k dθ . Note that U jn,k ⊂ ˜ U jn,k .8 Mareike Wolff Γ (cid:18) λ − , k − β | c j | (cid:19) Γ (cid:18) λ − , k β | c j | (cid:19) Γ (cid:18) λ − , k +1 β | c j | (cid:19) Γ (cid:18) λ − , k +2 β | c j | (cid:19) Im w = t jn,k ˜ T jn,k T jn,k Figure 4: An illustration of the sets T jn,k and ˜ T jn,k . Lemma 9.3.
There is n ∈ N such that for all n ∈ Z with | n | ≥ n , k ∈ N and j ∈ { , ..., d } , we have ˜ U jn,k ⊂ H (cid:18) λ, c ∗ , ν (cid:19) with c ∗ = max l | c l | as defined in (8.9) .Proof. Let v ∈ ˜ U jn,k . Note that r jn,k → ∞ as | n | → ∞ uniformly in k . In particular,Im v > π (cid:18) k β d (cid:19) d r jn,k dθ ≥ ν if | n | is sufficiently large. Also,Im v < (cid:18) k +1 β d (cid:19) d r jn,k dθ = 245 2 d dθ · (cid:18) k β d (cid:19) d r jn,k <
245 2 d dθ Re v, and hence | v | ≤ | Re v | + | Im v | < (cid:18) d dθ (cid:19) Re v. Thus, Re v ≥
11 + (24 / d dθ | v | ≥ λ log | v | − log 1 c ∗ if | n | and hence r jn,k and | v | are sufficiently large. Lemma 9.4.
There are β , ν > such that for all j ∈ { , ..., d } , k ∈ N and n ∈ Z with | t jn,k | > ν + 4 d +1 πθ , we have h j ( T jn,k ) ⊂ U jn,k and h j ( ˜ T jn,k ) ⊃ ˜ U jn,k . Proof.
First suppose that w ∈ T jn,k . Then by Lemma 9.2 and the fact that θ < / (2 · d +1 dπ ) arccos(11 / < / (2 d ) arccos(11 / , we haveRe h j ( w ) > (cid:18) k β d (cid:19) d r jn,k cos(2 dθ ) > (cid:18) k β d (cid:19) d r jn,k , ulia sets of measure zero h j ( w ) < (cid:18) k +1 β d (cid:19) d r jn,k , Im h j ( w ) > π (cid:18) k β d (cid:19) d r jn,k dθ , Im h j ( w ) < (cid:18) k +1 β d (cid:19) d r jn,k dθ . Hence, h j ( T jn,k ) ⊂ U jn,k .Also, Lemma 9.2 yields the following. If w ∈ Γ( λ − , k − β / | c j | ) with t jn,k − d +1 πθ ≤ Im w ≤ t jn,k − / (10 · d π ) θ , thenRe h j ( w ) < (cid:18) k − β d (cid:19) d r jn,k ≤ (cid:18) k β d (cid:19) d r jn,k . If w ∈ Γ( λ − , k +2 β / | c j | ) with t jn,k − d +1 πθ ≤ Im w ≤ t jn,k − / (10 · d π ) θ , thenusing that θ < / (2 · d +1 dπ ) arccos(11 / h j ( w ) > (cid:18) k +2 β d (cid:19) d r jn,k cos(2 d d +1 πθ ) > (cid:18) k +1 β d (cid:19) d r jn,k . If w ∈ H ( λ − , k +2 β / | c j | , ν ) \ H ( λ − , k − β / | c j | , ν ) and Im w = t jn,k − d +1 πθ ,then Im h j ( w ) > (cid:18) k +1 β d (cid:19) d r jn,k dθ . If w ∈ H ( λ − , k +2 β / | c j | , ν ) \ H ( λ − , k − β / | c j | , ν ) and Im w = t jn,k − / (10 · d π ) θ ,then Im h j ( w ) < π (cid:18) k β d (cid:19) d r jn,k dθ . Thus, h j ( ∂ ˜ T jn,k ) ∩ ˜ U jn,k = ∅ . Since T jn,k ⊂ ˜ T jn,k and h j ( T jn,k ) ⊂ U jn,k ⊂ ˜ U jn,k , we obtainthat h j ( ˜ T jn,k ) ⊃ ˜ U jn,k . Next, we show that the density of q ( F j ) in ˜ T jn,k is bounded below by a positiveconstant. Lemma 9.5.
There are δ > and n ∈ N such that for all j ∈ { , ..., d } , k ∈ N and n ∈ Z with | n | ≥ n , we have dens( q ( F j ) , ˜ T jn,k ) ≥ δ. Proof.
We only sketch the proof, since it is similar to the one of Lemma 8.5. By Lemma9.3, ˜ U jn,k ⊂ H (cid:18) λ, c ∗ , ν (cid:19) . By Lemma 9.4, h j ( ˜ T jn,k ) ⊃ ˜ U jn,k and h j ( T jn,k ) ⊂ U jn,k . Let V ⊂ ˜ T jn,k be the componentof h − j ( ˜ U jn,k ) containing T jn,k . As in the proof of Lemma 8.5, we get that f ( ϕ j ( V )) ⊂ S l for some l ∈ { , ..., d } , and thatdens( q ( F j ) , ˜ T jn,k ) ≥ c dens( q ( F l ) , U jn,k ) · dens( T jn,k , ˜ T jn,k )0 Mareike Wolff for some c > n, k and j . If | n | and hence r jn,k is sufficiently large, thenby Lemma 7.14, dens( q ( F l ) , U jn,k ) ≥ η . Also, the density of T jn,k in ˜ T jn,k is bounded below independent of n, k and j , whencethe claim follows.The final result of this section says that the density of q ( F j ) in large rectangles in G ν \ H ( λ − , β / | c j | , ν ) is bounded below. Lemma 9.6.
There are β , ν, D , η > such that for all j ∈ { , ..., d } and any rectan-gle R ⊂ G ν \ H ( λ − , β / | c j | , ν ) with sides parallel to the real and imaginary axis andside lengths at least D , we have dens( q ( F j ) , R ) ≥ η . Proof.
This is proved the same way as Lemma 8.6, using Lemma 9.5.
10 The set q ( F ( f )) : conclusions In this section, we combine the results of Sections 7-9 to show that q ( F j ) has positivedensity in large bounded subsets of C . Lemma 10.1.
There are
D, η > such that for all j ∈ { , ..., d } and any square R ⊂ C with sides parallel to the real and imaginary axis and side lengths at least D ,we have dens( q ( F j ) , R ) ≥ η . Proof.
Let E := H (cid:18) λ, β | c j | , ν (cid:19) \ H (cid:18) λ, | c j | , ν (cid:19) and E := H (cid:18) λ − , β | c j | , ν (cid:19) \ H (cid:18) λ − , α | c j | , ν (cid:19) . Also, let γ and γ be the left boundary curves of E and E , respectively, parametrisedby y = Im z . Justified by Lemma 5.2, we suppose that ν is so large that | γ (cid:48) k ( y ) | <
110 for | y | ≥ ν and k ∈ { , } . (10.1)Using the notation of Lemmas 7.14, 8.6 and 9.6, suppose that D > ν + 5 max { D , D , D } (10.2)and D >
20 max (cid:26) log β , log β α (cid:27) . (10.3)For S ⊂ C , let diam x ( S ) := sup {| Re( z − w ) | : z, w ∈ S} and diam y ( S ) := sup {| Im( z − w ) | : z, w ∈ S} . Define R + := R ∩ { z : Im z ≥ ν } , R − := R ∩ { z : Im z ≤ − ν } , ulia sets of measure zero R := (cid:40) R + if diam y ( R + ) ≥ diam y ( R − ) R − otherwise.By (10.2), diam y ( R ) > max { D , D , D } . We now divide R into 5 rectangles, R , , ..., R , , with diam y ( R ,k ) = diam y ( R )and diam x ( R ,k ) = diam x ( R ) for all k ∈ { , ..., } (see Figure 5). R , R , R , R , R , R Figure 5: The rectangle R , bounded by the solid line, is divided into five smallerrectangles by the dashed lines.By (10.2), diam x ( R ,k ) > max { D , D , D } . By (10.1), Lemma 5.2, (10.3) and thefact that R is a square of side length at least D , we havediam x ( E l ∩ R ) <
110 diam y ( R ) + 2 max (cid:26) log β , log β α (cid:27) <
110 diam y ( R ) + 110 D ≤
15 diam x ( R )for l ∈ { , } . Thus, E and E each intersect at most two of the rectangles R ,k . Hence,there exists l ∈ { , ..., } such that R ,l does not intersect E ∪ E . This implies that R ,l satisfies the hypothesis of one of Lemmas 7.14, 8.6 and 9.6. Hence,dens( q ( F j ) , R ,l ) ≥ min { η , η , η } and dens( q ( F j ) , R ) ≥ dens( q ( F j ) , R ,l ) · dens( R ,l , R ) ≥ min { η , η , η } ·
110 diam x ( R )(diam y ( R ) − ν )(diam x R ) ≥ min { η , η , η } · (cid:18) − νD (cid:19) . The following corollary is an immediate consequence of Lemma 10.1.
Corollary 10.2.
There are r , η > such that for all z ∈ C , all r ≥ r and all j ∈ { , ..., d } , we have dens( q ( F j ) , D ( z, r )) ≥ η. Remark 10.3.
Corollary 10.2 says that C \ q ( F j ) is thin at ∞ .
11 Proof of Theorem 1.1
Proof of Theorem 1.1.
We will verify the assumptions of Theorem 1.3. By Lemma 6.2,the set P ( f ) ∩ J ( f ) is finite, so it remains to prove that there exists R > J ( f ) is uniformly thin at ( P ( f ) ∩ C ) \ D (0 , R ) and that J ( f ) is thin at ∞ . Let r > Mareike Wolff (a) | q (cid:48) ( z ) | ≥ ( d/ | z | d − for all z ∈ C with | z | ≥ r ;(b) each z ∈ P ( f ) with | z | ≥ r is a zero of g and hence a superattracting fixed pointof f . Justified by Corollary 6.6, we also assume that there is j ∈ { , ..., d } with z ∈ S j , and dist( z , ∂ S j ) ≥
3. Moreover, suppose that the conclusion of Lemma7.5 holds for | z | ≥ r .Let r be the constant from Corollary 10.2. First, we will show that there exists η > j ∈ { , ..., d } , all z ∈ S j with | z | ≥ r and all r > r / ( d | z | d − ) with D ( z, r ) ⊂ S j , we have dens( F ( f ) , D ( z, r )) ≥ η . (11.1)Recall that q is injective in S j . By Koebe’s theorems, D (cid:18) q ( z ) , | q (cid:48) ( z ) | r (cid:19) ⊂ q ( D ( z, r )) ⊂ D ( q ( z ) , | q (cid:48) ( z ) | r ) . By (a) and the assumption on r , we have (1 / | q (cid:48) ( z ) | r ≥ r . Hence, by Corollary 10.2,dens (cid:18) q ( F j ) , D (cid:18) q ( z ) , | q (cid:48) ( z ) | r (cid:19)(cid:19) ≥ η. Thus,dens( q ( F j ) , q ( D ( z, r ))) ≥ dens (cid:18) D (cid:18) q ( z ) , | q (cid:48) ( z ) | r (cid:19) , q ( D ( z, r )) (cid:19) · dens (cid:18) q ( F j ) , D (cid:18) q ( z ) , | q (cid:48) ( z ) | r (cid:19)(cid:19) ≥ dens (cid:18) D (cid:18) q ( z ) , | q (cid:48) ( z ) | r (cid:19) , D ( q ( z ) , | q (cid:48) ( z ) | r ) (cid:19) · η = 1256 η. By the Koebe distortion theorem,dens( F ( f ) , D ( z, r )) ≥ (cid:18) min ζ ∈D ( z,r ) | q (cid:48) ( ζ ) | max ζ ∈D ( z,r ) | q (cid:48) ( ζ ) | (cid:19) dens( q ( F j ) , q ( D ( z, r ))) ≥ · η. This implies (11.1) with η = η/ (3 · . Let us now prove that there exists R > J ( f ) is uniformly thin at( P ( f ) ∩ C ) \ D (0 , R ). Let δ ∈ (0 , z ∈ P ( f ) with | z | > r + 1 and z ∈ D ( z , δ ) . By (b), D ( z, δ ) ⊂ S j . Also, | z | ≥ r . If | z − z | ≥ r / ( d | z | d − ), then by (11.1),dens( F ( f ) , D ( z, | z − z | )) ≥ η . Now suppose that | z − z | < r d | z | d − . (11.2)By Lemma 7.5, we have D ( z , / (3 d | z | d − )) ⊂ F ( f ). Hence,dens( F ( f ) , D ( z, | z − z | )) ≥ dens (cid:18) D (cid:18) z , d | z | d − (cid:19) , D ( z, | z − z | ) (cid:19) . The expression on the right hand side is bounded below independent of z and | z | ,provided (11.2) is satisfied. So J ( f ) is uniformly thin at ( P ( f ) ∩ C ) \ D (0 , r + 1).It remains to prove that J ( f ) is thin at ∞ . Let R be as in Section 3 and let r > max { R /d , r } . If r is sufficiently large, then Lemma 3.2 yields that (cid:83) dj =1 ∂ S j \D (0 , r ) ulia sets of measure zero d pairwise disjoint halfstrips, T , ..., T d , of width 1. We can assume that r is so large that dist( T k , T l ) ≥ k (cid:54) = l . Then for | z | ≥ r +3, the set D ( z, \ (cid:83) dj =1 T j contains a disk, D , of radius 1 /
2. There is j ∈ { , ..., d } with D ⊂ S j . Let D (cid:48) be thedisk with the same center as D and radius 1 /
4. If r is sufficiently large, then by (11.1),we have dens( F ( f ) , D (cid:48) ) ≥ η , and hencedens( F ( f ) , D ( z, ≥ dens( F ( f ) , D (cid:48) ) · dens( D (cid:48) , D ( z, ≥ η . We now consider the case that | z | < r + 3. Let ζ , ..., ζ n ∈ D (0 , r + 3) such that D (0 , r + 3) ⊂ n (cid:91) k =1 D ( ζ k , . Then η := min ≤ k ≤ n dens( F ( f ) , D ( ζ k , > . For z ∈ D (0 , r + 3), let k ∈ { , ..., n } such that z ∈ D ( ζ k , D ( ζ k , ⊂ D ( z, F ( f ) , D ( z, ≥
19 dens( F ( f ) , D ( ζ k , ≥ η . Thus, J ( f ) is thin at ∞ . Hence, Theorem 1.3 yields that J ( f ) has Lebesgue measurezero. Acknowledgements
I would like to thank Walter Bergweiler for helpful suggestionsand for carefully reading the manuscript.
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