GROWTH CONDITIONS FOR ENTIRE FUNCTIONS WITH ONLY BOUNDED FATOU COMPONENTS

Transcription

1 GROWTH CONDITIONS FOR ENTIRE FUNCTIONS WITH ONLY BOUNDED FATOU COMPONENTS AIMO HINKKANEN AND JOSEPH MILES Abstract Let f be a transcendental entire function of order < 1/2 We denote the maximum and minimum modulus of f by Mr, f = max{ fz : z = r} and mr, f = min{ fz : z = r} We obtain a minimum modulus condition satisfied by many f of order zero that implies all Fatou components are bounded A special case of our result is that if log log Mr, f = Olog r/log log r K for some K > 1, then there exist α > 1 and C > 0 such that for all large R, there exists r R, R α ] with 1, log mr, f log MR, f α C log log R K and this in turn implies boundedness of all Fatou components The condition on mr, f is a refined form of a minimum modulus conjecture formulated by the first author We also show that there are some functions of order zero, and there are functions of any positive order, for which even refined forms of the minimum modulus conjecture fail Our results and counterexamples indicate rather precisely the limits of the method of using the minimum modulus to rule out the existence of unbounded Fatou components 1 Introduction Let f be a transcendental entire function In 1981, IN Baker [2] proved that if the growth of the maximum modulus of f does not exceed a certain rate, then all the components of the Fatou set of f are bounded He asked what would be the largest rate of growth that guarantees this conclusion, and showed that the best one can hope for is that the growth of f does not exceed order 1/2, minimal type We shall use the usual definitions of complex dynamics see, eg, [4], [5], [10], [14] We denote the iterates of f by f 1 = f, and f n = f f n 1 for n 2 We say that z C lies in the Fatou set Ff of f if there 2000 Mathematics Subject Classification Primary: 30D05; Secondary: 37F50 This material is based upon work supported by the National Science Foundation under Grant No

2 2 AIMO HINKKANEN AND JOSEPH MILES exists a neighborhood U of z such that the family {f n U : n 1} of the restrictions of the iterates of f to U is a normal family The Julia set J f of f is J f = C \ Ff The definition shows that the set Ff is open The set J f is non-empty and perfect, and is equal to C or is a nowhere dense subset of C We write Mr, f = max{ fz : z = r} for the maximum modulus of f and mr, f = min{ fz : z = r} for the minimum modulus of f The notation mr, f is also the standard notation for the proximity function of f in the Nevanlinna theory, and we will explicitly point out the one time where the notation mr, f is used in that meaning The order ρf and lower order λf of f are defined by ρf = lim sup r log log Mr, f, λf = lim inf log r r If 0 < ρf = ρ < +, we define the type of f by τf = lim sup r log Mr, f r ρ log log Mr, f log r If τf = 0, we say that f is of minimal type If 0 < τf < +, we say that f is of mean type If τf = +, we say that f is of maximal type IN Baker [2] asked in 1981 whether every component of Ff is bounded if the growth of f is sufficiently small The function fz = z 1/2 sin z + z + a is of order 1/2, mean type, and if a is a sufficiently large positive number, then Ff has an unbounded component D containing a segment [x 0, of the positive real axis, such that f n z as n, locally uniformly in D Baker noted [2], p 484 that it is possible to have a function of order 1/2 and of arbitrarily small type with the same properties Thus, while one might hope to prove that all components of Ff are bounded provided that the growth of f does not exceed order 1/2, minimal type, one cannot do better For further background, we refer to the survey paper [8] of the first author Here we only briefly mention the following results that have been previously obtained on this problem That all components of Ff are bounded for a transcendental entire function f was proved by Baker [2] under the assumption that log Mr, f = Olog r p where 1 < p < 3, and by Stallard [13] when log r 1/2 log log Mr, f = O log log r ε

3 GROWTH CONDITIONS 3 for some ε > 0 The latter is the best known condition so far based on growth alone When the regularity of growth of Mr, f is also taken into account, the following conditions are known to imply the boundedness of all components of Ff, where we assume throughout that the order of f is < 1/2 Stallard [13] proved that this is the case if there exists a real number c [1, such that log M2r, f lim r log Mr, f = c Anderson and the first author [1] proved that all components of Ff are bounded if there exists a positive constant c such that for all sufficiently large x, the increasing convex function ϕx = log Me x, f satisfies ϕ x/ϕx 1 + c/x This is true, in particular, if f is of positive lower order [8], Section 8 Stronger results have been proved for particular types of components of Ff Let U be a periodic or preperiodic component of Ff other than a Baker domain or a preimage of a Baker domain Thus there is a positive integer n such that the component V of Ff containing f n U belongs to an attracting, superattracting, or parabolic cycle of components of Ff, or to a cycle of Siegel disks Baker [2] proved that if the growth of f is at most order 1/2, minimal type, then U is bounded Zheng [15] extended this result to the case when V belongs to a cycle of Baker domains so that lim m f m z for z V and f q V V for some positive integer q Stallard [12] proved this for functions of order < 1/2 The only components of Ff not covered by these results are the wandering domains U, characterized by the property that for all distinct positive integers m and n, we have f m U f n U = If for an arbitrary transcendental entire function f, such a component U is multiply connected, then by a result of Baker [3], all components of Ff are bounded Thus we may assume that all wandering domains of f are simply connected The problem can thus be formulated as follows Let f be a transcendental entire function whose growth is at most order 1/2, minimal type Suppose that f has at least one wandering domain and that all wandering domains of f are simply connected What else, if anything, needs to be assumed of f to prove that all of its wandering domains are bounded all other components of Ff necessarily being bounded? There could be many ways of approaching this problem In this paper we consider this problem from the point of view of minimum

4 4 AIMO HINKKANEN AND JOSEPH MILES modulus estimates for f Our starting point is the following result of the first author [7] Theorem 11 Let f be a transcendental entire function of order < 1/2 Suppose that there exist positive numbers R 0, L, δ, and C with R 0 > e, MR 0, f > e, L > 1, and 0 < δ 1 such that for every R > R 0 there exists r R, R L ] with 1 log mr, f log MR, f L 1 C log R δ Then all the components of the Fatou set of f are bounded The idea of the proof of Theorem 11 is that to get a contradiction, we assume that f has an unbounded simply connected wandering domain U, and then show that U has a compact subset K whose images under the iterates of f persist in having a large radial spread This firstly rules out the possibility of any subsequence of the f n having a finite necessarily constant limit function in U, so that lim n f n z = locally uniformly for z U On the other hand, since each application of f will not increase the hyperbolic distance between the points of K, measured in the distinct domains containing f n U, one can show that in the long run, spreads of the kind obtained are not possible, which then yields the desired contradiction More precisely, using dynamics and the hyperbolic geometry, we deduce that once K is given, there is a constant C > 1 depending on K such that 1 2 C f n z f n w C whenever n 1 and z, w K This is a limitation on the ratio of the moduli of two points The lower estimate for the spread that we obtain from arguments that have nothing to do with dynamics applies, instead, to the ratio of the logarithms of the moduli of two points We deduce that if K is properly chosen to begin with which choice then determines C, so that C may be large then for each n there are z n, w n f n K such that 3 log fz n log fw n L 0 > 1 for a fixed L 0 > 1 It is clear that even if C may be large and L 0 may be very close to 1, 2 and 3 will be incompatible when n is so large that f n z and f n w will be sufficiently large as determined by C and L 0 This last situation will occur since lim n f n z = uniformly for z K This is how a contradiction is produced under

5 GROWTH CONDITIONS 5 these assumptions: dynamics limits the radial spread while the fact that the minimum modulus is large sufficiently often forces the radial spread to remain large In [7] the first author suggested that perhaps the assumptions of Theorem 11 are valid for all transcendental entire functions f of order < 1/2, which would then imply that for all such functions all components of Ff are bounded In this paper we give counterexamples to show that these assumptions do not hold provided that the growth of f is fast enough There are even counterexamples of zero order Thus, if the growth of f is sufficiently rapid, it will be necessary to use totally different methods, or other methods in addition to those provided by Theorem 11 and its proof, if one hopes to prove that all components of Ff are bounded However, we also prove that if the growth of f is not too fast, then conditions that are close to the assumptions of Theorem 11 are satisfied, and as a result, all components of Ff are bounded Some technical modifications in the assumptions of Theorem 11 are necessary as Theorem 11 was clearly only a tentative result, its proof being based on choosing a particular convergent series and formulating the assumption accordingly If one wishes to take this technique to its limit, one must consider an arbitrary convergent series with positive terms This provides a condition based on growth alone that is stronger than that obtained by Stallard [13] and that is sufficient to imply the boundedness of the components of Ff Perhaps of greater interest is the fact that our results, taken together, illustrate the limits of this method of proof based on the minimum modulus alone Thus further results should require a more careful study of factors other than the modulus of the function, such as, perhaps, the behavior of the argument of the function and the more precise structure, or shape, of the hypothetical unbounded simply connected wandering domains of f We hope to be able to return to this latter subject in another paper 2 Results 21 Growth rate guaranteeing the boundedness of all Fatou components First we present our positive results to the effect that if a transcendental entire function does not grow too fast, then a modified form of the minimum modulus conjecture of the first author [7] is valid, and as a consequence, all components of the Fatou set of the function are bounded This is achieved through the combination of the following three theorems

6 6 AIMO HINKKANEN AND JOSEPH MILES Theorem 21 Suppose that f is a transcendental entire function of order 0 For r > e, write { log + log + } Mt, f 4 βr = sup : t r log t Suppose that γ : e, + 0, + is such that i γr 0 as r, and ii γr βr as r Then for each α > 1 and all large R, there exists r R, R α ] such that 5 log mr, f log MR, f α1 3γR Theorem 22 Let f be a transcendental entire function of order < 1/2 Suppose that there exist α > 1, R 0 > 1, and a positive decreasing function γr such that for every R > R 0 there exists r R, R α ] with 6 log mr, f log MR, f α1 γr and n γe2n < Then all the components of the Fatou set of f are bounded Theorem 23 Suppose that f is a transcendental entire function of order 0 Let βr be defined as in Theorem 21 If βe 2n <, then f has no unbounded Fatou component n To satisfy the condition n βe2n < in Theorem 23 it suffices to have, for example, βr 1 log log rlog log log r log j r K for all large r and some j 2, where K > 1, log 1 r = log r and log j r = loglog j 1 r 22 Growth rate beyond which the minimum modulus conjecture is not valid Next we turn to a counterexample Theorem 24 enables us to determine a growth rate for entire functions of order zero beyond which our method is not effective

7 GROWTH CONDITIONS 7 Theorem 24 Suppose that βr and γr are positive functions defined for r > e 3 such that 1 7 βr > log log r 2 for all r > e 3, βr is decreasing, βr log r is increasing, and, as r, we have βr 0, γr 0, and βr/γr Suppose that α > 1 Then there exists an entire function f such that for all large r, we have { log + log + } Mt, f 8 sup : t r βr, log t and also such that there exist arbitrarily large R for which 9 log mr, f log MR, f < α1 γr, for all r with R < r Rα The assumptions of Theorem 24 that βr is decreasing while βr log r is increasing are natural since these properties hold if βr is replaced by the left hand side of 8 In view of 7, βr log r as r Application If the function β is as in Theorem 24 and n βe2n =, Theorem 24 and its proof imply that there is no function γr to which Theorem 22 can be applied to show for all entire f satisfying 8 that such f have no unbounded Fatou components For if n βe2n =, then there exists γ satisfying the assumptions of Theorem 24 in place of γ such that n γ e 2n = A cursory examination of the proof of Theorem 24 shows there is enough freedom in the construction that for all large R there exists an entire function f = f R satisfying 8 for all large r and 9 If γ is a function to which Theorem 22 can be applied for all f satisfying 8, we conclude from 6 and 9 that γr > γ R for all large R Since n γe2n <, this is a contradiction, Our method of proof is thus effectively limited to functions satisfying the hypotheses of Theorem 23 Remark After preparing this paper, we learned that results closely related to ours have also been obtained by Rippon and Stallard; see [11] 3 Proof of Theorem 21 We employ the usual definitions and notation of the Nevanlinna theory as given, for example, in [6], except that, as noted, mr, f denotes the minimum modulus of f unless otherwise stated Let f satisfy the assumptions of Theorem 21 We may find a 0, 1], b C\{0} and a non-negative integer k so that if Hz = faz/bz k,

8 8 AIMO HINKKANEN AND JOSEPH MILES then H0 = 1 and H has no zeros in the unit disk The functions βr defined for f and H are asymptotic to one another; hence H satisfies hypothesis ii of Theorem 21 as well Therefore we assume for the time being that f0 = 1 and that f has no zeros in the unit disk, and indicate later how to proceed in the general case We first note for all large r that 10 γr log r > βr log r log + log + Mr, f >> 1 and for all t > e that 11 nt, 0, f log t Nt 2, 0, f log Mt 2, f exp{2βt 2 log t} For large R, define R 0 = R 0 R > R by log R 0 = αlog R1 γr Denote the zeros of f by z n = r n e iθn, with multiple zeros repeated according to their multiplicity Write f = F G, where F z = 1 zzn and Gz = r n<r 0 /10 r n R 0 /10 1 zzn Suppose that R 0 +1 < r < R α First consider a zero z n of F Certainly { } min log 1 reiθ r : 0 θ 2π = log 1 r n and { max log For 1 x R 0 /10, set z n } 1 Reiθ : 0 θ 2π = log hx = log r x 1 log R x + 1 Claim: For 1 < x < R 0 /10, we have h x > 0 We now justify the claim We have z n 1 + Rrn h x = RRx + x2 1 log r 1 rrx x 2 1 log R + 1 x x log R + 1 2, x which has the same sign as R r R + x log x 1 r R r x log x + 1 = I II,

9 GROWTH CONDITIONS 9 say We analyze I II on three different intervals: Interval A: 10R x R 0 /10 For x in interval A we have and Hence and R R + x 10R 11x r x > R 0 x 10 I > 10R 11x log 9 r r x r r r 10 = 10 9 Thus II < 10R 10 R 9x < 11 log 9 x < I Hence h x > 0 Interval B: Since α > 1, we may choose ω such that 23 α < ω < 1 22 Let interval B be R ω x < 10R We have and r log x 1 R R + x R R + 10R = 1 11 > 1 2 log r x 1 η 1 log r log R log 10 > 2 2 for each η < α if R > R η Thus I > η 1 log R for η < α, R > 22 R η We have as well r r x r = 1 + o1, R r 10R We have Thus R log x + 1 = log R x + log 1 + x R log R ω log R + x R 1 ω log R + 10 II < 1 + o1 1 ω log R + 10 log R,

10 10 AIMO HINKKANEN AND JOSEPH MILES Since 1 ω < α 1, there exists η < α such that for all large R 22 II < η 1 log R 22 Combining, we conclude for large R that = 1 x r 1 + x R I II > 0 Thus h x > 0 for x in interval B Interval C: Suppose that 1 x R ω We have I = R r x log r 1 x II r R + x log R + 1 x 1 + log r 1 log R + 1 x x log R + 1 x x r logr x logr + x 1 + x logr + 1 R Suppose that 1 < η < α For large R we have r > R η Since x < R, and since we may assume that 2 < R η ω, we have logr x > η log R log 2 and logr + x < log R + log 2 Thus logr x logr + x η 1 log R 2 log > 1 + logr + 1 logr + 1 > η for some η > 1, if R > R η Since x/r and x/r are both o1 in interval C, we see that I/II > 1 This shows that h x = I II > 0 for x in interval C when R is large, and proves the claim For each zero z n = r n e iθn of F we have 1 r n < R 0 /10, and by the claim and 10 we conclude for R < r < R α that hr n h1 log R 0 logr + 1 = αlog R1 γr logr + 1 say, implying that r log mr, F log 1 = r n r n<r 0 /10 R 12 > α R log + 1 r n r n<r 0 /10 r n<r 0 /10 α1 2γR = α R, R hr n log + 1 r n > α R log MR, F,

11 GROWTH CONDITIONS 11 for R < r < R α We now consider the factor Gz Set P 4 = R α /R 0 Note that Thus 4 log P = α log R log R 0 = αlog RγR log P = αlog RγR 4 For n = 1, 2, 3,, define R n = R 0 P n, and note that R 4 = R α Set I n = [R n, R n+1 for n 0 Note that [R 0, R α = I 0 I 1 I 2 I 3 Case I: Suppose that G has no zeros of modulus less than R 2 Then log mr 1, G 1 R 1 r n r n R 2 log If r n I j for j 2, then log 1 R 1 log 1 R 1 > 2R 1 = 2 r n R j R j P j 1 = 2 exp j 1 log P { } αj 1log RγR = 2 exp 4 Let n j be the number of z n counted according to multiplicity with z n I j We have from 11 that n j < nr j+1, 0, G exp { } 2βRj+1 2 log R j+1 = exp { 2βRj+1 2 log R 0 + j + 1 log P } = exp { 2βR 2 j+1 log R 0 + j + 1αγR log R 4 } Thus, with and X j = 2βR 2 j+1 Y j = X j log R 0 + j + 1αγR log R 4 j 1αγR log R, 4

12 12 AIMO HINKKANEN AND JOSEPH MILES we have log mr 1, G log j=2 r n I j 2 2 j=2 1 Rrn { n j exp exp {Y j } 2, j=2 } j 1αγR log R 4 since this series is dominated by a rapidly converging geometric series with the first term and ratio both less than exp { αγr log R/2} Recall that βr 2 j+1 is much smaller than γr for all j 2, and, by 10, that γr log R is large for large R Also, with Z j = X j + 1 α log R + αγr log R j log P, we have 13 = log MR, G j=2 j=2 n j R R j j=2 r n Ij log 1 + Rrn exp {X j } exp {log R log R j } j=2 { } α 1/2 1 α 1 exp {Z j } < 2 exp log R = 2, 2 R since the last series is dominated by a rapidly converging geometric series with first term less than exp {1 α/2 log R} and ratio less than { exp } αγr log R 8 Note that again we use the facts that βr 2 j+1 is much smaller than γr for j 2 and that γr log R is large for large R Combining, we get log mr 1, f log mr 1, F + log mr 1, G > α R log MR, F 2

13 GROWTH CONDITIONS 13 Also, log MR, f log MR, F + log MR, G α 1/2 1 < log MR, F + 2 R Thus log mr 1, f log MR, f α R log MR, F 2 log MR, F R α 1/2 = α R 2α R 1 R α 1/2 + 2 log MR, F R > α R o1 log R > α 1 3γR =: α R for large R, α 1/2 where we have again used 10 Case II: Suppose that G has a zero in {z : R 0 /10 z < R 2 } Clearly then there exists t I 2 = [R 2, R 3 such that log Nt, 0, G > 0 Claim: For large R if t R 4, then 14 log Nt, 0, G log Nt, 0, G log t log t < 1 10 We now justify the claim To get a contradiction, suppose that the claim is false Then there exists t R 4 such that log Nt, 0, G log Nt, 0, G 1 10 log t log t

14 14 AIMO HINKKANEN AND JOSEPH MILES Now log t log t = = > log t log t log t log t log t 1 log t log t 1 log R 3 log R 4 log t = logr 4/R 3 log t log R 4 = log P log t log R 4 = αγrlog Rlog t 4 log R 4 Thus if there exists t R 4 such that 14 is false, then 10 log Nt, 0, G > 10 log Nt, 0, G 10 log Nt, 0, G αγr log R log t = γr log t, 4 log R 4 4 implying that γr log t < 40 log Nt, 0, G < 40βt log t, which is a contradiction for large R since t > R This establishes the claim It follows that there exists t 1 R 0 /10, R α such that 1/10 t log Nt, 0, G log Nt 1, 0, G, for t 1 R 0 10 < t < To see this, consider the graph of log Nt, 0, G as a function of log t, for t R 0 /10, and the line L on that graph of slope 1/10 passing through log t, log Nt, 0, G The claim asserts that the graph of log Nt, 0, G as a function of log t for t R α lies below L Now consider parallel translates of L translation upward until we obtain the highest translate L of L that intersects the graph; a point of intersection of the graph with this highest translate is a suitable point log t 1, log Nt 1, 0, G Any such t 1 lies in R 0 /10, R α, and we choose t 1 to be as large as possible If we write c m r, G = 1 2π e imθ log Gre iθ dθ, 2π 0

15 usual estimates see [9] give for all integers m GROWTH CONDITIONS 15 c m t 1, G Nt 1, 0, G m 2, implying, after an application of Jensen s Theorem, that log mt 1, G = min { log Gt 1 e iθ : 0 θ 2π } > 0 Claim: In fact t 1 R 0 + 1, R α We now justify the claim Note that dlog Nt, 0, G dlog t = nt 1, 0, G t=t1 Nt 1, 0, G = 1 10 But if R 0 /10 < t 1 R then for all r n < t 1 with Gr n e iθn = 0 for some θ n, we have logt 1 /r n < log 11 It follows that Nt 1, 0, G = Combining, we get R 0 /10 r n<t 1 log t 1 r n < nt 1, 0, G log 11 1 log 11 < nt 1, 0, G Nt 1, 0, G = 1 10, which is a contradiction This establishes the claim We have from 12 log mt 1, f log mt 1, F + log mt 1, G log mt 1, F α R log MR, F since R < t 1 < R α By estimates very analogous to those leading to 13 this time considering possible zeros of G with modulus between R 0 /10 and R 2 as well as those of modulus greater than R 2, we again have α 1/2 1 log MR, G < 2, for R > R, say R

16 16 AIMO HINKKANEN AND JOSEPH MILES Combining, we have log mt 1, f log MR, f α R log MR, F log MR, F + log MR, G log MR, F α R log MR, F + 1 α 1/2 R α 1/2 = α R α R 1 R log MR, F + 1 R α 1/2 = α R o1 log R > α R, for R > R Combined with the conclusion in Case I, this proves Theorem 21 when f0 = 1 and f has no zeros in the unit disk Consider then the general case and define Hz = faz/bz k, where a 0, 1], b C \ {0} and k is a non-negative integer so that H0 = 1 and H has no zeros in the unit disk The argument above proves Theorem 21 for H instead of f, and we write H = F G in what follows If we are in Case I, we recall that log R 1 = log R 0 + log P = αlog R1 γr + αlog RγR/4 = αlog R1 3/4γR and obtain for large R log mar 1, f = k log R 1 + log mr 1, H + log b k log R 1 + α R log MR, F 2 + log b k log R log b + α Rlog MR, H log MR, G α 1/2 1 > k log R log b + α R log MR, H 2 R α R log MaR, f + klog R 1 α R log R + O1 α R log MaR, f + kαlog R 1 34 γr 1 2γR + O1 = α R log MaR, f + 5/4kαlog RγR + O1 α R log MaR, f + O1 α R log MaR, f

17 GROWTH CONDITIONS 17 In Case II, we note that t 1 > R 0 and deduce for large R that log mat 1, f = k log t 1 + log mt 1, H + log b k log t 1 + α R log MR, F + log b k log t 1 + log b + α Rlog MR, H log MR, G α 1/2 1 > k log t 1 + log b + α R log MR, H 2 R k log t 1 + α Rlog MaR, f k log R + O1 α R log MaR, f + klog t 1 α R log R + O1 > α R log MaR, f + klog R 0 α R log R + O1 = α R log MaR, f + kαlog R1 γr 1 2γR + O1 = α R log MaR, f + kαγr log R + O1 α R log MaR, f In both cases 5 follows for f, with r = ar 1 or with r = at 1 and with R replaced by ar Note that the required inequality ar 1 ar α is equivalent to log R 1 α log R + α 1 log a 0, that is, 3/4γRαlog R + α 1 log a 0, which holds for all large R since γr log R as R by 4 and the condition ii of Theorem 21 For the inequality at 1 ar α we need to know that we can choose t 1 a α 1 R α < R α = R 4 since 0 < a 1 < α For this, we need to ensure that 14 holds for all t a α 1 R 4 if R is large enough If not, then there is t [a α 1 R 4, R 4 for which 14 fails, and then we obtain, as before, since t R 3, that log t log t loga α 1 R 4 log R 3 = α 1 log a + > αγr log R 5 αγrlog Rlog t 5 log R 4 for all large R since γr log R as R Thus 10 log Nt, 0, G > 10 log Nt, 0, G 10 log Nt, 0, G αγr log R log t = γr log t, 5 log R 4 5 implying that αγr log R 4 γr log t < 50 log Nt, 0, G 50 log Nt, 0, H < 51βt log t,

18 18 AIMO HINKKANEN AND JOSEPH MILES which is a contradiction to the condition ii of Theorem 21 for large R since t > R This completes the proof of Theorem 21 4 Proof of Theorem 22 Let the assumptions of Theorem 22 be satisfied Thus we assume that f is a transcendental entire function of order < 1/2 To get a contradiction, we assume that the Fatou set of f contains an unbounded component D that is a simply connected wandering domain As mentioned in the introduction, all other cases have already been settled The proof is divided into two parts The first part uses dynamics to show that since we are dealing with a component of the Fatou set, the radial spreads of the iterates cannot be too large in a certain sense The second part uses 6 and shows that the spread must be larger than that after all These two facts are incompatible, so we get a contradiction, which then completes the proof of Theorem 22 Suppose that K is a compact subset of D In this first part of the proof, our aim is to show that for a certain complex constant a depending only on D and for a possibly large positive number C > 1 depending on K, we have 15 1 C f j z a f j w a C for all z, w K and for all j 0 This is a consequence of standard estimates for the hyperbolic metric in simply connected domains This argument is well known see [1]; [8], Section 10 but we give the details for completeness To find a, note that since D is a wandering domain, it is disjoint from any of its inverse images Thus there is a disk Ba, r = { z C : z a < r } such that 16 Ba, r j=0f j D = Let D j be the component of the Fatou set of f containing f j D Note that D j is also an unbounded wandering domain of f and hence is simply connected Let L > 1 be a large constant, to be determined soon Pick j 0 and z, w K Suppose that f j z a / f j w a > L Let ζ D j be the point closest to a, so that in particular, ζ a < f j w a Let h Ω z 1, z 2 denote the hyperbolic distance between the points z 1, z 2 of the domain Ω, and let λ Ω z denote the density of the hyperbolic

19 metric of Ω at z Ω Thus GROWTH CONDITIONS 19 h Dj f j z, f j w h D z, w L 0 := max{h D z 1, z 2 : z 1, z 2 K} Since D j is simply connected, it follows from Koebe s one-quarter theorem that 1 λ Dj z 4 dist z, D j 1 4 z ζ 1 4 z a + ζ a for all z D j, where dist z, D j denotes the Euclidean distance of z from D j Hence L 0 h Dj f j z, f j w f j z a f j w a = 1 4 log f j z a + ζ a f j w a + ζ a 1 4 log f j z a + f j w a 2 f j w a = 1 log 1 + f j z a log 2 4 f j w a dr 4r + ζ a 1 4 log 1 + L log 2, which gives a contradiction if L is sufficiently large if compared to L 0 This proves 15 Next we observe that even though the constant C depends on K and may be large, we can control the radial spread of the set f j K better by using the logarithmic scale Suppose that C 0 is a preassigned constant subject only to C 0 > 1 Next we show that by 15 and 16, we have 17 1 C 0 log2 f j z a / r log2 f j w a / r C 0 for all z, w K and for all sufficiently large j j 0, say Having to restrict ourselves to j j 0 is one cost that we pay in order to get an estimate involving an arbitrary C 0 > 1 For if 17 does not hold, then there are sequences z j, w j K and integers n j such that that is, 18 log2 f n j z j a / r log2 f n j wj a / r > C 0, 2 f n j z j a r 2 f n j C0 w j a > r

20 20 AIMO HINKKANEN AND JOSEPH MILES By passing to a subsequence, we may assume that f n j z j a R 2 and f n j w j a R 1, say, where r R 1 < R 2 < or R 1 = R 2 = In the former case, we do not have f n j locally uniformly in D, so that by passing to a further subsequence, we may assume that f n j ω locally uniformly in D, where ω is a complex number with ω a r, by 16 Hence f n j z j ω and f n j w j ω as j, which contradicts 18 Thus R 1 = R 2 = But now, by 15, f n j z j a C f n j w j a < 2/ r C0 1 f n j w j a C 0 when f n j w j is large enough, which is a contradiction This completes the proof of 17 This also finishes off the first part of the proof of Theorem 22 We have now established an upper bound for radial spread, which is effective since the number C 0 > 1 is still at our disposal and so we may choose C 0 to be very close to 1 We proceed to show that if we choose K to be of large radial spread, as we may since we are choosing a compact subset of an unbounded domain D, then the large radial spread in fact persists under iteration, to the extent that we obtain a contradiction to 17 This contradiction then shows that the domain D with its defining properties could not exist at all, and the proof of Theorem 22 is complete Suppose that α, R 0, and γr satisfy the conditions of Theorem 22 Choose C 2 > 2 so large that 1 α < n=1 Next pick R so that 1 γe C n 2 C n 2 log + M1, f 19 r < r C 2 < Mr, f for all r R Choose R 1 max { R 0, R, exp {C 2 } } and, in addition, so that α1 γr > 1 for all r > R 1 Let K be a compact connected subset of D containing points z 0 and w 0 with w 0 > z 0 > R 1 and 20 log w 0 log z 0 > α2 Set K n = f n K We seek to prove that for each n 1, there are points z n, w n K n with w n > z n > R 1

21 and log w n log z n > α2 n k=1 GROWTH CONDITIONS 21 1 γe Ck 2 C k 2 log + M1, f > α Since K is connected and 20 holds, there is ζ 0 K with w 0 = ζ 0 α Thus ζ 0 > z 0 By 6, there is t ζ 0, w 0 ] with We have log mt, f log M ζ 0, f α1 γ ζ 0 fz 0 M z 0, f Take any point u 0 K with u 0 = t This is possible since K is connected We have log fu 0 log mt, f log mt, f log M ζ 0, f = log fz 0 log M z 0, f log M ζ 0, f log M z 0, f We next find a lower bound for log M ζ 0, f log M z 0, f We write ϕx = log Me x, f If 1 < r 1 < r 2 and x j = log r j for j = 1, 2, and if r 1 = z 0 and r 2 = ζ 0, we have Since ϕ is convex, we have so that hence If ϕ0 0, we get In general, if ϕ0 > 0, ϕx 2 ϕx 1 x 2 x 1 log M ζ 0, f log M z 0, f = ϕx 2 ϕx 1 ϕx 1 x 2 x 1 x 2 ϕ0 + x 1 x 2 ϕx 2 ϕx 2 x 2 ϕx 1 x 2 x 1 ϕ0, x 1 x 1 ϕx 2 ϕx 1 x 2 x2 ϕ0 1 x 1 x 1 ϕx 1 { 1 ϕx 2 ϕx 1 x 2 x 1 1 x } 1 ϕ0 x { 2 1 ϕ0 } x 2 ϕx 1 x 1 ϕx 1

22 22 AIMO HINKKANEN AND JOSEPH MILES Thus, whether ϕ0 0 or ϕ0 > 0 we have log M ζ 0, f log M z 0, f = ϕx 2 ϕx 1 log ζ 0 log z 0 We conclude that log fu 0 log M z 0, f αlog ζ 0 = log w 0 log z 0 1 γ ζ 0 log w 0 log z 0 log w 0 log z 0 log w 0 log z 0 log z 0 1 γ ζ 0 1 log+ M1, f log M z 0, f 1 γ ζ 0 log+ M1, f log M z 0, f 1 γ z 0 log+ M1, f log M z 0, f { 1 log+ M1, f log M z 0, f } 1 log+ M1, f log M z 0, f 1 γ z0 C 1 2 log + M1, f > α 2 1 γe C 2 C 1 2 log + M1, f > α > 1, where in the last inequality we use z 0 > exp {C 2 } Since fz 0 M z 0, f and fk = K 1 is connected, the set K 1 contains a point z 1 with z 1 = M z 0, f We set fu 0 = w 1 and note that log w 1 log z 1 > α2 1 γe C 2 C2 1 log + M1, f > α To extend this argument from K 1 to K n, we prove the following lemma Lemma 41 Suppose that n 1 and that for all m with 1 m n, there exist z m, w m K m with for 1 m n, and κ m log w m log z m > α2 w m > z m > R 1, z m z m 1 C 2 z 0 Cm 2 m k=1 1 γe Ck 2 C k 2 log + M1, f > α Then K n+1 contains points z n+1 and w n+1 such that w n+1 > z n+1 > R 1, z n+1 z n C 2 z 0 Cn+1 2,

23 and κ n+1 log w n+1 log z n+1 > α2 GROWTH CONDITIONS 23 n+1 k=1 1 γe Ck 2 C k 2 log + M1, f > α Proof of Lemma 41 Since K n is connected and log w n log z n > α, there is ζ n K n with log w n log ζ n = α Now by 6, find t ζ n, w n ] = ζ n, ζ n α ] with 21 log mt, f log M ζ n, f α 1 γ ζ n > 1 Then choose u n K n with u n = t Note that z n < ζ n We have, as before, log fu n log mt, f log M ζ n, f log M z n, f log M ζ n, f log M z n, f > α 1 γ ζ n log ζ n 1 log+ M1, f log z n log M z n, f log w n 1 γ ζ n C n+1 2 log + M1, f log z n Note that log ζ n > log z n C2 n log z 0 C2 n+1 Choose z n+1 K n+1 with z n+1 = M z n, f This is possible since K n+1 is connected, fz n K n+1, fz n M z n, f, and fu n K n+1 while by 21, fu n mt, f M ζ n, f > M z n, f We get, with w n+1 = fu n, that κ n+1 = log w n+1 log z n+1 > κ n 1 γe Cn+1 2 C n+1 2 log + M1, f n+1 > α 2 1 γe Ck 2 C k 2 log + M1, f > α k=1 Also z n+1 = M z n, f z n C 2 > z n > R 1 by 19 This completes the proof of Lemma 41 We continue with the proof of Theorem 22 We have previously shown that the hypothesis of Lemma 41 holds for n = 1 Hence

24 24 AIMO HINKKANEN AND JOSEPH MILES induction on n together with Lemma 41 shows that for every n 1 there are z n, w n K n with log w n log z n > α The distinction between log w n and log2 w n a / r is immaterial and is easily handled by taking R 1 even larger; we omit the details Taking C 0 in 17 with 1 < C 0 < α and choosing the appropriate j 0, we obtain a contradiction as soon as n j 0 This completes the proof of Theorem 22 5 Proof of Theorem 23 Let the assumptions of Theorem 23 be satisfied Let γ : e, 0, be a decreasing function such that γr/βr as r and γe 2n < n Since γr satisfies the hypothesis of Theorem 21, 3γr satisfies the hypothesis of Theorem 22 Thus Theorem 23 now follows from Theorem 22 6 Proof of Theorem 24 Let the assumptions of Theorem 24 be satisfied Define γ 1 r = sup{γt : t r} for r > e 3 Then the function γ 1 is decreasing and γ 1 r γr for all r > e 3 We claim that βr/γ 1 r as r Suppose that r n Let t n r n be such that γt n 1/2γ 1 r n Then βr n /γ 1 r n βt n /γ 1 r n βt n /2γt n, which tends to infinity by assumption For r > e 6 define βr = 1/2β r and Γr = γ 1 r Note that βr log r = 1/2β r log r = β r log r is a increasing function tending to infinity as r We also have βr = 1/2β r > 2 log log r log 2 > 1 2 4log log r 2 for large r by 7 Let γr be a function satisfying γr Γr, βr/ γr, and γr log r as r

25 GROWTH CONDITIONS 25 We choose an increasing sequence R k, for k 1, satisfying for all k 2 23 γr k log R k > 3 log R k 1 and 24 { } { } k 1 βrj log R j βrk log R k exp < exp 2 64 j=1 By choosing R 1 sufficiently large, we may and will assume that furthermore, 25 for all k 1 We set 1 < α1 3 γr k and α1 3 γr k+1 log R k+1 > 1 + α1 + γr k log R k 26 ρ k = βr k 8α γr k Evidently, ρ k as k Associate with R k a finite sequence defined as follows We set log r 0k = α1 3 γr k log R k, log r 1k = α1 2 γr k log R k, log r 2k = α1 γr k log R k, log r 3k = α log R k, log r 4k = α1 + γr k log R k Our assumptions 25 guarantee that R k < r 0k and r 4k < R k+1 < r 0,k+1 For large k, the ratio R k+1 /r 4k is large For a real number x, we denote by x the largest integer not exceeding x For k 1, let G k be the finite product of Weierstrass factors of genus 0 with positive zeros satisfying ρk t nt, 0, G k =, r 0k t r 4k r 0k If m > ρ k + 1, the m th Fourier coefficient of log G k re iθ is see [9], with nt = nt, 0, G k, c m r, G k := 1 2π = 1 2 2π 0 r 0 e imθ log G k re iθ dθ m t nt dt 1 r t 2 r r t m nt t dt

26 26 AIMO HINKKANEN AND JOSEPH MILES We suppose that r 0k < r r 3k Elementary calculations yield 1 r m ρk t nt 1 r dt 2 r t 2m + ρ k 0 Since nt t/r 0k ρ k 1, it is easily seen that 1 r m nt dt 1 r4k r m nt dt 2 r t t 2 r t t ρk m ρk 1 r r 1 1 2m ρ k 2m r 0k We now specify a choice of m We choose m = ρ k 1 + δ k where 2/ρ k < δ k < 3/ρ k We note that 1 < m ρ k < 3 For this m, we have uniformly for r 0k < r < r 3k as k that m ρk r 0 r 4k r 0k r 4k and thus for all large k and for all r r 0k, r 3k ], 1 r m nt dt > 1 ρk r 2 t t 8 Combining, we get 27 c m r, G k > 1 16 r r r 0k r 0k ρk We have the elementary estimate for r 0k < r r 3k that r ρk t dt Nr, 0, G k r 0k t = 1 ρk r 1 ρ k r 0k r 0k Evidently Nr, 0, G k < 8δ k c m r, G k From Jensen s Theorem we now conclude for r 0k < r r 3k and with mr, 1/G k having its usual Nevanlinna theory meaning as the proximity function of 1/G k that This implies that mr, 1/G k = T r, G k Nr, 0, G k 1 2 c mr, G k Nr, 0, G k 1 > c m r, G k 2 8δ k 28 log min { G k re iθ : 0 θ 2π } < c m r, G k 1 2 8δ k

27 We make the following observations: i nr 4k, 0, G k = r 4k /r 0k ρ k, or Thus GROWTH CONDITIONS 27 log nr 4k, 0, G k ρ k log r 4k log r 0k 29 nr 4k, 0, G k exp ii For r 1k < r r 3k, we have = ρ k 4α γr k log R k = βr k log R k 2 { βrk log R k log c m r, G k ρ k log r log r 0k log by 26 and 27 βr k log R k log 16 > βr k log R k 8 16 For k 2, define the polynomial F k by k 1 F k = G j For r 1k < r r 3k we have r log Mr, F k = log 1 + r 31 dnt, 0, F k 0 t r < e log r dnt, 0, F k = enr, 0, F k t Thus by 24 and 29 0 j=1 log log Mr, F k < 1 + log Nr, 0, F k { } βrj log R j < 1 + log exp log r 2 j<k { } βrk log R k < 1 + log exp log r 64 2 } = 1 + βr k log R k 64 + log log r < βr k log R k 32, 32 where in the last step we have applied 22

28 28 AIMO HINKKANEN AND JOSEPH MILES We have from 28, 30, and 32 that for large k log mr, F k G k log Mr, F k + log mr, G k { } { } βrk log R k < exp 1 βrk 32 4 exp log R k < 0 16 We define f = j=1 where the convergence of the infinite product is easily checked and we omit the details; for each k 2 we may write f = F k G k H k where H k = G j The above certainly implies that G j j=k+1 33 log mr, f < 0, for r 1k < r r 3k Recall that we have required that γr k log R k > 3 log R k 1 For 1 j k 1 this guarantees that which is equivalent to or 34 γr k log R k > 2 + γr k log R j, log R k < For R k < r r 1k, we have γr k log R k log R j, log R k log R k log R j < γr k log Mr, F k log MR k, F k < 1 + Nr, 0, F k NR k, 0, F k = 1 + nr k, 0, F k log r R k Thus log Mr, F k log MR k, F k 1 < nr k, 0, F k log MR k, F k log r 1 + R k log MR k, F k nr k, 0, F k log MR k, F k log R 1 kα1 2 γr k 1 + log MR k, F k

29 GROWTH CONDITIONS 29 Now, with n j denoting the total number of zeros of G j, we have nr k, 0, F k log MR k, F k log R j<k k n j log R k j<k n j logr k /R j < γr k by 34 Thus log Mr, F k log MR k, F k 1 < γr k {α1 2 γr k 1} + < α1 γr k γr k + Thus for R k < r r 1k, we have log Mr, F k log MR k, F k < α1 γr k, 1 log MR k, F k 1 log MR k, F k where we have used the fact that γr log r as r For R k < r r 1k, we have 35 log mr, f log MR k, f < log mr, F k log MR k, F k < log Mr, F k log MR k, F k < α1 γr k α1 ΓR k = α1 γ 1 R k α1 γ 1 R k α1 γr k, where we note from 28 that log mr, f/f k < 0 The combination of 33 and 35 establishes 9 with R = R k It remains only to show that log log Mr, f < βr log r for all sufficiently large values of r For then, if t r, we have log log Mt, f/ log t < βt βr since the function βr is decreasing, and 8 follows First suppose that 2r 4k < r < R 2/3α k+1 We have log Mr, F k G k < enr, 0, F k G k by an argument analogous to that establishing 31 For j > k, integration by parts yields r4j log Mr, G j = log 1 + r dnt, 0, G j r 0j t ρ ρj 1 j r r4j ρ j 1 r 0j r 0j

30 30 AIMO HINKKANEN AND JOSEPH MILES We note that log r log R2/3α j < r 0j r 0j 1 log R j 8 and ρ j 1log r 4j log r 0j < 4ρ j α γr j log R j = 1 2 βr j log R j Combining, we conclude that log Mr, G j < 2 exp In view of 23, we deduce for large r that log Mr, H k = From 24 and 29 we have implying that { 1 2 βr j 1 } log R j 8 j=k+1 log Mr, G j < 1 log Mr, f < enr, 0, f + 1 < enr, 0, f log r + 1 βrk log R k < 4 exp log r + 1, 2 36 log log Mr, f < log 8 + βr k log R k + log log r 2 < βr log r = β r log r βr log r, where we have used 22 and the monotonicity of βr log r Now suppose that R 2/3α k r 2r 4k It follows from 23 that for all large k we have 2r 4k < R 4/3α k r 2 4r 2 4k < R 2/3α k+1 Applying an intermediate inequality in 36 to r 2, we obtain log log Mr, f < log log Mr 2, f < βr log r This completes the proof of Theorem 24

31 GROWTH CONDITIONS 31 7 Concluding remarks Theorem 24 shows that our minimum modulus approach to the study of the existence of unbounded components of the Fatou set of an entire function cannot be effective for entire functions of order zero whose growth is faster than that specified in Theorem 23 A simpler construction shows that this approach cannot be effective for entire functions of order ρ with 0 < ρ 1/2 We now outline a construction that shows that no reasonable analogue of Theorem 21 exists for entire functions of positive order less than or equal to 1/2 This construction can be modified to produce a function of order 1/2, minimal type Theorem 71 Suppose that 0 < ρ 1/2 There exists an entire function f of order ρ such that if α > 1, there exist α 0, α and an unbounded sequence R k such that for all k 37 log mr, f log MR k, f < α, whenever R k r R α k Proof of Theorem 71 We sketch the construction of the required f Let a k be a sequence of real numbers greater than 1 that is dense in 1, Let α n be a sequence each of whose members is among the a k and such that each a k appears infinitely often in the sequence α n We choose ρ 0, ρ, and set β n = α n ρ, α n = α n 2α n 2β n 2α n β n We note that β n < α n and α n < α n We next choose a rapidly increasing sequence r k with r 1 > 1, specifically requiring for k 2 that 38 log r k 1 < β k log r 2αk 2 k = ρ log r k 2α k and 39 k 1 r ρ j log r k < rρ ρ r j k4 j=1 We set n k = r ρ k and define fz = k=1 1 z r k nk It is easy to verify that the infinite product converges and defines an entire function of order ρ We note that f has zeros of large multiplicity

32 32 AIMO HINKKANEN AND JOSEPH MILES at widely-spaced points on the positive real axis For each r > 0 it is evident that Mr, f = f r and mr, f = fr For k = 1, 2, 3,, define and k 1 F k z = 1 z nj, r j G k z = H k z = j=1 1 z r k nk, j=k+1 1 z r j nj For k 1, define R k by R α k k = r k Note that rk 1 4 < R k First suppose that R k r R α k β k k It is elementary that k 1 log mr, f < log mr, F k = n j log 1 r r j Elementary estimates yield j=1 < α k β k log R k nr k 1, 0, f k 1 log MR k, f > log MR k, F k = n j log 1 + R k r j k 1 > n j log R k log r k 1 > j=1 j=1 1 β k 2α k where we have used 38 We conclude that log R k nr k 1, 0, f, 40 log mr, f log MR k, f < α k β k = α 1 β k k, whenever R k r R αk βk k 2α k Now suppose that R α k β k k < r R α k k We have log mr, G k H k < log mr, G k = n k log 1 rrk < n k r n k Rα k β k k r k R α k k < 1 R α k ρ k 2 R β k k = 1 R α kρ ρ k 2

33 We also have for such r that GROWTH CONDITIONS 33 log Mr, F k log MR α k k, F k 1 k = k 1 < 2 n j log r k Rαkρ ρ k <, r j 2 j=1 j=1 n j log 1 + Rα k k r j where we have used 39 We conclude that 41 log mr, f < log Mr, F k + log mr, G k H k < 0, R α k β k k < r R α k k The combination of 40 and 41 yields 42 log mr, f log MR k, f < α k, whenever R k r R α k k Now suppose that α > 1 is given There is a constant subsequence, say α mp, of the sequence α k such that for all p 1 we have α mp = γ α For each p, the quantity α m p takes a certain constant value γ 0, γ Considering how γ depends on γ via the fixed choice of ρ, it is clear that we may choose γ so that then γ < α Now considering 42 only when k is in the subsequence m p, we obtain 37, with the sequence R mp playing the role of the sequence R k required for 37 and with γ playing the role of α This completes the proof of Theorem 71 Remark We see from the proof of Theorem 71 that the problematic situation where the minimum modulus is small throughout a long interval occurs immediately before the radii corresponding to the zeros of f of high multiplicity In the proof of Theorem 24 we do not use sparsely occurring multiple zeros, but simple zeros spread in a suitable way within a zone, occurring in sparse zones This allows us to get a better bound in Theorem 24 than would otherwise be possible For functions of positive order, such refinements are not needed, and the use of multiple zeros is sufficient for our purposes References [1] JM Anderson and A Hinkkanen, Unbounded domains of normality, Proc Amer Math Soc , [2] IN Baker, The iteration of polynomials and transcendental entire functions, J Austral Math Soc Series A , [3] IN Baker, Wandering domains in the iteration of entire functions, Proc London Math Soc Series , [4] AF Beardon, Iteration of rational functions, Springer-Verlag, Berlin New York 1991

34 34 AIMO HINKKANEN AND JOSEPH MILES [5] L Carleson and T Gamelin, Complex dynamics, Universitext: Tracts in Mathematics, Springer, New York, 1993 [6] WK Hayman, Meromorphic functions, Clarendon Press, Oxford, 1964 [7] A Hinkkanen, Entire functions with no unbounded Fatou components, pp in Complex analysis and dynamical systems II, Contemporary Mathematics, vol 382, AMS, 2005 [8] A Hinkkanen, Entire functions with bounded Fatou components, to appear in Transcendental Dynamics and Complex Analysis, Cambridge Univ Press [9] J Miles and DF Shea, An extremal problem in value-distribution theory, Quart J Math , [10] J Milnor, Dynamics in one complex variable, Introductory lectures, Friedr Vieweg & Sohn, Braunschweig, 1999 [11] PJ Rippon and GM Stallard, Functions of small growth with no unbounded Fatou components, preprint [12] G Stallard, Some problems in the iteration of meromorphic functions, PhD thesis, Imperial College, London, 1991 [13] G Stallard, The iteration of entire functions of small growth, Math Proc Cambridge Philos Soc , [14] N Steinmetz, Rational iteration, Complex analytic dynamical systems, de Gruyter Studies in Mathematics, 16, Walter de Gruyter & Co, Berlin, 1993 [15] Jian-Hua Zheng, Unbounded domains of normality of entire functions of small growth, Math Proc Cambridge Philos Soc , University of Illinois at Urbana Champaign, Department of Mathematics, 1409 West Green Street, Urbana, IL USA address: aimo@uiucedu, joe@mathuiucedu