Iterating the minimum modulus: functions of order half, minimal type

For a transcendental entire function $f$, the property that there exists $r>0$ such that $m^n(r)\to\infty$ as $n\to\infty$, where $m(r)=\min \{|f(z)|:|z|=r\}$, is related to conjectures of Eremenko and of Baker, for both of which order $1/2$ minimal type is a significant rate of growth. We show that this property holds for functions of order $1/2$ minimal type if the maximum modulus of $f$ has sufficiently regular growth and we give examples to show the sharpness of our results by using a recent generalisation of Kjellberg's method of constructing entire functions of small growth, which allows rather precise control of $m(r)$.


Introduction
Let f be a transcendental entire function and denote by f n , n ∈ N, the n-th iterate of f . The results in this paper address a question in complex dynamics concerning the escaping set of a transcendental entire function f , defined as namely, whether all the components of I(f ) are unbounded. This question is known as Eremenko's conjecture [3] and despite much work it remains open. 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 > 0 and the function M(r) is strictly increasing and unbounded. On the other hand, the function m(r) is alternately increasing and decreasing between adjacent values of r for which m(r) = 0 and eventually decreases to 0 in the case that f has only finitely many zeros. For certain functions, however, m(r) is comparable 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 > 0 such that m n (r) → ∞ as n → ∞, may or may not hold, depending on the function f .
2010 Mathematics Subject Classification. Primary 37F10, Secondary 30D05. The last two authors were supported by the EPSRC grant EP/R010560/1. 1 It has been known for some time that the sequence M n (r) is of importance in relation to work on Eremenko's conjecture, since it plays a key role in the definition of a subset of I(f ) called the fast escaping set, all of whose components are unbounded; see, for example, [12]. More recently, it has been observed that property (1.2), when it is true, can also play an important role in relation to this conjecture; see [9], [10] and [8]. For example, in [8] we obtained the following result, which gives a family of transcendental entire functions for which Eremenko's conjecture holds in a particularly strong way. Theorem 1.1. Let f be a real transcendental entire function of finite order with only real zeros for which there exists r > 0 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 , the set 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 entire function f are 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 functions satisfy property (1.2); in particular, which amongst those that satisfy the hypotheses of this theorem. In [8] we proved several results in relation to this question, including the following.
(a) For a real transcendental entire function f of finite order with only real zeros: In this paper, we show that (1.2) does indeed hold for many families of such functions, but does not hold for all. Our main positive result is the following which, roughly speaking, shows that (1.2) holds whenever (log M(r))/r 1/2 tends to 0 (as r → ∞) in a sufficiently regular manner.
In Section 2, we give a number of conditions on the maximum modulus and on the zeros of f which imply the hypotheses of Theorem 1.2 and are convenient for applications.
In the second half of the paper we use a recent generalisation of a method of Kjellberg [13] to construct examples of functions of order 1/2, minimal type, for which 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 cannot deduce that property (1.2) holds. Then there exists a real transcendental entire function f of order 1/2, minimal type, with only real zeros, and r 0 > 0 such that (1.5) log M(r, f ) ≤ r 1/2−δ(r) , for r ≥ r 0 , for which property (1.2) does not hold.
Next we show that making the additional assumption of positive lower order, or even of lower order 1/2, is insufficient to ensure that property (1.2) holds.
There exists a real transcendental entire function f of order 1/2, minimal type, and of lower order 1/2, with only real zeros such that property (1.2) does not hold.
Finally, we remark that consideration of the iterates of the minimum modulus first arose in connection with the conjecture of Baker that entire functions of order 1/2, minimal type, have no unbounded Fatou components; see [7] for the most recent progress on this conjecture.
The structure of the paper is as follows. In Section 2 we prove our positive results, 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.

Proof of Theorem 1.2 and some special cases
In this section we prove Theorem 1.
In order to work with property (1.2), the function

was introduced in [10] and it was shown that property (1.2) is true if and only if
(2.2) there exists R > 0 such that 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 property (1.2) does not hold, then by (2.2) there exist arbitrarily large r > r 0 such that m(r) ≤ r; that is, for arbitrarily large r we have For such an r, there exists by hypothesis s < r such that log M(s) ≥ 2 log r and (1.4) holds. However, we deduce from (2.3) by applying Lemma 2.1 to which contradicts condition (1.4).
For applications of Theorem 1.2, it is often useful to express condition (1.4) in terms of the functions ε(r), r > 0, and k(r), r > 0, defined as follows: log M(r) = r 1/2−ε(r) and k(r) = ε(r) log r; Note that if f has order at most 1/2, minimal type, then for all sufficiently large values of r we have 0 < ε(r) < 1/2 and hence 0 < k(r) = ε(r) log r < 1 2 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) → 0 as r → ∞, We now give two special cases of Theorem 1.2 which are convenient for applications. Theorem 2.2, part (a), shows that property (1.2) holds whenever the function f has positive lower order and ε(r) = k(r)/ log r does not tend to 0 too quickly. Example 1.3 shows that positive lower order alone is not sufficient here.
Therefore, the condition M(s) ≥ r 2 is satisfied, for sufficiently large r, by taking s = (log r) For any transcendental entire function f we define n(r) = n(r, f ) to be the number 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 in addition 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 where n(r, f ) ∼ r 1/2−ε(r) as r → ∞. Then Proof of Theorem 2.3. Since n(r) ∼ r 1/2−ε(r) as r → ∞, where ε(r) is positive and decreasing to 0, and k(r) = ε(r) log r is increasing, we have and Since ε(r) is decreasing, we have Part (a) follows by Lemma 2.4 and the above inequalities for N(r) and Q(r).

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 functions of order less than 1/2 by approximating certain continuous subharmonic functions by functions of the form log |f | where f is a transcendental entire function. In this section we summarise the results from [13] which are needed to construct our examples.
Kjellberg's method is a two stage process: 1. a continuous subharmonic function u with the required properties is obtained by using a positive harmonic function defined in the complement of a particular sequence of radial slits, on which u vanishes; 2. the Riesz measure of u is discretised to produce an entire function f such that 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 slits to be chosen more flexibly than in [6].
We now recall some of the key definitions and terminology needed to state the results from [13]. Remark For each closed subset of the negative real axis, there is exactly one corresponding function u ∈ K up to positive scalar multiples, by a result of Benedicks [1,Theorem 4].
Recall that for a set S ⊂ R + and r > 1, we define the upper logarithmic density Theorem 3.2. Let u ∈ K, with E the corresponding closed subset of the negative real axis. Then u has the following properties.
where 0 ≤ a 0 < b 0 < a 1 < b 1 < · · · , and a n → ∞ as n → ∞. Put Then there exists an entire function f with only negative zeros, all lying in the set E, such that Moreover, if we also have then there exists R = R(u) > 0 such that Theorem 3.4 will enable us to approximate subharmonic functions u ∈ K by functions of the form log |f |, where f is a transcendental entire function of the same order, lower order and type class as u.

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 construct entire functions that approximate suitable subharmonic functions. In each case we show that property (1.2) is false by arranging that the minimum modulus of 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 Example 1.3. The proof is rather long as the set E has to be chosen carefully so that the function u ∈ K does not grow too quickly on the positive real axis and takes large values on the negative real axis only relatively rarely. and let a n and b n , n ≥ 0, be chosen to satisfy (4.1) a n+1 ≥ b 10 n , for n ≥ 0, Then the unique subharmonic function u ∈ K corresponding to the set E = n≥0 [−b n , −a n ], with u(1) = 1, satisfies ρ(u) = 1/2 and u(r)/r 1/2 → 0 as r → ∞, and also (a) there exists r 0 > 0 such that (4.4) u(r) ≤ r 1/2−δ(r) , for r ≥ r 0 ; (b) we have u(−r) = 0, for r ∈ [a n , b n ], n ≥ 0, and, for n sufficiently large, Proof. Let u ∈ K be the unique subharmonic function with u(1) = 1 corresponding to the set E = n≥0 [−b n , −a n ], where a n and b n satisfy (4.1), (4.2) and (4.3).
Also, by (4.1) and Theorem 3.3, we have u(r) r 1/2 → 0 as r → ∞. In the proof, we often use the following property of u, from Theorem 3.2, part (a): 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 ).
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), for n ≥ 0, 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 ≥ 0.
Next we claim that B n < 0 for n sufficiently large. We do this by using the fact that 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 and in view of (4.10) and the fact that u has order 1/2.
and, by the convexity of φ, we have for n sufficiently large. Hence there exists a positive integer N 1 such that (4.13) B n < 0, for n ≥ N 1 , as claimed.
Let u be the subharmonic function constructed in Lemma 4.1 and let D = C\E = C \ n≥0 [−b n , −a n ]. We apply Theorem 3.4 to the function u to obtain an entire function f 1 with zeros in the set E such that  In view of the symmetry of u in the real axis, we can assume that arg f 1 (x) = 0 for x > 0.
We deduce from (4.17) and (4.18) that f 1 has order 1/2 minimal type, since u has these properties. In particular we can represent f 1 as an infinite product of the form (4.19) f 1 (z) = c ∞ n=0 1 + z t n , where c > 0 and t n ∈ E, n ≥ 0.