Iterating the minimum modulus: functions of order half, minimal type
aa r X i v : . [ m a t h . C V ] F e b ITERATING THE MINIMUM MODULUS: FUNCTIONS OFORDER HALF, MINIMAL TYPE
D. A. NICKS, P. J. RIPPON, AND G. M. STALLARD
Abstract.
For a transcendental entire function f , the property that thereexists r > m n ( r ) → ∞ as n → ∞ , where m ( r ) = min {| f ( z ) | : | z | = r } , is related to conjectures of Eremenko and of Baker, for both of which order1 / / f hassufficiently regular growth and we give examples to show the sharpness of ourresults by using a recent generalisation of Kjellberg’s method of constructingentire functions of small growth, which allows rather precise control of m ( r ). Dedicated to the memory of Walter Hayman. Introduction
Let f be a transcendental entire function and denote by f n , n ∈ N , the n -thiterate of f . The results in this paper address a question in complex dynamicsconcerning the escaping set of a transcendental entire function f , defined as I ( f ) = { z : f n ( z ) → ∞ as n → ∞} , namely, whether all the components of I ( f ) are unbounded. This question isknown as Eremenko’s conjecture [3] and despite much work it remains open.For any transcendental entire function f , the maximum modulus and minimummodulus of f are defined as follows, for r ≥ M ( r ) = M ( r, f ) = max | z | = r | f ( z ) | and m ( r ) = m ( r, f ) = min | z | = r | f ( z ) | , respectively. There is a huge literature about the relationship between m ( r ) and M ( r ) for various types of transcendental entire functions. Clearly m ( r ) < M ( r )for all r > M ( r ) is strictly increasing and unbounded. On theother hand, the function m ( r ) is alternately increasing and decreasing betweenadjacent values of r for which m ( r ) = 0 and eventually decreases to 0 in thecase that f has only finitely many zeros. For certain functions, however, m ( r ) iscomparable in size to M ( r ) for an unbounded set of values of r .We let M n ( r ) and m n ( r ) be defined by iterating the real functions M ( r ) and m ( r ) respectively. For any transcendental entire function f we have(1.1) M n ( r ) → ∞ as n → ∞ , for r ≥ R = R ( f ) say, but for the iterated minimum modulus the property:(1.2) there exists r > m n ( r ) → ∞ as n → ∞ , may or may not hold, depending on the function f . Mathematics Subject Classification.
Primary 37F10, Secondary 30D05.The last two authors were supported by the EPSRC grant EP/R010560/1.
It has been known for some time that the sequence M n ( r ) is of importancein relation to work on Eremenko’s conjecture, since it plays a key role in thedefinition of a subset of I ( f ) called the fast escaping set , all of whose componentsare unbounded; see, for example, [12]. More recently, it has been observedthat property (1.2), when it is true, can also play an important role in relationto this conjecture; see [9], [10] and [8]. For example, in [8] we obtained thefollowing result, which gives a family of transcendental entire functions for whichEremenko’s conjecture holds in a particularly strong way. Theorem 1.1.
Let f be a real transcendental entire function of finite order withonly real zeros for which there exists r > such that m n ( r ) → ∞ as n → ∞ . Then I ( f ) is connected and hence Eremenko’s conjecture holds for f .Remark In [8, Theorem 1.1] we showed moreover that, for such functions f , theset I ( f ) has the structure of an (infinite) spider’s web.Recall that the order ρ ( f ) and the lower order λ ( f ) of a transcendental entirefunction f are ρ ( f ) = lim sup r →∞ log log M ( r )log r and λ ( f ) = lim inf r →∞ log log M ( r )log r , respectively, and f is said to be real if f (¯ z ) = f ( z ) , for z ∈ C . In view of Theorem 1.1, it is natural to ask which transcendental entire func-tions satisfy property (1.2); in particular, which amongst those that satisfy thehypotheses of this theorem. In [8] we proved several results in relation to thisquestion, including the following.(a) For a real transcendental entire function f of finite order with only realzeros:(i) if 0 ≤ ρ ( f ) < /
2, then property (1.2) holds;(ii) if ρ ( f ) >
2, then property (1.2) does not hold.(b) For any ρ ∈ [1 / ,
2] there are examples of real transcendental entire func-tions with only real zeros of order ρ for which property (1.2) holds andalso examples of such functions for which (1.2) does not hold.Actually, property (1.2) holds for all entire functions of order less than 1 /
2; see[10, Theorem 1.1]. This follows from the cos πρ theorem which shows that forfunctions of order less than 1 / r ; see [5, Theorem 6.13].For functions of order 1 /
2, minimal type, weaker results on the size of the mini-mum modulus are known; in particular, Wiman showed that, for such functions, m ( r ) is unbounded on (0 , ∞ ); see [5, Theorem 6.4]. In view of Wiman’s result,one might hope that property (1.2) holds for real entire functions with only realzeros of order 1/2, minimal type, that is, when(1.3) lim sup r →∞ log log M ( r )log r = 12 and lim r →∞ log M ( r ) r / = 0 . In this paper, we show that (1.2) does indeed hold for many families of suchfunctions, but does not hold for all. Our main positive result is the followingwhich, roughly speaking, shows that (1.2) holds whenever (log M ( r )) /r / tendsto 0 (as r → ∞ ) in a sufficiently regular manner. TERATING THE MINIMUM MODULUS 3
Theorem 1.2.
Let f be a transcendental entire function of order at most 1/2,minimal type, and suppose that there exists r > such that, for r > r , (1.4) log M ( r ) r / ≤
14 log M ( s ) s / , for some s ∈ (0 , r ) which satisfies M ( s ) ≥ r . Then there exists r > such that m n ( r ) → ∞ as n → ∞ . Remarks
1. The second hypothesis of Theorem 1.2 implies the first hypothesis,that f has order at most 1/2, minimal type, though this is not quite obvious.2. Theorem 1.2 implies the result stated earlier that, if ρ ( f ) < /
2, then prop-erty (1.2) holds. Indeed, if ρ ( f ) < / δ := 1 / − ρ ( f ) ∈ (0 , / , then (1.4) holds with s = r δ for r sufficiently large, since this choice of s gives r / /s / = r ρ ( f )+ δ/ and M ( r δ ) ≥ r for r sufficiently large.In Section 2, we give a number of conditions on the maximum modulus and onthe zeros of f which imply the hypotheses of Theorem 1.2 and are convenient forapplications.In the second half of the paper we use a recent generalisation of a method ofKjellberg [13] to construct examples of functions of order 1 /
2, minimal type, forwhich property (1.2) does not hold. First, we show that no matter how slowly log M ( r ) grows, consistent with f having order 1/2, minimal type, we cannotdeduce that property (1.2) holds. Example 1.3.
Let δ : (0 , ∞ ) → (0 , ) be a decreasing function such that δ ( r ) → as r → ∞ . Then there exists a real transcendental entire function f of order / , minimaltype, with only real zeros, and r > such that (1.5) log M ( r, f ) ≤ r / − δ ( r ) , for r ≥ r , for which property (1.2) does not hold. Next we show that making the additional assumption of positive lower order, oreven of lower order 1 /
2, is insufficient to ensure that property (1.2) holds.
Example 1.4.
There exists a real transcendental entire function f of order / ,minimal type, and of lower order / , with only real zeros such that property (1.2) does not hold. Finally, we remark that consideration of the iterates of the minimum modulusfirst arose in connection with the conjecture of Baker that entire functions oforder 1 /
2, minimal type, have no unbounded Fatou components; see [7] for themost recent progress on this conjecture.The structure of the paper is as follows. In Section 2 we prove our positiveresults, including Theorem 1.2, and in Section 3 we recall some results from [13]needed for the proofs of Examples 1.3 and 1.4, which are given in Section 4.
D. A. NICKS, P. J. RIPPON, AND G. M. STALLARD Proof of Theorem 1.2 and some special cases
In this section we prove Theorem 1.2, our positive result about property (1.2),and give some special cases of it which we use to obtain examples of functionsthat satisfy the hypotheses of Theorem 1.2.The proof of Theorem 1.2 depends on the following lemma of Beurling [2, page 95].For any subharmonic function u we write B ( r, u ) = max | z | = r u ( z ) , where r > . Lemma 2.1. If u is subharmonic in C , < r < r , and E ( r , r ) = { r ∈ [ r , r ] : inf | z | = r u ( z ) ≤ } , then (2.1) B ( r , u ) >
12 exp (cid:18) Z E ( r ,r ) t dt (cid:19) B ( r , u ) . In particular, if E ( r , r ) = [ r , r ] , then B ( r , u ) ≥ (cid:18) r r (cid:19) / B ( r , u ) . In order to work with property (1.2), the function e m ( r ) := max { m ( s ) : 0 ≤ s ≤ r } , for r ∈ [0 , ∞ ) , was introduced in [10] and it was shown that property (1.2) is true if and only if(2.2) there exists R > e m ( r ) > r, for r ≥ R. We use this equivalent property to prove Theorem 1.2.
Proof of Theorem 1.2.
Suppose the hypotheses of Theorem 1.2 hold. If prop-erty (1.2) does not hold, then by (2.2) there exist arbitrarily large r > r suchthat e m ( r ) ≤ r ; that is, for arbitrarily large r we have(2.3) m ( t ) ≤ r, for 0 < t ≤ r. For such an r , there exists by hypothesis s < r such that log M ( s ) ≥ r and (1.4) holds. However, we deduce from (2.3) by applying Lemma 2.1 to u ( z ) = log( | f ( z ) | /r ) thatlog M ( r ) > log M ( r ) r > (cid:16) rs (cid:17) / log M ( s ) r ≥ (cid:16) rs (cid:17) / log M ( s ) , which contradicts condition (1.4). (cid:3) For applications of Theorem 1.2, it is often useful to express condition (1.4) interms of the functions ε ( r ) , r > , and k ( r ) , r > , defined as follows:log M ( r ) = r / − ε ( r ) and k ( r ) = ε ( r ) log r ;equivalent forms of condition (1.4) are then(2.4) s ε ( s ) ≤ r ε ( r ) and k ( r ) ≥ k ( s ) + log 4 , for r > r , where M ( s ) ≥ r as before. TERATING THE MINIMUM MODULUS 5
Note that if f has order at most 1/2, minimal type, then for all sufficiently largevalues of r we have 0 < ε ( r ) < / < k ( r ) = ε ( r ) log r < log r .It is also useful to note that if f has order 1/2 then:(a) f has lower order 1/2 if and only if ε ( r ) → r → ∞ ,(b) f is of minimal type if and only if k ( r ) → ∞ as r → ∞ .We now give two special cases of Theorem 1.2 which are convenient for appli-cations. Theorem 2.2, part (a), shows that property (1.2) holds whenever thefunction f has positive lower order and ε ( r ) = k ( r ) / log r does not tend to 0 tooquickly. Example 1.3 shows that positive lower order alone is not sufficient here. Theorem 2.2.
Let f be a transcendental entire function of order 1/2, minimaltype, and let ε ( r ) , r > , and k ( r ) , r > , be defined as above. Then there exists r > such that m n ( r ) → ∞ as n → ∞ if either of the following statementsholds: (a) there exist δ ∈ (0 , / , C > and r > such that (2.5) δ log r ≥ k ( r ) ≥ C δ / − δ log log r, for r > r ;(b) ε ( r ) → as r → ∞ and there exist C, d > and r > such that (2.6) k ( r ) ≥ Ck ((log r ) d ) , for r > r . Proof.
By Theorem 1.2, it is sufficient to show that, for all sufficiently large r ,condition (1.4) holds for some s ∈ (0 , r ) with M ( s ) ≥ r .(a) The inequality k ( r ) ≤ δ log r can be written as ε ( r ) ≤ δ , that is,log M ( r ) ≥ r / − δ , so the condition M ( s ) ≥ r is satisfied by taking s = (2 log r ) / (1 / − δ ) . With thischoice of s the required condition (1.4), or equivalently (2.4), can be written as(2.7) k ( r ) ≥ k ((2 log r ) / (1 / − δ ) ) + log 4 , for r > r . The hypothesis k ( r ) ≤ δ log r for r > r implies that k ((2 log r ) / (1 / − δ ) ) ≤ δ log (cid:0) (2 log r ) / (1 / − δ ) (cid:1) = δ / − δ (log 2 + log log r ) , for (2 log r ) / (1 / − δ ) > r , so a condition of the form k ( r ) ≥ δ / − δ (log 2 + log log r ) + log 4 , for r > r , is sufficient to imply that (2.7) and hence (1.4) hold for some r ≥ r . Thisproves part (a).(b) We first choose δ > / (1 / − δ ) < d . The hypothesis that ε ( r ) → r → ∞ implies that there exists r ( C, d ) > ε ( r ) ≤ δ/C for r > r ( C, d ), so log M ( r ) ≥ r / − δ/C , for r > r ( C, d ) . D. A. NICKS, P. J. RIPPON, AND G. M. STALLARD
Therefore, the condition M ( s ) ≥ r is satisfied, for sufficiently large r , by taking s = (log r ) / (1 / − δ ) , since C >
1. With this choice of s the required condition(1.4), or equivalently (2.4), can be written as k ( r ) ≥ k ((log r ) / (1 / − δ ) ) + log 4 , for r > r , so a condition of the form k ( r ) ≥ Ck ((log r ) / (1 / − δ ) ) , for r > r , is sufficient to imply that (1.4) holds for some r ≥ max { r , r ( C, d ) } . This provespart (b). (cid:3) Theorem 2.2 can be used to give many examples of transcendental entire func-tions of order 1/2, minimal type, for which property (1.2) holds, including onesfor which we can deduce that I ( f ) is connected by using Theorem 1.1. We givehere some examples of this type, derived from Theorem 2.2, part (b), in whichthe functions have a very regular distribution of zeros.For any transcendental entire function f we define n ( r ) = n ( r, f ) to be thenumber of zeros of f in { z : | z | ≤ r } , counted according to multiplicity. Theorem 2.3.
Let ε ( r ) satisfy the hypotheses of Theorem 2.2, part (b) and inaddition suppose that ε ( r ) is decreasing and k ( r ) = ε ( r ) log r is increasing.Let f be a transcendental entire function of the form f ( z ) = Y n ∈ N (cid:18) − za n (cid:19) , < | a | < | a | < · · · , where n ( r, f ) ∼ r / − ε ( r ) as r → ∞ . Then (a) there exist constants < A < B and R > such that Ar / − ε ( r ) ≤ log M ( r, f ) ≤ Br / − ε ( r ) , for r > R, (b) there exists r > such that m n ( r ) → ∞ as n → ∞ .Remark A family of functions that satisfy the hypotheses of Theorem 2.2,part (b), and Theorem 2.3 is given by k ( r ) = α (log n r ) β , α, β > , n ≥ , where log n denotes the n -th iterated logarithm.For the proof of Theorem 2.3, we require a result about the relationship between M ( r ) = M ( r, f ) and the following quantities: N ( r ) = Z r n ( t ) t dt and Q ( r ) = r Z ∞ r n ( t ) t dt. We use the following estimates; see [11, Lemma 3.3], for example.
Lemma 2.4.
Let f be a transcendental entire function of order less than with f (0) = 1 . Then, for r > , N ( r ) ≤ log M ( r ) ≤ N ( r ) + Q ( r ) . TERATING THE MINIMUM MODULUS 7
Proof of Theorem 2.3.
Since n ( r ) ∼ r / − ε ( r ) as r → ∞ , where ε ( r ) is positiveand decreasing to 0, and k ( r ) = ε ( r ) log r is increasing, we have N ( r ) = Z r o (1) t / ε ( t ) dt ≥ (1 + o (1)) r ε ( r ) Z r dtt / = 2 r / − ε ( r ) (1 + o (1))and Q ( r ) = r Z ∞ r o (1) t / ε ( t ) dt ≤ (1 + o (1)) r − ε ( r ) Z ∞ r dtt / = 2 r / − ε ( r ) (1 + o (1)) , so Q ( r ) ≤ N ( r )(1 + o (1)) as r → ∞ . Since ε ( r ) is decreasing, we have N ( r ) = Z r (1 + o (1)) t / ε ( t ) dt ≤ (1 + o (1)) Z r dtt / ε ( r ) = 1 + o (1)1 / − ε ( r ) r / − ε ( r ) , so N ( r ) + Q ( r ) ≤ r / − ε ( r ) (1 + o (1)) as r → ∞ . Part (a) follows by Lemma 2.4 and the above inequalities for N ( r ) and Q ( r ).It follows from part (a) thatlog M ( r, f ) = r / − ˆ ε ( r ) , where ˆ ε ( r ) = ε ( r ) + O (1) / log r as r → ∞ , so ˆ ε ( r ) → r → ∞ and ˆ k ( r ) = ˆ ε ( r ) log r satisfiesˆ k ( r ) = k ( r ) + O (1) as r → ∞ . Since k ( r ) satisfies (2.6), so in particular k ( r ) → ∞ as r → ∞ , it follows that ˆ k ( r )also satisfies (2.6). Hence f satisfies the hypotheses of Theorem 2.2, part (b), sothe proof is complete. (cid:3) Constructing entire functions of small order
Our method of proving Examples 1.4 and 1.3 uses a recent generalisation [13]of Kjellberg’s method [6, Chapter 2] for constructing transcendental entire func-tions of order less than 1 / | f | where f is a transcendental entire func-tion. In this section we summarise the results from [13] which are needed toconstruct our examples.Kjellberg’s method is a two stage process:1. a continuous subharmonic function u with the required properties is ob-tained by using a positive harmonic function defined in the complementof a particular sequence of radial slits, on which u vanishes;2. the Riesz measure of u is discretised to produce an entire function f suchthat log | f | is close to u away from the zeros of f .The paper [13] gives a generalisation of Kjellberg’s method which allows the slitsto be chosen more flexibly than in [6].We now recall some of the key definitions and terminology needed to state theresults from [13]. D. A. NICKS, P. J. RIPPON, AND G. M. STALLARD
Definition 3.1.
A subharmonic function u is in the class K if u is continuousin C and positive harmonic in D = C \ E , where E ⊂ ( −∞ ,
0] is a closed set onwhich u vanishes. We assume that each point of E is regular for the Dirichletproblem in D . Remark
For each closed subset of the negative real axis, there is exactly onecorresponding function u ∈ K up to positive scalar multiples, by a result ofBenedicks [1, Theorem 4].Recall that for a set S ⊂ R + and r >
1, we define the upper logarithmic density of S , Λ( S ) = lim sup r →∞ r Z S ∩ (1 ,r ) dtt , and the lower logarithmic density of S ,Λ( S ) = lim inf r →∞ r Z S ∩ (1 ,r ) dtt . When Λ( S ) = Λ( S ) we speak of the logarithmic density of S , denoted by Λ( S ).Recall next that, for a continuous subharmonic function u in C , A ( r ) = A ( r, u ) = min | z | = r u ( z ) and B ( r ) = B ( r, u ) = max | z | = r u ( z ) , and the order and lower order of u are ρ ( u ) = lim sup r →∞ log B ( r )log r and λ ( u ) = lim inf r →∞ log B ( r )log r , respectively. For all u ∈ K we have 0 ≤ λ ( u ) ≤ ρ ( u ) ≤ / K ;see [13, Theorem 1.2]. Theorem 3.2.
Let u ∈ K , with E the corresponding closed subset of the negativereal axis. Then u has the following properties. (a) Monotonicity properties: for all r > , u ( re iθ ) is decreasing as a function of θ, for ≤ θ ≤ π, so, in particular, B ( r, u ) = u ( r ) and A ( r, u ) = u ( − r ) for all r > . Also, u ( r ) r / = B ( r, u ) r / is decreasing for r > , so, in particular, ρ ( u ) ≤ / . (b) Bounds for order and lower order: ρ ( u ) ≥ Λ( E ∗ ) a nd λ ( u ) ≥ Λ( E ∗ ) , where E ∗ = { x : − x ∈ E } . The next result, again taken from [13, Theorem 1.5], concerns the behaviour ofcertain functions in the class K . TERATING THE MINIMUM MODULUS 9
Theorem 3.3.
Let u ∈ K , with E the corresponding closed subset of the negativereal axis. If E c ⊃ [ n ≥ ( − d n , − c n ) , where ≤ c < d < c < d < · · · , and lim sup n →∞ d n /c n > , then u ( r ) r / → as r → ∞ . The final result we need from [13, Theorem 1.6] describes the approximationused in the second stage of Kjellberg’s process, showing how we can approximatea function u ∈ K by log | f | , where f is entire. This generalises the result givenby Kjellberg for a particular type of set E ; see [6, Chapter 4]. Theorem 3.4.
Suppose that u ∈ K and E = [ n ≥ [ − b n , − a n ] , where ≤ a < b < a < b < · · · , and a n → ∞ as n → ∞ . Put D = C \ { z : dist( z, E ) ≤ } . Then there exists an entire function f with only negative zeros, all lying in theset E , such that (3.1) log | f ( z ) | − u ( z ) = O (log | z | ) as z → ∞ , for z ∈ D . Moreover, if we also have (3.2) b n /a n ≥ d > , for n ≥ , then there exists R = R ( u ) > such that (3.3) log | f ( z ) | ≤ u ( z ) + 4 log | z | , for | z | ≥ R. Theorem 3.4 will enable us to approximate subharmonic functions u ∈ K byfunctions of the form log | f | , where f is a transcendental entire function of thesame order, lower order and type class as u .4. Proofs of Examples 1.3 and 1.4
We prove Examples 1.3 and 1.4 by using Theorems 3.2, 3.3 and 3.4 to constructentire functions that approximate suitable subharmonic functions. In each casewe show that property (1.2) is false by arranging that the minimum modulusof the entire function is relatively small on intervals that are relatively long,consistent with the required growth of the maximum modulus.We begin by proving the following result, needed in the construction of Exam-ple 1.3. The proof is rather long as the set E has to be chosen carefully so thatthe function u ∈ K does not grow too quickly on the positive real axis and takeslarge values on the negative real axis only relatively rarely. Lemma 4.1.
Let δ : (0 , ∞ ) (0 , / be a decreasing function such that δ ( r ) → as r → ∞ , and let a n and b n , n ≥ , be chosen to satisfy (4.1) a n +1 ≥ b n , for n ≥ , (4.2) δ ( a n +1 ) ≤ log b n b n , for n ≥ , and (4.3) log b n +1 = b n log b n log a n +1 , for n ≥ . Then the unique subharmonic function u ∈ K corresponding to the set E = S n ≥ [ − b n , − a n ] , with u (1) = 1 , satisfies ρ ( u ) = 1 / and u ( r ) /r / → as r → ∞ , and also (a) there exists r > such that (4.4) u ( r ) ≤ r / − δ ( r ) , for r ≥ r ;(b) we have u ( − r ) = 0 , for r ∈ [ a n , b n ] , n ≥ , and, for n sufficiently large, (4.5) u ( − r ) < log b n +1 , for b n ≤ r ≤ a n +1 . Proof.
Let u ∈ K be the unique subharmonic function with u (1) = 1 correspond-ing to the set E = S n ≥ [ − b n , − a n ], where a n and b n satisfy (4.1), (4.2) and (4.3).The property (4.3) implies that Λ( E ∗ ) = 1, so u has order 1/2 by Theorem 3.2,part (b).Also, by (4.1) and Theorem 3.3, we have u ( r ) r / → r → ∞ . In the proof, we often use the following property of u , from Theorem 3.2, part (a):(4.6) u ( r ) r / = B ( r, u ) r / is decreasing as r → ∞ . We also need further properties of u ( r ) and of the average I ( r ) = 12 π Z π u ( re iθ ) dθ, r > , namely that(4.7) u ( r ) and I ( r ) are positive increasing convex functions of log r, and(4.8) I ( r ) = A n log r + B n , for r ∈ [ b n , a n +1 ] , n ≥ , for some constants A n > B n , n ≥
0; see [4, Section 2.7]. Also,(4.9) u ( − r ) < I ( r ) < u ( r ) , for r > , by Theorem 3.2, part (a), and(4.10) u ( r ) ≤ I (2 r ) , for r ≥ , by using the Poisson integral of u to majorise u in { z : | z | ≤ r } ; see [4, Theo-rem 2.5] for example.Taken together, these properties will give us good control over the behaviour of u ( r ) and u ( − r ) in terms of the sequences ( a n ) and ( b n ). TERATING THE MINIMUM MODULUS 11
We write u ( r ) = r / − ε ( r ) , r > , where 0 ≤ ε ( r ) < /
2, for r >
1, by (4.6). In order to verify (4.4), we need toshow that ε ( r ) ≥ δ ( r ) for r sufficiently large.To do this, we obtain bounds on the coefficients A n and B n , n ≥
0, in (4.8).First, we obtain an upper bound for A n . By (4.6), (4.8) and (4.9), A n log( b n ) + B n b n = I ( b n ) b n < u ( b n ) b n ≤ u ( b n ) b / n , for n ≥ , so(4.11) A n log b n + ( A n log b n + B n ) < u ( b n ) b / n ≤ b n , for n ≥ , by (4.6) again. Since A n log b n + B n = I ( b n ) >
0, we deduce from (4.11) that(4.12) A n < b n log b n , for n ≥ . Next we claim that B n < n sufficiently large. We do this by using the factthat φ ( t ) = I ( e t ) is a positive increasing convex function with the property that φ ( t ) /t → ∞ as t → ∞ . This last property holds because φ is convex andlim sup t →∞ log φ ( t ) t = lim sup r →∞ log I ( r )log r = lim sup r →∞ log u ( r )log r = 1 / , in view of (4.10) and the fact that u has order 1 / φ ( t ) t = A n + B n t , for log b n ≤ t ≤ log a n +1 , and, by the convexity of φ , we have A n ≥ φ (log b n ) − φ (1)log b n − > φ (log b n )log b n = A n + B n log b n , for n sufficiently large. Hence there exists a positive integer N such that(4.13) B n < , for n ≥ N , as claimed.The estimate (4.12) implies that there exists N ≥ N such that A n < b n log b n ≤ r / log r / < r / r , for r ≥ b n , n ≥ N . Therefore, we deduce from (4.10), (4.13) and (4.8) that1 r ε ( r ) = u ( r ) r / ≤ I (2 r ) r / < A n log 2 rr / < r / , for b n ≤ r ≤ a n +1 , n ≥ N . Hence(4.14) ε ( r ) > ≥ δ ( r ) , for b n ≤ r ≤ a n +1 , n ≥ N . To complete the proof of (4.4), we show that ε ( r ) ≥ δ ( r ) for r ∈ [ a n , b n ] and n ≥ N + 1 by using the fact that u ( r ) r / ≤ u ( a n )( a n ) / , for r ≥ a n , n ≥ , by (4.6) once again, that is, ε ( r ) ≥ ε ( a n ) log a n log r , for r ≥ a n , n ≥ . Therefore, for r ∈ [ a n , b n ] and n ≥ N + 1, we have, by (4.14), (4.3) and (4.2), ε ( r ) ≥ ε ( a n ) log a n log r > log a n r (4.15) ≥ log a / n
20 log b n = log b n − b n − ≥ δ ( a n ) ≥ δ ( r ) , since δ is a decreasing function. Combining (4.14) and (4.15), we obtain (4.4).Finally, (4.5) follows immediately from the fact that, for r ∈ [ a n , b n ], we have u ( − r ) = 0 and for r ∈ [ b n , a n +1 ], n ≥ N , we have u ( − r ) < I ( r ) = A n log r + B n ≤ A n log a n +1 ≤ b n log b n log a n +1 = log b n +1 , by (4.9), (4.13), (4.12) and (4.3). (cid:3) We now give the proof of Example 1.3. Here we use again the fact, mentionedin Section 2, that property (1.2) holds if and only if(4.16) there exists
R > e m ( r ) > r, for r ≥ R, where e m ( r ) := max { m ( s ) : 0 ≤ s ≤ r } , for r ∈ [0 , ∞ ) . Proof of Example 1.3.
Without loss of generality we can assume that δ ( r ) :(0 , ∞ ) → (0 , /
4) is a decreasing function such that δ ( r ) → r → ∞ .Let u be the subharmonic function constructed in Lemma 4.1 and let D = C \ E = C \ S n ≥ [ − b n , − a n ]. We apply Theorem 3.4 to the function u to obtain an entirefunction f with zeros in the set E such that(4.17) log | f ( z ) | − u ( z ) = O (log | z | ) as z → ∞ , z ∈ D , where D = D \ { z : dist( z, ∂D ) ≤ } , and also such that(4.18) log | f ( z ) | ≤ u ( z ) + 4 log | z | , for | z | ≥ R, for some R = R ( u ) > u in the real axis, we can assume that arg f ( x ) = 0for x > f has order 1 / u has these properties. In particular we can represent f as an infinite product ofthe form(4.19) f ( z ) = c ∞ Y n =0 (cid:18) zt n (cid:19) , where c > t n ∈ E, n ≥ . TERATING THE MINIMUM MODULUS 13
We also deduce from (4.18) that there exist r > m ( r, f ) ≤ exp( u ( − r ) + 4 log r ) , for r ≥ r , and hence, since u vanishes on the set E ,(4.21) m ( r, f ) ≤ r , for r ≥ r , − r ∈ E. Now we consider the function f defined by(4.22) f ( z ) = c ∞ Y n =5 (cid:18) zt n (cid:19) , and prove that f satisfies (1.5). For some constant K >
0, we have, by (4.18)and (4.4), M ( r, f ) ≤ M ( r, f ) Kr ≤ exp( u ( r ) + 4 log r ) Kr ≤ exp( u ( r )) ≤ exp (cid:0) r / − δ ( r ) (cid:1) , provided that r is sufficiently large.Finally, we show that property (2.2) does not hold for this function f . For r ∈ [ a n +1 , b n +1 ], we have m ( r, f ) ≤ m ( r, f ) Kr < ≤ b n +1 , provided that n is sufficiently large, by (4.22) and (4.21). Next, for r ∈ [ b n , a n +1 ],we have, by (4.18) and (4.5), m ( r, f ) ≤ m ( r, f ) Kr ≤ exp ( u ( − r ) + 4 log r ) Kr ≤ exp( u ( − r )) ≤ b n +1 , provided that n is sufficiently large.Therefore, for n sufficiently large, e m ( b n +1 , f ) = max { m ( r ) : 0 ≤ r ≤ b n +1 } ≤ b n +1 , so property (4.16) does not hold. Hence, for the function f , there is no valueof r such that m n ( r ) → ∞ as n → ∞ .This completes the proof of Example 1.3. (cid:3) Proof of Example 1.4.
Here we follow a similar procedure to the previous proof,starting this time with a subharmonic function u ∈ K , where E = S n ≥ [ − b n , − a n ],such that u (1) = 1, b n /a n ր ∞ as n → ∞ and a n +1 = 2 b n for n ≥ u has order 1/2 minimal type and lowerorder 1/2, so the entire functions f and f resulting from applying Theorem 3.4(as in the previous proof) also have these properties, by (4.17) and (4.18). Finally, to show that property (4.16) fails, we need to choose the sequences ( a n )and ( b n ) iteratively so that(4.23) u ( − r ) < log b n +1 , for b n ≤ r ≤ a n +1 , n ≥ , also holds, in order that m ( r, f ) ≤ exp( u ( − r )) ≤ b n +1 , for b n ≤ r ≤ a n +1 , n ≥ . Arranging for (4.23) to hold is clearly possible sincemax { u ( − r ) : b n ≤ r ≤ a n +1 } ≤ max { u ( r ) : b n ≤ r ≤ a n +1 } ≤ a / n +1 , by Theorem 3.2, part (a). (cid:3) Acknowledgement
Phil Rippon expresses his deep gratitude to Walter Hayman,his PhD adviser, for Walter’s inspiring research and teaching over many years.In particular, several aspects of the work in this paper can be traced to an earlysupervision at Imperial College when Walter passed on a copy of Bo Kjellberg’sthesis [6], which contains the origins of the ideas that we use for constructingour examples and also the key Lemma 2.1, due to Beurling, that we use to proveour positive theorems.
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TERATING THE MINIMUM MODULUS 15
School of Mathematical Sciences, The University of Nottingham, UniversityPark, Nottingham NG7 2RD, UK
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