Annales Academiæ Scientiarum Fennicæ Series A. I. Mathematica Volumen 20, 1995, 149–153

WANDERING DOMAINS IN THE ITERATION OF COMPOSITIONS OF ENTIRE FUNCTIONS

I.N. Baker and Anand P. Singh

Imperial College, Department of Mathematics Huxley Building, 180 Queen’s Gate, London SW7 2BZ, England; i.baker@ic.ac.uk University of Jammu, Department of Mathematics Jammu-180004, India

Abstract. If p is entire, g(z) = a + b exp(2πi/c), where a, b, c are non-zero constants and the normal set of g(p) has no wandering components, then the same is true for the normal set of p(g).

Let f be a rational function of degree at least 2 or a nonlinear entire function. Let f^{n}, for n ∈ N denote the nth iterate of f. Denote the set of normality by N(f) and the Julia set by J(f). Thus

N(f) = {z : (f^{n}) is normal in some neighbourhood of z}, J(f) = C − N(f).

By definition N(f) is open (and possibly empty) and it is well known (see for example [8], [9]) that J(f) is nonempty and perfect and J(f) is completely invariant under f, that is, z ∈ J(f) implies f(z) ∈ J(f) and z_{0} ∈ J(f) for any z_{0} such that f(z_{0}) = z. Consequently N(f) is completely invariant.

If U is a component of N(f) then f(U) lies in some component V of N(f). In fact V \f(U) is either ∅ or a single point, by an unpublished result of M. Herring. By a slight abuse of language we write V = f(U) even when V \f(U) is a singleton. If all f^{n}(U) with n ∈ N are different components of N(f) then U is called a wandering domain.

D. Sullivan [13] proved that the set of normality of a rational function has no wandering domain, thus solving a problem open since the papers of Fatou and Julia. On the other hand this is not so for transcendental entire functions. In [1] the first author constructed an entire function f such that N(f) has wandering domains. Since then several entire functions which have wandering domains with various different properties have been constructed, see for instance [3], [7]. Also at the same time there has been a move to classify those entire functions which do not have wandering domains [2], [6], [11]. In particular this is the case for functions which have only a finite number of asymptotic or critical values. Such functions are denoted as having finite type. In this paper we shall identify a class of composite entire functions which have no wandering domains and a class of composite entire functions which have wandering domains. We shall prove

1991 Mathematics Subject Classification: Primary 30D05.

Theorem 1. Let p(z) be a nonconstant entire function and let g(z) = a + be^{2πiz/c} where a, b, c are nonzero constants. If h = g(p) has no wandering domains then neither does p(g).

In particular for a polynomial p(z) it is known [2] e^{p(z)} has no wandering domain and consequently it follows immediately that p(e^{z}) has no wandering domain (also proved in [2]). As another application of the above theorem we shall show that e^{e^{z}} − e^{z} has no wandering domain. Also e^{e^{z}} − e^{z} is not of finite type, and so provides an example of an entire function which is of not finite type and having no wandering domain.

Proof of Theorem 1. Suppose f = p(g) has a wandering domain say U_{1} . Then U_{n} = f^{n−1}(U_{1}) are distinct for all n = 1, 2, …. Since g(z + c) = g(z) we have N(f) = N(f) + c. Now suppose U_{j} ∩ (U_{k} + cl) ≠ ∅ for some j, k, l then U_{j+1} ∩ U_{k+1} ≠ ∅ and so j = k. Thus for j ≠ k, U_{j} ∩ (U_{k} + cl) = ∅ and consequently g(U_{j}) ∩ g(U_{k}) = ∅.

For each k, let V_{k} = g(U_{k}). Then for j ≠ k, V_{j} ∩ V_{k} = ∅ and h(V_{k}) = h(g(U_{k})) = g(f(U_{k})) ⊆ V_{k+1} . Thus h^{n}(V_{k}) ⊆ V_{k+n} and so does not meet V_{k}, n > 1. Thus (h^{n}) is normal in each V_{k} and so V_{k} belongs to a component of N(h).

We finally show that V_{k} is a component of N(h). We first show that if β ∈ ∂V_{k} has the form β = g(α), α ∈ ∂U_{k} then β ∈ J(h). Indeed since U_{k} is a component of f, ∂U_{k} ⊆ J(f), and so α is a limit point of repelling periodic points z_{n} (≠ α) of f say f^{ν_{n}}(z_{n}) = z_{n}. Since g(f^{n}) = h^{n}(g) for all n, one obtains h^{ν_{n}}(g(z_{n})) = g(z_{n}). Thus g(z_{n}) are periodic points of h (of arbitrarily large order). Also g(z_{n}) → g(α) = β (but g(z_{n}) ≠ g(α) for large n). Thus β ∈ J(h).

To complete the proof we assume that there exists β ∈ ∂V_{k} with β ∉ J(h). Then β is (by the above) not a limit of points of J(h), hence not a limit of points in g(∂U_{k}). Thus there is a disc D = D(β, r), r > 0 which contains no points of g(∂U_{k} \ {∞}). Since β ∈ ∂g(U_{k}) = ∂V_{k}, there exists w′ ∈ D(β, r) with w′ = g(z′), z′ ∈ U_{k} and without any loss of generality we assume w′ ≠ a. We can continue g^{−1}(w) = c/(2πi) log((w − a)/b) from w′ to β along a path γ in D which avoids a and the values of g^{−1} lie in U_{k} since they can never meet ∂U_{k}. Since β ∉ g(U_{k}) the only possibility is that g^{−1}(γ) → ∞ in U_{k} and hence β is an asymptotic value of g on this path, i.e. β = a.

Summing up, ∂V_{k} ⊆ J(h) except perhaps for a single isolated point. If there is such an isolated point we add it to V_{k} and then have, since the V_{k} are distinct, that the V_{k} are wandering components of h with h(V_{k}) = V_{k+1} . This contradicts the hypothesis and the proof is complete.

As an application of Theorem 1 we shall show that e^{e^{z}} − e^{z} has no wandering domain. For its proof we shall need the following lemma.

Lemma 1. Let f and g be entire functions having a finite number of asymptotic values. Then so does f(g).

Wandering domains in the iteration of compositions of entire functions 151

Proof. Let c be an asymptotic value for f(g). Thus there exists a curve Γ → ∞ on which f(g) → c. Associated with this curve Γ we have either g(Γ) → ∞ or g(Γ) ↛ ∞. If g(Γ) → ∞ then c is an asymptotic value for f. Since f has only finitely many asymptotic values, such c’s must be finite in number.

We next consider the case when g(Γ) ↛ ∞. Since Γ is a curve tending to ∞, we can find a sequence z_{n} → ∞ on this curve for which lim_{n→∞} g(z_{n}) = w_{0} for some finite w_{0} . Thus f(w_{0}) = lim_{n→∞} f(g(z_{n})) = c. Consider ϱ > 0 fixed, but arbitrarily small. Then |f(w)−c| > ε > 0 for w ∈ (|w−w_{0}| = ϱ). Next, as c is an asymptotic value for f(g), |f(g(z))−c| < ε for all |z| > A on Γ where A is some constant. In particular if |z_{n}| are sufficiently large on Γ then |f(g(z)) − c| < ε for all z beyond z_{n} on Γ and |g(z_{n}) − w_{0}| < ϱ. Thus |g(z) − w_{0}| < ϱ for all sufficiently large z on Γ. Thus w_{0} is an asymptotic value for g(z) on Γ, where f(w_{0}) = c. But the number of asymptotic values of g is finite. This completes the proof.

Lemma 2. If f and g are entire functions of finite order then f(g) has at most a finite number of asymptotic values.

The proof is immediate from Lemma 1 and the fact that an entire function of order k has at most 2k different asymptotic values [12, p. 307].

Lemma 3 [6]. Let ℐ denote the collection of transcendental entire functions of finite type. Functions in ℐ have no wandering domains.

Theorem 2. The function e^{e^{z}} − e^{z} has no wandering domain.

Proof. Set g(z) = e^{z} and p(z) = e^{z} − z, then by Lemma 2 f(z) = g(p(z)) has a finite number of asymptotic values. Also clearly f(z) has e as the only critical value. Thus e^{e^{z}−z} is of finite type and so by Lemma 3, e^{e^{z}−z} has no wandering domain. We now apply Theorem 1 to conclude p(g(z)) = e^{e^{z}} − e^{z} has no wandering domain.

We next prove the following theorem.

Theorem 3. Let g be a transcendental entire function having at least one fixed point. Then there exists an entire function f such that g(f) has a wandering domain.

The proof of this theorem is based on the proof of theorems in [4], [5] and so on the method of construction of wandering domain first introduced by A. Eremenko and M. Lyubich [7]. We first recall the following facts: If F denotes a closed subset of C and C_{a}(F) the functions which are continuous on F and analytic in F^{0} then F is called a Carleman set (for C) if, for any g in C_{a}(F) and for any positive continuous function ε on F, there is an entire function f such that |g(z) − f(z)| < ε(z), z ∈ F. By Arakelyan’s theorem (e.g. [10, p. 137]) we have (i) Ĉ \ F must be connected and also locally connected at ∞. If in addition to (i) we have (ii) for each compact K the union W (K) of those components of F^{0} which meet K is relatively compact in C, then F is indeed a Carleman set [10, p. 157].

Proof of Theorem 3. Without any loss of generality let the fixed point z_{0} of g satisfy Re(z_{0}) < −2 and Im(z_{0}) = 0. Let

\begin{aligned}B & = \{z : |z − z_{0}| ≤ 1\}, \\ L_{m} & = \{z : Re(z) = 4m\},  m ≥ 10, \\ A_{m} & = \{z : |z − (4m + 2)| ≤ 1\},  m ≥ 10. \end{aligned}

Then clearly F = B ⋃{⋃_{m=10}^{∞}{A_{m} ∪ B_{m}}} is a Carleman set. As g is continuous at the fixed point z_{0} , we can choose δ > 0 so small that |g(z) − z_{0}| < \frac{1}{2} whenever |z − z_{0}| < δ. Now consider a branch value g^{−1}(4m + 2) where m ≥ m_{0} . Then again by the continuity of g, there exist δ_{m} > 0 such that |g(z) − (4m + 2)| < \frac{1}{2} for all |z − g^{−1}(4m + 2)| < δ_{m}. By the above remark it follows that there exists an entire function f such that

\begin{aligned}|f(z) − z_{0}| & < δ,  z ∈ B, \\ |f(z) − z_{0}| & < δ,  z ∈ L_{m}, m ≥ 10, \\ |f(z) − g^{−1}(4m + 6)| & < δ_{m+1},  z ∈ A_{m}. \end{aligned}

And clearly g(f) = h is an entire function with h(A_{m}) ⊆ A_{m+1} . Also h^{n}(z) → ∞ in each A_{m} and so A_{m} ∈ N(h). On the other hand h maps B into a smaller disc |z − z_{0}| < \frac{1}{2} and so h contains an attractive fixed point ξ such that h^{n} → ξ in B. Finally h maps L_{m} (m ≥ 10) into B similarly. Further L_{m} belongs to a component of N(h) different from a component of G_{m} to which A_{m} belongs. Thus each G_{m} is a wandering domain mapping to G_{m+1} under z → h(z).

Acknowledgement. The work of the second author was supported by a grant of the Association of Commonwealth universities.

References

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Wandering domains in the iteration of compositions of entire functions 153

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Received 3 November 1993