Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 23, 1998, 225–250

DYNAMICS OF TRANSCENDENTAL MEROMORPHIC FUNCTIONS

P. Domínguez

Imperial College of Science, Technology and Medicine, Department of Mathematics London SW7 2BZ, United Kingdom; p.dominguez@fcfm.bvap.mx

Abstract. The paper examines some properties of the dynamics of entire functions which extend to general meromorphic functions and also some properties which do not. For a transcendental meromorphic function f(z) whose Fatou set F (f) has a component of connectivity at least three, it is shown that singleton components are dense in the Julia set J(f). Some problems remain open if all components are simply or doubly connected.

Let I(f) denote the set of points whose forward orbits tend to ∞ but never land at ∞. For a transcendental meromorphic function f(z) we have J(f) = ∂I(f), I(f) ∩ J(f) ≠ ∅. However in contrast to the entire case, the components of \overline{I(f)} need not be unbounded, even if f(z) has only one pole.

If f(z) has finitely many poles then, as in the entire case, F (f) has at most one completely invariant component.

1. Introduction

Let f(z): C → Ĉ denote a meromorphic function, f^{n}, n ∈ N, the nth iterate of f(z), and f^{−n} the set of inverse functions of f^{n}.

Let A be the set of poles of f(z), B = ⋃_{n=1}^{∞} f^{−n}(A) the set of preimages of ∞, and B′ the derived set of B.

The Fatou set F (f) is defined to be the set of those points z ∈ Ĉ such that the sequence (f^{n})_{n∈N} is well defined, is meromorphic, and forms a normal family in some neighbourhood of z. The complement J(f) of F (f) is called the Julia set of f(z).

We define E(f) to be the set of exceptional values of f(z), that is, the points whose inverse orbit O^{−}(z) = {w : f^{n}(w) = z for some n ∈ N} is finite. There can be at most two such points and if ∞ ∈ E(f) then either f(z) is entire or has a single pole α which is a Picard value of f(z), so that

(1)  f(z) = α + (z − α)^{−k}e^{g(z)},

for some k ∈ N and some entire function g(z). The meromorphic functions thus fall into four disjoint classes.

1991 Mathematics Subject Classification: Primary 30D05, 58F08.

(i) The rational functions (where we assume the degree to be at least two). (ii) The class E of transcendental entire functions. (iii) The class P of transcendental functions of the type (1), self maps of the

punctured plane which have a meromorphic extension to C. (iv) M, the class called general meromorphic functions in [11], of transcendental

meromorphic functions, for which ∞ ∉ E(f).

The iteration of the class (i) and (ii) is sufficiently well known. The class P has been studied in [7], [15], [17], [23]–[25], [32]–[38], [41] and M in [9], [10], [11]. See [14] for a survey of the main results. For the first three classes the iterates are automatically defined in the plane or punctured plane and the study of the normality of {f^{n}} is the main feature. For f(z) ∈ M the set B, defined above, is infinite and it is shown in [9] that J(f) =\overline{B} = B′.

Many properties of J(f) and F (f) are much the same for M as for the other classes but different proofs are needed and some discrepancies arise. We recall (see e.g. [14]) that for f(z) in any of the four classes:

1. F (f) is open and completely invariant under f(z), that is, z ∈ F (f) if and only if f(z) ∈ F (f).

2. For z_{0} ∉ E(f), J(f) ⊂ \overline{O^{−}(z_{0})}.

3. Repelling periodic points are dense in J(f).

4. If U is a periodic component of F (f) of period p, that is if f^{p}(U) ⊂ U, there

are the following possible cases:

– U contains an attracting periodic point z_{0} of period p. Then f^{np}(z) → z_{0} for z ∈ U as n → ∞, and U is called the immediate attractive basin of z_{0} .

– ∂U contains a periodic point z_{0} of period p and f^{np}(z) → z_{0} for z ∈ U as n → ∞. Then (f^{p})′(z_{0}) = 1 if z_{0} ∈ C. (For z_{0} = ∞ we have (g^{p})′(0) = 1 where g(z) = 1/f(1/z).) In this case, U is called a Leau domain.

– There exists an analytic homeomorphism ψ: U → D where D is the unit disc such that ψ(f^{p}(ψ^{−1}(z))) = e^{2πiα}z for some α ∈ R\Q. In this case, U is called a Siegel disc.

– There exists an analytic homeomorphism ψ: U → A where A is an annulus A = {z : 1 < |z| < r}, r > 1, such that ψ(f^{p}(ψ^{−1}(z))) = e^{2πiα}z for some α ∈ R \ Q. In this case, U is called a Herman ring.

– There exists z_{0} ∈ ∂U such that f^{np}(z) → z_{0} , for z ∈ U as n → ∞, but

f^{p}(z_{0}) is not defined. In this case, U is called a Baker domain.

5. If p = 1 in 4, that is if U is an invariant component of F (f), then U has connectivity 1, 2, or ∞. If the connectivity is two, then U is a Herman ring. The purpose of this paper is to examine whether some properties of the sets

J(f) or F (f), which hold for transcendental entire functions, extend to transcendental meromorphic functions or perhaps at least to transcendental meromorphic functions with finitely many poles.

It is known that the Fatou set F (g) of a transcendental entire function g(z) may have a multiply-connected component U [1], that any such component is bounded [3], and that g^{n} → ∞ in U so that each g^{n}(U) belongs to a different component of F (g). It is a corollary that J(g) is not totally disconnected but a recent result [20] shows that if U is multiply connected, then singleton components are everywhere dense in J(g).

The existence of singleton components of J(f) for transcendental meromorphic functions follows from the following general result.

Theorem A. Suppose that f(z) is a transcendental meromorphic function and that F (f) has a component H of connectivity at least three. Then singleton components are dense in J(f).

We note that in [10] there are examples of functions in M with Fatou components of any prescribed connectivity.

If R is a doubly-connected component of F (f), then its complement consists of two components, one bounded and the other unbounded. The methods used to prove Theorem A also give Theorem B.

Theorem B. Suppose that f(z) is a transcendental meromorphic function and that F (f) has three doubly-connected components U_{i} such that either,

(a) each component lies in the unbounded component of the complement of the other two or

(b) two of the components U_{1} , U_{2} lie in the bounded component of U_{3}^{c} but U_{1} lies in the unbounded component of U_{2}^{c} and U_{2} lies in the unbounded component of U_{1}^{c}.

Then singleton components are dense in J(f).

The preceding results show that if for the transcendental meromorphic function f(z) the Julia set has no singleton components, then any multiply-connected component of F (f) has connectivity two and moreover, the distribution of these doubly-connected components is subject to severe restrictions. On the other hand we have not been able to show that such components cannot occur. For functions f(z) of class P it is known [7] that F (f) can have at most one multiply-connected component and the connectivity of such component is 2. Thus J(f) has no singleton components.

W. Bergweiler in [14, p. 164] mentions the problem of whether a meromorphic function f(z) in the special class P can have a Herman ring. This would necessarily be invariant. The question is answered by the following theorem.

Theorem C. If f(z) ∈ P, then f(z) has no Herman ring.

If we drop the requirement that f(z) is transcendental the rational function R(z) = z^{2}+λ/z^{3} , λ > 0, has infinitely many doubly-connected Fatou components, while J(R) is a Cantor set of circles and in particular has no singleton components, see [13, p. 266] for the details.

Theorem D. Let f(z) be a transcendental meromorphic function and suppose that F (f) has multiply-connected components A_{i}, i ∈ N, all different, such that each A_{i} separates 0, and ∞ and f(A_{i}) ⊂ A_{i+1} for i ∈ N. Then J(f) has a dense set of singleton components.

There are examples of f(z) ∈ E which satisfy the assumptions of Theorem D and in this case A_{i} → ∞ [4], [5]. It is not easy to determine the connectivity of A_{i} although examples are known [6] where this is infinite. It seems to be an open problem as to where the connectivity of A_{i} can be 2 for f(z) ∈ E. Similar constructions would give examples of functions in M with a finite number of poles and A_{i} which tend to ∞. More interesting examples can be given by other methods as shown by Theorem E.

Theorem E. There is a transcendental meromorphic function such that F (f) has a sequence of multiply-connected components A_{i}, i ∈ N, all different, such that each A_{i} separates 0 and ∞ and f(A_{i}) ⊂ A_{i+1} , i ∈ N. Moreover A_{2i} → ∞ as i → ∞ and A_{2i+1} → 0 as i → ∞.

It will follow from the later Theorem G that any function which satisfies the assumption of Theorem E has infinitely many poles.

Functions with finitely many poles. We shall examine whether a number of results about the dynamics of transcendental entire functions extend to meromorphic functions with finitely many poles. A useful tool here is a theorem of H. Bohr (Theorem 5.1) which was used already in dynamics by Pólya [40] to estimate the growth of composed functions such as iterates. We make a slight extension of the result in Theorem 5.2. From it we shall deduce the following

Theorem F. If f(z) is a transcendental meromorphic function with finitely many poles, then there is some S > 0 such that no invariant component G of F (f) can contain a curve γ which lies in {z : |z| > S} and satisfies n(γ, 0) ≠ 0.

Corollary. If f is a transcendental meromorphic function with finitely many poles, then J(f) cannot be totally disconnected.

The result is known for f(z) ∈ E [3] but is untrue for meromorphic functions without restriction to finitely many poles. For example, the function f(z) = λ tan z, 0 < λ < 1, has a Julia set which is a Cantor subset of the real line, see [10] and [19]. For f(z) ∈ E any unbounded invariant component is simply connected. This is no longer true for functions with even one pole. We show in Section 9 that for ε = 10^{−2} , f(z) = z + 2 + e^{z} + ε(z − (1 + iπ))^{−1} has a multiply-connected invariant component in which f^{n}(z) → ∞.

Orbits which tend to ∞. For f(z) ∈ E Eremenko [21] studied the set

I(f) = {z ∈ C : f^{n}(z) → ∞ as n → ∞}.

He proved that for f(z) ∈ E we have (a) I(f) ≠ ∅, (b) J(f) = ∂I(f),

(c) I(f) ∩ J(f) ≠ ∅, and (d) the closure \overline{I(f)} of I(f) has no bounded components. It remains open as to whether I(f) itself may have bounded components. Fatou [27] had already noted that in the examples studied by him, I(f) contains unbounded analytic curves and asked whether this was a general phenomenon.

For meromorphic functions f(z) we generalise Eremenko’s definition by writing

I(f) = {z ∈ C : f^{n}(z) → ∞ as n → ∞ and f^{n}(z) ≠ ∞}.

We show in Section 6 that for transcendental meromorphic functions I(f) has the properties (a), (b), and (c) listed above but (d) does not hold in general, even for functions with finitely many poles. These results form the following five theorems.

Theorem G. For a transcendental meromorphic function f(z) with only finitely many poles the set I(f) is not empty. Indeed for any curve γ with n(γ, 0) ≠ 0 and such that d(0, γ) is sufficiently large we have I(f) ∩ γ ≠ ∅.

Theorem H. For a transcendental meromorphic function f(z) with infinitely many poles the set I(f) is not-empty. Indeed in any neighbourhood of a pole there are points of I(f).

Theorem I. If f(z) is a transcendental meromorphic function, then J(f) = ∂I(f).

Theorem J. If f(z) is a transcendental meromorphic function, then I(f) ∩ J(f) ≠ ∅.

Theorems G and J follow from the same argument as Theorem F, based on our extension of Bohr’s theorem. This gives an alternative to Eremenko’s proof in the special case of entire functions. His proof was based on Wiman–Valiron theory.

The assertion of Theorem G is not in general true for meromorphic functions with infinitely many poles. For example, if f(z) = λ tan z, 0 < λ < 1, then \overline{I(f)} ⊂ J(f), and J(f) is a real Cantor set which fails to meet γ = C(0, r) for many arbitrarily large values of r. However, we shall have I(f) ≠ ∅. Further (d) fails for this choice of f(z), since every component of \overline{I(f)} is bounded. The example f(z) = λ sin z − ε/(z − π), ε > 0, 0 < λ < 1, considered in Section 9 shows that if g(z) has even one pole, then \overline{I(g)} may have bounded components.

Completely invariant components. For transcendental functions it has been known since the early work of Fatou and Julia that F (f) can have at most two completely invariant components. In [8] it was shown that if f(z) ∈ E, then there is at most one completely invariant component of F (f). In Section 7 we show that this last result extends to transcendental meromorphic functions with finitely many poles.

Theorem K. Let f(z) be a meromorphic function with at most finitely many poles. Then there is at most one completely invariant Fatou domain.

The theorem fails for functions with infinitely many poles. For example if f(z) = λ tan z, λ > 1, then F (f) has two completely invariant components namely the upper and lower half-plane (each of which contains an attracting fixed point). It is an open problem for general transcendental meromorphic functions as to whether the Fatou set can have more than two completely invariant components. However for a transcendental function whose inverse is singular at only finitely many points the maximum number of completely invariant Fatou components is two [11].

Functions with one pole. The final section contains two examples. Recall that for f(z) ∈ E any multiply-connected component G of F (f) is bounded with f^{n} → ∞ in G. This is no longer true for meromorphic functions with even one pole.

We show in Example 1 that for ε = 10^{−2} and a = 1 + iπ the function f(z) = z + 2 + e^{z} + ε(z − a)^{−1} has a multiply-connected invariant component in which f^{n}(z) → ∞.

In a similar way we show that for 0 < λ < 1 and for sufficiently small positive ε the function f(z) = λ sin z − ε/(z − π) has a single completely invariant Fatou component. This is unbounded, multiply connected and the iterates f^{n} converge to a finite limit (an attracting fixed point) in F (f). Singleton components of J(f) are dense in J(f), which also contains unbounded continua.

We precede the examples by stating some results about the relation between the orbits of singular points of f^{−1} and the possible limit functions of iterates. These are used in the discussion of Example 2.

The author wishes to thank Professor I.N. Baker for many helpful suggestions, CONACyT (Consejo Nacional de Ciencia y Tecnología) and Universidad Autonoma de Puebla for their financial support.

2. Proof of Theorem A

In this section our aim is to prove Theorem A. We shall start this section by giving the following lemma.

Lemma 1. If D is a domain in Ĉ whose complement contains at least p components H_{1}, …, H_{p}, then there are p disjoint simple polygons γ_{i}, 0 ≤ i ≤ p, in D such that γ_{i} separates H_{i} from the remaining H_{j}, j ≠ i. Thus there is a component D_{i} of the complement of γ_{i} such that H_{i} ⊂ D_{i} and \overline{D}_{i} ∩\overline{D}_{j} = ∅ for j ≠ i.

A proof may be obtained from results in [39]. Firstly there is a partition of the complement of D into p disjoint closed sets F_{i}, such that H_{i} ⊂ F_{i}, 1 ≤ i ≤ p. The method of Theorem 3.1 and Theorem 3.2 in Chapter VI of [39] gives a construction of a simple polygon γ_{i}, uniformly as close as we wish to a subset of F_{i} which satisfies the requirements of the lemma.

In order to prove Theorem A we require a result from Ahlfors’ theory of covering surfaces. If a function g(z) is meromorphic in C and D is a simplyconnected domain in Ĉ, whose boundary is a sectionally analytic Jordan curve γ, then an island (with respect to g(z)) over D is a bounded component G of g^{−1}(D) so that g(G) = D and the map g: G → D is a finite branched cover. Thus we have g(∂G) = γ.

The following result is given as a corollary of Theorem VI.8 in [42]. It also follows from [30, Theorem 5.5].

Lemma 2. If g(z) is a transcendental and meromorphic function in C, then given any three simply-connected domains D_{i}, 1 ≤ i ≤ 3, with sectionally analytic boundaries, such that \overline{D}_{i} are mutually disjoint, there is at least one value of i such that g(z) has infinitely many simply-connected islands over D_{i}.

Proof of Theorem A. Since the theorem is known for f(z) ∈ E and since for f(z) ∈ P there are no components of F (f) whose connectivity is greater than or equal to three, we can suppose that f(z) ∈ M and that F (f) has a component D whose connectivity is at least three. Thus D^{c} has at least three components H_{1} , H_{2} , H_{3} . It follows from Lemma 1 that there are simple polygons γ_{i} in D, 1 ≤ i ≤ 3, and (simply-connected) complementary domains D_{i} of γ_{i} such that \overline{D}_{i} ∩\overline{D}_{j} = ∅, j ≠ i, and each D_{i} contains points of the Julia set.

Now pick any point ξ_{1} in the Julia set and any open neighbourhood V_{1} which contains ξ_{1} . Then there is some k ∈ N such that f^{k} has a pole β in V_{1} .

There is a neighbourhood U of β, \overline{U} ⊂ V_{1} which is mapped by f^{k} locally univalently (except perhaps at β) onto a neighbourhood N of ∞.

It follows from Lemma 2 that f(z) has a simply-connected island G ⊂ N ∩C, which lies over one of the D_{i}, say D_{1} . The branches of (f^{k})^{−1} which take values in U for z ∈ G are each univalent in G. We take one such branch, say h, which maps G univalently onto the simply-connected domain V_{2} ⊂ U \ {β}. Then f^{k+1} maps V_{2} onto D_{1} and since D_{1} meets J(f) it follows that V_{2} contains a point ξ_{2} of J(f). From f^{k+1}(∂V_{2}) = γ_{1} it follows that ∂V_{2} ⊂ F (f).

We may now replace V_{1} by V_{2} , ξ_{1} by ξ_{2} and deduce that V_{2} contains an arbitrarily small simply-connected neighbourhood V_{3} such that \overline{V}_{3} ⊂ V_{2} , ∂V_{3} ⊂ F (f), and ξ_{3} ∈ V_{3} ∩ J(f) ≠ ∅. By continuing inductively we obtain a sequence of nested simply-connected domains V_{n}, each of which contains a point ξ_{n} ∈ J(f). The sets \overline{V}_{n} may be assumed to shrink to a single point ξ. Then ξ = lim ξ_{n} ∈ J(f). Further ∂V_{n} is constructed to be in F (f), and so ξ is a singleton component of J(f).

As an application of the preceding theorem we note that for any meromorphic function f(z) with a completely invariant multiply-connected component G of F (f), the connectivity of G is infinite, by item 5 of the introduction, and hence J(f) has dense singleton components.

Further examples are given by the meromorphic functions constructed in [10] which have wandering Fatou components of arbitrary finite connectivity.

3. Proofs of Theorems B and C

Proof of Theorem B. We may suppose that f(z) ∈ M since f(z) ∈ P can have at most one doubly-connected Fatou component.

Let γ_{i}, 1 ≤ i ≤ 3, be sectionally analytic curves such that γ_{i} ⊂ U_{i} ⊂ F (f) and γ_{i} cannot be deformed to a point in U_{i}.

In case (a) we take D_{i} to be the interior of γ_{i}, 1 ≤ i ≤ 3, that is, the bounded component of γ_{i}^{c}. Now γ_{1} ⊂ F (f) is separated by the boundary of the unbounded component of U_{1}^{c} from γ_{2} and γ_{3} . Hence \overline{D}_{1},\overline{D}_{2},\overline{D}_{3} are disjoint and each D_{i} contains points of J(f). The proof now proceeds precisely as for Theorem A but using this interpretation of D_{i}.

In case (b) we take D_{1} and D_{2} as in case (a) but take D_{3} to be the exterior of γ_{3} . See Figure 1.

Figure 1. The interiors D_{i}, 1 ≤ i ≤ 2, and the exterior D_{3} .

Proof of Theorem C. Suppose that f(z) is a function in P with an invariant Herman ring R. Thus f(z) has the form (1) for some α and k.

Let γ be the closure of an orbit in R, where we choose γ to avoid the countable set of algebraic singularities of f^{−1} . Then γ is an analytic Jordan curve. Denote the interior of γ by I. We show that I contains the pole α of f(z).

Suppose f(z) is analytic in I. Then f(I) is bounded and ∂(f(I)) ⊂ f(∂I) = f(γ) = γ. It follows that f(I) = I which implies that f^{n}(I) = I and I ⊂ F (f), but I contains points of ∂R by assumption. This contradiction shows that I contains the pole α.

Now I contains an orbit γ′ of f(z) in R which is different from γ. Hence f(I) meets I and so I ⊂ f(I). Hence there is some point β ∈ I such that f(β) = α ∈ I. This contradicts the assumed form of f(z) and the theorem is proved.

4. Proofs of Theorems D and E

Proof of Theorem D. The assertion of the Theorem D follows from Theorem A unless all A_{i} have connectivity two. The result is already known for f(z) ∈ E. Further, it is known [7] that if f(z) ∈ P the set F (f) has at most one multiply-connected component. Thus we may assume that f(z) ∈ M and that A_{i} are all bounded doubly-connected domains which separate 0 from ∞.

Now A_{1} is a wandering component of F (f) so that the limit functions of convergent sequences f^{n_{j}} in A_{1} are necessarily constant [11]. We note that each A_{i} is bounded and thus the map f: A_{i} → A_{i+1} is a (possibly branched) covering. In fact it follows from the Riemann–Hurwitz relation that f(z) is an unbranched k_{i} to 1 covering map for some k_{i} ∈ N. Thus if the path γ ∈ A_{1} is a generator of the homotopy group of A_{1} the image γ_{n} = f^{n−1}(γ) is a path in A_{n} which is not homotopic to a constant. Since all limit functions of subsequences of f^{n} are constant in A_{1} , it follows that the spherical diameter of γ_{n} → 0 as n → ∞. For any convergent subsequence f^{n_{j}} we have either γ_{n_{j}} → ∞ or γ_{n_{j}} → 0. Thus at least one of ∞ or 0 is a singleton component of J(f) cut off from any other point of J(f) by certain arbitrarily small curves γ_{n_{j}} ⊂ A_{n_{j}} ⊂ F (f). The boundaries of A_{n_{j}} are in J(f) and converge to ∞ or 0. If ∞ is a singleton component of J(f), then, since f(z) ∈ M, the preimages of ∞, which are also singleton components, are dense in J(f).

For the remaining case we may assume that no sequence f^{n_{j}} → ∞ in A_{1} and hence that the whole sequence f^{n} → 0 in A_{1} , that is the domains A_{n} → 0 as n → ∞. Hence 0 ∈ J(f) and therefore A_{n+1} = f(A_{n}) → f(0) so that f(0) = 0. Then f(z) is analytic in some neighbourhood U of 0. Further, A_{n} is in U for n > n_{0} and A_{n} → 0 as n → ∞. This implies that there is some n > n_{0} such that A_{n+1} is inside A_{n}, that is, in the component of A_{n}^{c} which contains 0. If α_{n} is the outer boundary of A_{n} and I is the component of α_{n}^{c} which contains 0, then f(I) is a bounded domain whose boundary belongs to f(α_{n}) ∈ ∂A_{n+1} which is inside I. Thus f(I) ⊂ I and (f^{n}) is analytic and normal in I, but this contradicts the fact that 0 ∈ J(f). The proof of Theorem D is complete.

In order to prove Theorem E we need the following results.

Lemma 3 (Runge, see for example [28]). Suppose that K is compact in C and f(z) is holomorphic on K; further, let ε > 0. Let E be a set such that E meets every component of Ĉ \ K. Then there exists a rational function R with poles in E such that

|f(z) − R(z)| < ε, z ∈ K.

Lemma 4 [10]. Suppose that D is an unbounded plane domain with at least two finite boundary points and that g is analytic in D and satisfies (i) g(D) ⊂ D and (ii) g^{n}(z) → ∞ as n → ∞, z ∈ D. Then for any z_{0} in D there is a constant k > 0 such that

lim sup \frac{1}{n} log log |g^{n}(z_{0})| ≤ k.

Further, there is a path γ in D and a constant A > 1 such that γ leads to ∞ and |z|^{1/A} ≤ |g(z)| ≤ |z|^{A} for all large z on γ.

Proof of Theorem E. Choose ε > 0 so that

\frac{3}{4} < ∏_{1}^{∞}(1 − ε^{n}) < ∏_{1}^{∞}(1 + ε^{n}) < \frac{3}{2},

and a sequence r_{n}, n ≥ 2, so that r_{n+1} > 16r_{n}, r_{n+1} > e^{r_{n}} and r_{2} > 16. Thus the disc \overline{D} =\overline{D}(0, 1) and the annuli \overline{B}_{n} : \frac{1}{3}r_{n} ≤ |z| ≤ 3r_{n}, n ≥ 2, are all disjoint. Then δ_{n} = r_{n}/r_{n+1} , n ≥ 2, satisfies 0 < δ_{n} < 1/16 and δ_{n} → 0 as n → ∞.

Now choose positive numbers ε_{1} and ε_{n+1} , n ∈ N, so small that

(2)  (1 − 6r_{n+1}ε_{n+1}δ_{n+1}^{−1})^{−1} < (1 + ε^{n})^{1/2},
(3)  r_{n+2} + ε_{1} < (1 + ε^{n})^{1/2}r_{n+2},
(4)  (1 + 6r_{n+1}ε_{n+1}δ_{n+1}^{−1})^{−1} > (1 − ε^{n})^{1/2},
(5)  r_{n+2} − ε_{1} > (1 − ε^{n})^{1/2}r_{n+2},
(6)  ε_{n+1} < \frac{1}{2}ε_{n},  ε_{1} < \frac{1}{4}.

Since |(1 ± ε^{n})^{1/2} − 1|r_{n+2} ∼ \frac{1}{2}ε^{n}r_{n+2} → ∞ as n → ∞ we may choose ε_{1} to satisfy (3) and (5). Then we choose ε_{n+1} , n ∈ N, to satisfy (2), (4), and (6).

Now define f_{1}(z) = z^{−1} . It follows from Lemma 3 that there is a rational function f_{n+1} , n ≥ 1, with poles only at \frac{1}{4}r_{n+1} and ∞, such that

(7)  |f_{n+1}(z)| < ε_{n+1}  in \overline{D}(0, \frac{1}{5}r_{n+1})

and

(8)  |f_{1}(z) + ⋯ + f_{n+1} − δ_{n+1}z^{−1}| < ε_{n+1}  in B_{n+1}.

Then f defined by f(z) = ∑_{1}^{∞} f_{n}(z) is meromorphic, by (6) and we have, using (7) and (8) that

(9)  |f(z) − z^{−1}| ≤ ∑_{2}^{∞} ε_{n} < ε_{1}  in \overline{D},

(10)  |f(z) − δ_{n+1}z^{−1}| ≤ ∑_{j=n+1}^{∞} ε_{j} < 2ε_{n+1}  in B_{n+1}.

Thus if 1 < λ ≤ 3 and r_{n+1}λ^{−1} < |z| < λr_{n+1} , n ∈ N, we have from (9) and (10) that

f(z) = δ_{n+1}z^{−1} + η_{n}, |η_{n}| < 2ε_{n+1},

so that f(z) ∈ D and

(11)  f^{2}(z) = \frac{z}{δ_{n+1}} (1 + η_{n}zδ_{n+1}^{−1})^{−1} + η′,  |η′| < ε_{1}.

So it follows from (2), (3), (4), and (5) that

λ^{−1}r_{n+1}δ_{n+1}^{−1}(1 + 2λr_{n+1}ε_{n+1}δ_{n+1}^{−1})^{−1} − ε_{1} < |f^{2}(z)|
< λr_{n+1}δ_{n+1}^{−1}(1 − 2λr_{n+1}ε_{n+1}δ_{n+1}^{−1})^{−1} + ε_{1},

and hence

(12)  λ^{−1}r_{n+2}(1 − ε^{n}) < |f^{2}(z)| < λr_{n+2}(1 + ε^{n}).

Thus if we set A_{2} = \{z : \frac{1}{2}r_{2} < |z| < 2r_{2}\} ⊂ B_{2} we find that

\frac{3}{8}r_{n+2} < \frac{1}{2}r_{n+2} ∏_{1}^{∞}(1 − ε^{n}) < |f^{2n}(z)| < 2r_{n+2} ∏_{1}^{∞}(1 + ε^{n}) < 3r_{n+2}.

Hence f^{2n}(z) is in B_{n+2} and so f^{2n}(z) → ∞ in A_{2} . Further, f^{2n}(A_{2}) is in some component Ã_{2n+2} of F (f).

Again, using (2) in (11) we see that for z in B_{n+1}

f^{2}(z) = \frac{z}{δ_{n+1}} (1 + μ) + η′

where

\begin{aligned}|μ| & = |(η_{n}zδ_{n+1}^{−1})(1 + η_{n}zδ_{n+1}^{−1})^{−1}| ≤ |(6r_{n+1}ε_{n}δ_{n+1}^{−1})(1 − 6r_{n+1}ε_{n}δ_{n+1}^{−1})^{−1}| \\ & ≤ (1 + ε^{n})^{1/2} − 1 \end{aligned}

and |η| < ε_{1} . It follows inductively that each f^{2n}(A_{2}) is a region in B_{n+2} which separates 0 and ∞. Clearly then Ã_{2n+2} which contains f^{2n}(A_{2}) is multiply connected. Also the components Ã_{2n+2} are all different since otherwise we have an unbounded component of F (f) which is invariant under f^{2} in which f^{2n} → ∞. It follows from Lemma 4 that there is then a constant K and a curve Γ which tends to ∞ on which |f^{2}(z)| < |z|^{K} for large z. But in any f^{2n}(A_{2}) we have |z| < 3r_{n+2} and |f^{2}(z)| > \frac{3}{8}r_{n+3} > \frac{3}{8}e^{r_{n+2}} , so no such Γ exists. Finally Ã_{2n+1} = f(Ã_{2n}) is multiply connected. Further, Ã_{2n+1} → 0 and Ã_{2n+2} → ∞.

5. Functions with finitely many poles

We will start this section by giving a theorem of H. Bohr.

Theorem 5.1 [18]. If f(z) = ∑_{n=1}^{∞} a_{n}z^{n} is analytic in D(0, 1) and for some ϱ, 0 < ϱ < 1, M_{f}(ϱ) = Max {|f(z)| : |z| = ϱ} satisfies M_{f}(ϱ) > 1, then f(D) contains the circle C_{r} = {z : |z| = r} for some r > C(ϱ), where C(ϱ) is a positive constant which depends only on ϱ.

We wish to have a similar result to Bohr’s theorem for functions f(z) which are analytic in 0 < R ≤ |z| < ∞.

Theorem 5.2. If f(z) is analytic in 0 < R ≤ |z| < ∞ and M_{f}(r) → ∞ as r → ∞, then for all sufficiently large ϱ the image f(A_{ϱ}) of A_{ϱ} = {z : R ≤ |z| ≤ ϱ} contains some circle C_{r} = \{z : |z| = r\} where r ≥ cM_{f}(\frac{1}{2}ϱ2) and c is a positive absolute constant.

We shall use a consequence of Schottky’s theorem, see e.g. [12, Lemma 1].

Lemma 5. If g(z) is regular in the annulus A = {z : α ≤ |z| ≤ β}, β/α = γ > 1, if g(z) takes neither value 0 nor 1 in A, and if for some |z_{0}| = \sqrt{αβ} one has |g(z_{0})| ≤ μ, then there is a constant K depending only on γ, μ such that |g(z)| ≤ K uniformly on |z| = \sqrt{αβ}.

Proof of Theorem 5.2. I. Let K denote the constant in Lemma 5 when γ = 4 and μ = 2. Since log M_{f}(r) is a convex function of log r, it follows that M_{f}(r) can have at most one minimum in [R, ∞), at s say, and is increasing for r > s.

Since M_{f}(r) → ∞ as n → ∞ we see that there is some T ≥ R such that M_{f}(r) > M_{f}(R) for r > T.

Now choose ε = 1/2(1+3K). Then there is a positive ϱ_{0} such that for ϱ ≥ ϱ_{0} we have

\frac{1}{4}εM_{f}(\frac{1}{2}ϱ) > M_{f}(R)  and  ϱ > 4R.

Suppose that for ϱ ≥ ϱ_{0} and for every r ≥ (\frac{1}{4}ε)M_{f}(\frac{1}{2}ϱ) there is some point on C_{r} which is not in f(A_{ϱ}). In particular there is w′ with

|w′| = (\frac{1}{4}ε)M_{f}(\frac{1}{2}ϱ)

and w′′ with

|w′′| = (\frac{1}{2}ε)M_{f}(\frac{1}{2}ϱ)

such that f(z) ≠ w′, w′′ in A_{ϱ}.

There is some r′ in R ≤ r′ < \frac{1}{2}ϱ where M_{f}(r′) = |w′| and there is some z′ with |z′| = r′ and |f(z′)| = M_{f}(r′). The point z′ lies on a level curve

Γ = {z : |f(z)| = |w′|}.

Now Γ cannot meet |z| = R since (\frac{1}{4}ε)M_{f}(\frac{1}{2}ϱ) > M_{f}(R). If the component L of {z : |f(z)| < |w′|} which contains {z : R < |z| < r′} lies inside A_{ϱ}, then we can take Γ to be one of the boundary curves of L which closes in \overline{A}_{ϱ}. Since arg f(z) is monotone on Γ, it then follows that arg f(z) changes by at least 2π on Γ, that is f(z) would take every value of modulus |w′| at some point on Γ. Thus Γ meets \{z : |z| = \frac{1}{2}ϱ\} and there is some point z_{0} such that |z_{0}| = \frac{1}{2}ϱ and |f(z_{0})| = |w′|.

II. Now consider

g(z) = \frac{f(z) − w′}{w′′ − w′}

which is analytic in \{z : \frac{1}{4}ϱ < |z| < ϱ\} ⊂ A_{ϱ} and g(z) ≠ 0, 1. Using the results in I we have

|g(z_{0})| ≤ \frac{|f(z_{0})| + |w′|}{|w′′| − |w′|} = \frac{2|w′|}{|w′′| − |w′|} = 2.

It follows from Lemma 5, with α = \frac{1}{4}ϱ, β = ϱ, that |g(z)| ≤ K uniformly on |z| = \sqrt{αβ} = \frac{1}{2}ϱ.

But on |z| = \frac{1}{2}ϱ we have uniformly

\begin{aligned}|f(z)| & ≤ |w′| + K|w′′ − w′| ≤ |w′| + K(|w′′| + |w′|) \\ & = M_{f}(\frac{1}{2}ϱ)(1 + 3K)(\frac{1}{4}ε) = \frac{1}{8}M_{f}(\frac{1}{2}ϱ) \end{aligned}

so that |f(z)| < M_{f}(\frac{1}{2}ϱ), which is a contradiction. Thus the theorem holds with c = \frac{1}{4}ε where ε = \frac{1}{2}(1 + 3K).

We shall prove Theorem F by using Theorem 5.2.

Proof of Theorem F. (i) Take R > 0 such that D(0, R) contains the finite set of poles of f(z). Choose ϱ so large that ϱ > M_{f}(R) and that cM_{f}(\frac{1}{2}σ) > 2σ for every σ ≥ ϱ, where c is the constant of Theorem 5.2. If the theorem is false there is an invariant component G of F (f) and a curve γ which lies in G ∩ {z : |z| > ϱ} and satisfies n(γ, 0) = 1. Then the annulus A_{ϱ} = {z : R ≤ |z| ≤ ϱ} is in the interior of γ.

Let B denote the region between C(0, R) and γ so that A_{ϱ} ⊂ B, see Figure 2.

Figure 2. B is the region between C(0, R) and γ.

The function f(z) is analytic in B, so

∂f(B) ⊂ f(∂B) = f(C(0, R)) ∪ f(γ)

where f(C(0, R)) ⊂ \overline{D}(0, M_{f}(R)) with M_{f}(R) < ϱ. Since f(A_{ϱ}) ⊂ f(B) it follows from Theorem 5.2 that f(B) contains some circle C(0, r) such that r ≥ cM_{f}(\frac{1}{2}ϱ) > 2ϱ.

The boundary of f(B), which is compact in C, contains the boundary of the unbounded component of the complement of f(B). Thus the latter, say γ_{1} is part of f(γ) and is a continuum which lies in |z| > 2ϱ, and belongs to G.

We may now repeat the above argument but with B replaced by the region between |z| = R and γ_{1} , say B_{1} and with ϱ replaced by 2ϱ. Thus f(B_{1}) has an ‘outer boundary’ γ_{2} which is in |z| > 4ϱ and belongs to G.

Repeating this argument inductively, there is a simple closed curve γ_{n} which is part of f^{n}(γ) and contains the disc D(0, 2^{n}ϱ) in its interior. Then there is a compact K ⊂ γ such that f^{n}(K) ⊂ γ_{n}, for n ∈ N, therefore f^{n} → ∞ on K. Since f^{n} → ∞ at some points of γ, it follows that f^{n} → ∞ in G.

(ii) We may assume without loss of generality that 0 and 1 are not in G. Let H = Ĉ − {0, 1, ∞}, γ_{n} be as above and let h = max d_{G}(z, w) < ∞ where z ∈ γ, w ∈ γ_{1} , and d_{G}(z, w) denotes the hyperbolic distance in G.

For any w ∈ γ_{n} we have that w = f^{n}(z) for some z ∈ γ therefore

d_{G}(w, f(w)) = d_{G}(f^{n}(z), f^{n+1}(z)) ≤ d_{G}(z, f(z)) ≤ h.

Now f^{n}(z) → ∞ as n → ∞, thus for w ∈ γ_{n} there is some path δ from w to f(w) such that

h ≥ d_{G}(w, f(w)) = ∫_{δ} ϱ_{G}(z) |dz|

where ϱ_{G} is the Poincaré density for G_{1} . Since G ⊂ H we have ϱ_{G} ≥ ϱ_{H} and ϱ_{H} ∼ c′/|z| log |z| for some constant c′, as z → ∞. Now

(13) h ≥ ∫_{δ} \frac{c′ |dz|}{|z| log |z|} ≥ c′ ∫_{|w|}^{|f(w)|} \frac{d|z|}{|z| log |z|} = c′(log log |f(w)| − log log |w|)

so

(14)  |f(w)| ≤ |w|^{p},  p = e^{h/c′}.

The inequality (14) holds for w ∈ γ_{n}, n large. The function f(z) is analytic in the annulus R ≤ |z| < ∞, then there are numbers a_{j}, j ∈ Z, such that

f(z) = ∑_{−∞}^{∞} a_{j}z^{j} in R < |z| < ∞.

For an integer k > p, the function

F (z) = \frac{f(z)}{z^{k}} = ∑_{−∞}^{∞} a_{j}z^{j−k}

is analytic in R ≤ |z| < ∞. It follows from (14) that Sup_{γ_{n}} |F (z)| → 0 as n → ∞. Now the maximum principle implies that F (z) is bounded, say b, thus |F (z)| < b in a neighbourhood of ∞ (outside γ_{1} ). Thus ∞ is a removable singularity of F (z) and a pole of f(z) contrary to the assumption that f(z) is transcendental.

6. Orbits which tend to ∞

Proof of Theorem G. This is just part (i) of the proof of Theorem F. We note that we do not use the hypothesis that γ ⊂ F (f), made in Theorem F until we have proved that f^{n}(γ) contains a simple closed curve γ_{n} which winds around zero and satisfies d(0, γ_{n}) → ∞. Moreover γ_{n+1} ⊂ f(γ) and A_{n} = γ ∩ f^{−n}(γ_{n}) is a decreasing family of compact sets whose intersection K is in γ ∩ I(f).

Proof of Theorem H. Let p_{1} be a pole of f(z) of order k and let D_{1} = D(p_{1}, r_{1}), r_{1} > 0. It is easy to see that f(D_{1}) is a neighbourhood of ∞. Since p_{1} is a k -fold pole of f(z), for r_{1} sufficiently small there is a simply-connected neighbourhood V_{1} of p_{1} in D_{1} which is mapped locally univalently except at p_{1} onto |f| > R_{1} , R_{1} > 1, for some R_{1} (\overline{V}_{1} to |f| ≥ R_{1} ). Thus we can find a pole p_{2} in {|z| > R_{1}} such that |p_{2}| > |p_{1}|, and a disk D_{2} = D(p_{2}, r_{2}) ⊂ {|z| > R_{1}}, r_{2} very small. By the preceding argument p_{2} ∈ V_{2} , where V_{2} is a simply-connected neighbourhood in D_{2} which is mapped locally univalently by f, except at p_{2} , to \{|f| > R_{2}\} for some R_{2} > 2R_{1} . We may take a branch of f^{−1} analytic in \overline{V}_{2} so that U_{1} = f^{−1}(\overline{V}_{2}) ⊂\overline{V}_{1} .

Using the same argument we can take a pole p_{3} in {|z| > R_{2}} and the disk D_{3} = D(p_{3}, r_{3}), r_{3} very small, such that p_{3} ∈ V_{3} where V_{3} is a simply-connected neighbourhood in D_{3} which is mapped locally univalently by f, except at p_{3} , to \{|f| > R_{3}\} where R_{3} > 2R_{2} . We take a branch of f^{−1} analytic in \overline{V}_{3} so that f^{−1}(\overline{V}_{3}) ⊂\overline{V}_{2} .

Figure 3. The process.

Repeating this process, see Figure 3, inductively we have D_{n} = D(p_{n}, r_{n}) → ∞ and f^{−n}(\overline{V}_{n+1}) = U_{n} ⊂ \overline{V}_{1} ⊂ D_{1} , n ∈ N. Then ∩U_{n} ≠ ∅ and β ∈ ∩U_{n} implies f^{n}(β) ∈ D_{n+1} . Thus D_{1} contains β ∈ I(f). The theorem is proved.

Remark In the above construction we have p_{3} ∈ V_{3} so that U_{2} = f^{−2}(\overline{V}_{3}) contains a point f^{−2}(p_{3}) = f^{−3}(∞). In general U_{n} contains a preimage f^{−n}(∞). We may choose the r_{n} so small that ∩U_{n} is a single point β. Since preimages of poles are in J(f) we then have β ∈ J(f).

In the introduction we defined the set E(f) as the set of exceptional points of f(z).

Lemma 6 [9, Lemma 1]. Let f(z) be a transcendental meromorphic function. For any q ∈ J(f) and any p not in E(f), q is an accumulation point of O^{−}(p).

Proof of Theorem I. It follows from Theorem G and H that I(f) ≠ 0. It is easy to see that I(f) is infinite since for any z ∈ I(f) all f^{n}(z), n ∈ N, are different and in I(f). Thus we can choose three different points β, γ, δ in I(f) such that f(β) = γ and f(γ) = δ. Now take a point α in the Julia set and let V be a neighbourhood of α. Since there are at most two exceptional points in the sense of Lemma 6 there exists a pre-image α^{∗} in V of one of the points β, γ, δ. The point α^{∗} belongs to I(f), so J(f) ⊂ \overline{I(f)}. Now periodic points are dense in J(f) and do not belong to I(f). Hence the interior of I(f) belongs to F (f) so that J(f) ⊂ ∂I(f).

To prove the opposite inclusion, take a point α ∈ ∂I(f) and suppose that α ∈ F (f). Let V ⊂ F (f) be a neighbourhood of α. Then V contains points of I(f) and all f^{n} are analytic in V and f^{n} → ∞ in V . This implies that V ⊂ I(f) and contradicts the assumption that α ∈ ∂I(f).

Proof of Theorem J. (i) First suppose that f(z) has only finitely many poles. Choose positive ϱ so large that γ = C(0, ϱ) satisfies the assumption of Theorem G. Let γ_{n}, A_{n}, K be as in the proof of Theorem G, so that K ⊂ I(f). If K meets J(f) we are finished, so assume that K ⊂ F (f). Since K is compact it is covered by some finite set D_{1}, D_{2}, …, D_{p} of components of F (f). Since K = ⋂^{∞} A_{n} with A_{n} compact we have A_{n} ⊂ ⋃_{i=1}^{p} D_{i} ⊂ F (f), n > n_{0} . But f^{n}(A_{n}) = γ_{n} so that γ_{n} ⊂ F (f).

Let G_{n} denote the component of F (f) which contains γ_{n} so that f(G_{n}) = G_{n+1} , since f(γ_{n}) meets γ_{n+1} . If some G_{n} = G_{n+1} , then G_{n} is an unbounded invariant component of F (f) which contains arbitrarily large curves γ_{m}, m ≥ n > n_{0} , which wind round zero. By Theorem F this is impossible.

It follows that all G_{n} are different and hence bounded components of F (f). Since d(0, γ_{n}) → ∞ as n → ∞ we see that f^{n} → ∞ in each G_{k} and also that G_{n} → ∞ as n → ∞. But f(∂G_{n}) = ∂f(G_{n}) so that each ∂G_{k} ⊂ I(f) and of course belongs to J(f). The proof is complete in this case.

(ii) If f(z) has infinitely many poles the proof follows from the remark at the end of the proof of Theorem H, where β ∈ I(f) ∩ J(f).

7. Completely invariant domains

Proof of Theorem K. Suppose that F (f) has two mutually disjoint completely invariant domains G_{1} and G_{2} (possibly others too). Then G_{i}, 1 ≤ i ≤ 2 are simply connected and infinite [11, Lemma 4.3].

Take a value α in G_{1} such that f(z) = α has infinitely many simple roots z_{i} (f′(z) = 0 at only countably many z so we have to avoid only countably many choices of α). All z_{i} are in G_{1} .

Similarly take β in G_{2} such that f(z) = β has infinitely many simple roots z′_{i} in G_{2} . By Gross’ star theorem [29] we can continue all the regular branches g_{i} of f^{−1} such that g_{i}(α) = z_{i}, along almost every ray to ∞ without meeting any singularity (even algebraic). Thus we can move β ∈ G_{2} slightly if necessary so that all g_{i} continue to β analytically along the line l, which joins α and β. The images g_{i}(l) are disjoint curves joining z_{i} to z′_{i}. Denote g_{i}(l) = γ_{i}. Note that γ_{i} is oriented from z_{i} to z′_{i}.

The branches f^{−1} are univalent so γ_{i} are disjoint simple arcs. Different γ_{i} are disjoint since γ_{i} meets γ_{j} at say w_{0} only if two different branches of f^{−1} become equal with values w_{0} which can occur only if f^{−1} has branch point at f(w_{0}) in l, but this does not occur.

Take γ_{1} and γ_{2} . Since G_{1} is a domain we can join z_{1} to z_{2} by an arc δ_{1} in G_{1} and similarly z′_{1} to z′_{2} by an arc δ′_{1} in G_{2} . If δ′_{1} is oriented from z′_{1} to z′_{2} , let p′ be the point where, for the first time, γ_{1} meets δ′_{1} and q′ be the point where, for the first time, γ_{2} meets δ′_{1} . If δ_{1} is oriented from z_{1} to z_{2} , let p be the point where, for the last time, γ_{1} meets δ_{1} and q be the point where, for the last time, γ_{2} meets δ_{1} . These might look like Figure 4.

Figure 4. The arcs γ_{1} , γ_{2} , δ_{1} and δ′_{1} .

Denote by β_{1} the part of δ_{1} which joins the points p and q, by β′_{1} the part of δ′_{1} which joins the points p′ and q′, by γ̂_{1} the part of γ_{1} which joins the points p and p′, oriented from p to p′, and by γ̂_{2} the part of γ_{2} which joins the points q and q′, oriented from q to q′. Then γ̂_{1}β′_{1}γ̂_{2}^{−1}β_{1}^{−1} is a simple closed curve Γ_{1} with an interior A_{1} . We assert that the function f(z) has a pole in A_{1} .

Suppose that f(z) is analytic in A_{1} , then f(A_{1}) is a bounded set and the frontier of f(A_{1}) is contained in f(Γ_{1}) and hence in f(β_{1}) ∪ f(β′_{1}) ∪ l.

The the curves f(β_{1}) and f(β′_{1}) are closed curves lying in G_{1} and G_{2} respectively. Thus f(β_{1}) and f(β′_{1}) are mutually disjoint.

Consider the unbounded component H of the complement of f(β_{1}) ∪ f(β′_{1}). The component H meets l and in fact if a is the last point of intersection of l with f(β_{1}) and b the first point of the intersection of l with f(β′_{1}) the segment ab of l is a cross cut of H whose end points belong to different components of the frontier of H. It follows that ab does not disconnect H. Now in fact a point w of ab (≠ a or b) is the image of f(z) of an interior point z in the arc γ̂_{1} of Γ_{1} . In the neighbourhood of z and inside Γ_{1} the function takes an open set of values near w, some of which lie off l and in H \ (ab). Then since the frontier of A_{1} is contained in f(β_{1}) ∪ f(β′_{1}) ∪ l we see that f(A_{1}) must contain the whole of the unbounded set H \ (ab). This contradiction proves that there is a pole in A_{1} . So far the argument closely resembles that of [8].

Now for large j, γ_{j} does not meet A_{1} . Pick another pair, say γ_{3} and γ_{4} , which do not meet A_{1} ∪ Γ. Thus the points z_{3} , z_{4} are in G_{1} and the points z′_{3} , z′_{4} are in G_{2} such that γ_{3} joins z_{3} to z′_{3} , oriented from z_{3} to z′_{3} , and γ_{4} joins z_{4} to z′_{4} , oriented fromz_{4} to z′_{4} .

It follows from Theorem 7.1 in [39, p. 151] that if z_{3} , z_{4} are separated in G_{1} by Γ_{1} , then they are separated in G_{1} by a component of G_{1} ∩ Γ_{1} , that is by a cross cut ϱ, part of Γ_{1} , in G_{1} . Thus z_{3} , z_{4} belong to different components of G_{1} \ ϱ one inside Γ_{1} , in A_{1} say P_{1} , and the other outside Γ_{1} , say P_{2} , so we can take z_{3} ∈ P_{1} and z_{4} ∈ P_{2} . But z_{3} is outside A_{1} by assumption. Hence we may join z_{3} to z_{4} by an arc δ_{2} in G_{1} which does not meet Γ_{1} or A_{1} . Similarly we can join z′_{3} to z′_{4} by an arc δ′_{2} in G_{2} without meeting Γ_{1} or A_{1} .

Repeating the argument given before to construct the Jordan curve Γ_{1} , we can construct another Jordan curve Γ_{2} , composed of arcs of γ_{3}, γ_{4}, δ_{2}, δ′_{2} , which does not meet Γ_{1} or A_{1} . Then the interior A_{2} of Γ_{2} satisfies A_{1} ∩ A_{2} = ∅ and A_{2} contains a pole of f(z).

By induction we can construct a sequence of Jordan curves Γ_{i} of the same type as Γ_{1} , with interior A_{i} such that the sets (Γ_{i} ∪A_{i}) are disjoint for different i. Having constructed Γ_{1}, …, Γ_{i} we choose γ_{2i+1}, γ_{2i+2} disjoint from \overline{A}_{1}∪⋯∪\overline{A}_{i} and can join the ends z_{2i+1}, z_{2i+2} in G_{1} without meeting Γ_{1} ∪⋯∪Γ_{i} and z′_{2i+1}, z′_{2i+2} in G_{2} without meeting Γ_{1} ∪ ⋯ ∪ Γ_{i}. Each A_{i} contains a pole. Thus if F (f) contains more than one completely invariant component, then f(z) has infinitely many poles. The theorem is proved.

8. Singularities of the inverse function of a transcendental meromorphic function

We recall that the singular values of the inverse f^{−1} of a meromorphic function consist of algebraic branch points or critical values together with the asymptotic values along paths which tend to ∞. The latter are known as transcendental singularities.

Let B_{j} = {z : f^{j} is not meromorphic at z}, so that B_{0} = ∅, B_{1} = {∞}, and B_{j} = {∞} ∪ f^{−1}(∞) ∪ ⋯ ∪ f^{−(j−1)}(∞) in general.

Denote E_{n}(f) = {singularities of f^{−n}}, n ∈ N, so that E_{1}(f) is the set of singularities of f^{−1}(z). Then as shown in [31, Theorem 7.1.2],

f^{n−1}(E_{1}(f) \ B_{n−1}) ⊆ E_{n}(f) ⊆ ⋃_{j=0}^{n−1} f^{j}(E_{1}(f) \ B_{j}).

Thus the set E(f) = {w ∈ Ĉ : for some n ∈ N, f^{−n} has a singularity at w} is given by E(f) = ⋃_{j=0}^{∞} f^{j}(E_{1}(f) \ B_{j}).

Define by E′(f) the set of points which are either accumulation points of E(f) or are singularities of some branch of f^{−n} for infinitely many values of n.

Theorem 8.1 [31, Theorem 7.1.3]. Let f(z) be a meromorphic function. Any constant limit function of a subsequence of f^{n} in a component of F (f) belongs to the set E(f) ∪ E′(f).

For rational functions E(f)∪E′(f) may be replaced by E′(f). The result was proved by Fatou [26, Chapter 4, p. 60], for rational maps and modified by Baker [2] for transcendental entire functions, Bergweiler et al. [16] obtained a stronger result for entire functions. A more general result, which applies to certain classes of functions meromorphic outside a compact totally disconnected set of essential singularities, was proved by Herring [31].

Theorem 8.2 [31, Theorem 7.1.4]. If f(z) is a meromorphic function and C = {U_{0}, U_{1}, … U_{p−1}} is a cycle of Siegel discs or Herman rings of F (f), then for each j, ∂U_{j} ⊂ E(f) ∪ E′(f). Further, in any attracting or parabolic cycles the limit functions belong to E′(f).

This was proved by Fatou [26, §§30–31] for rational functions and the proof in fact remains valid in our case.

We remark as a corollary (see e.g. [31, Theorem 7.1.7], if E(f) ∪ E′(f) has an empty interior and connected complement, then no subsequence of f^{n} can have a non-constant limit function in any component of F (f).

As we shall see in the example of the next section, these results can be useful in proving the non-existence of wandering components of F (f) in cases where the obvious generalisation of Sullivan’s theorem does not apply.

9. Examples

Example 1. If f ∈ E is such that F (f) has an unbounded component then every component of F (f) is simply connected and J(f) has no singleton components. These results no longer hold if f(z) is allowed to have even one pole.

We show that if ε = 10^{−2} , a = 1 + iπ, and f(z) = z + 2 + e^{−z} + ε/(z − a), then F (f) has a multiply-connected unbounded invariant component H′, which contains H (to be defined below), in which f^{n} → ∞.

It follows from [11] that the connectivity of H′ is infinite and from [20] that singleton components are everywhere dense in J(f). We can show explicitly that for some x_{0} < 0 the line S = {z : z = x + iπ, −∞ < x ≤ x_{0}} is outside H′, which verifies the assertion of Theorem F.

Let V_{δ} = \{z : |z −a| ≤ δ\} with δ = 0.1 and let G = \{z : | Re z| ≥ \frac{1}{2}\}. We can define the domain H as follows: H = G∩V_{δ}^{c}, see Figure 5. Thus H is unbounded and multiply connected.

Figure 5. The domain H.

We shall prove that H is an invariant domain, that is, f(H) ⊂ H. Let z be in H and choose ε = 0.01. If we look at |e^{−z} + ε/(z − a)| we have

(15)  |e^{−z} + \frac{ε}{z − a}| ≤ |e^{−z}| + |\frac{ε}{z − a}| ≤ e^{−1/2} + \frac{ε}{δ}.

Thus

Re\{z+2+e^{−z}+ \frac{ε}{z − a}\} ≥ Re(z+2)−|e^{−z}+ \frac{ε}{z − a}| ≥ \frac{1}{2} +2−(e^{−1/2}+ \frac{ε}{δ}) ≈ 1.793.

Hence f(H) ⊂ H ⊂ F (f). By similar arguments we can see that f^{n} → ∞. Let z be in H as before and take f(z) − z = 2 + e^{−z} + ε/(z − a). It follows from (15) that

Re\{f(z) − z\} ≥ 2 − e^{−1/2} − \frac{ε}{z − a} ≈ 1.29 > 1.

Thus Re{f^{n}(z) − z} > n → ∞, n ∈ N, and f^{n}(z) → ∞.

The component H′ of F (f) which contains H thus has the properties claimed. Since f(x + iπ) = x + iπ + 2 − e^{−x} + ε/(x − 1) we see that the segment

S = {z : z = x + iπ, −∞ < x ≤ x_{0}} is invariant under f(z), for some x_{0} < −3. Suppose that z = x′ + iπ ∈ H′ ∩ S. Then it follows from Lemma 4 that there is some constant K such that log log |f^{n}(z)| < Kn. For a constant A > e^{2K} we have | Re f(x + iπ)| > |x|^{A} for x < x_{1} < x_{0} say. Since f^{n}(x′) → ∞ we have for some p ∈ N that Re f^{q}(x′ + iπ) < −x_{1} , q ≥ p. It follows that for n > p

| Re f^{n}(x′ + iπ)| > x_{1}^{A^{n−p}} > e^{e^{(n−p) log A}}.

Thus log log |f^{n}(z)| > Kn if n is sufficiently large. This contradiction proves that H′ does not meet S.

Example 2. For the function f(z) = λ sin z − ε/(z − π), 0 < λ < 1, ε > 0, we wish to show that, for sufficiently small ε, the Fatou set F (f) is a single completely invariant domain of infinite connectivity. It will follow from Theorem A that singleton components are everywhere dense in J(f) but that J(f) also contains unbounded continua.

We take λ′ so that λ < λ′ < 1 and α so that 0 < α < 1 and that λ| cos z| < λ′ for z ∈ S = {z : | Im z| < α}.

Figure 6. T = S ∩ {z : |z − π| > ϱ}.

Let ϱ satisfy ϱ < α and assume that ε is so small that λ′α + εϱ^{−1} < α. Let T = S ∩ {z : |z − π| > ϱ}, see Figure 6. For z = x + iy ∈ T we have

|f(z) − λ sin x| ≤ λ| sin z − sin x| + \frac{ε}{|z − π|} ≤ λ′α + εϱ^{−1} < α < 1.

Since λ sin x ∈ I = [−λ, λ], it follows that f(z) belongs to a compact subset of T. Thus T belongs to an invariant component H of F (f) in which f^{n}(z) → β where β is an attracting fixed point of f(z).

Now we consider the singular points of f^{−1} . On any path Γ which tends to ∞, ε(z − π)^{−1} → 0 and f(z) has a limit l if and only if λ sin z → l. This is possible only for l = ∞. Thus apart from ∞ all singularities of f^{−1} are finite critical values of f(z).

If f′(z) = 0 and |z − π| > t = \frac{1}{4}π, then

| cos z| < ελ^{−1}t^{−2} < 2ελ^{−1}, sin z = ±(1 + η)

where |η| < 2ε^{2}λ^{−2} (if ε was originally chosen small enough). For any such z we have |f(z) − λ sin z| < 4ε/π and so |f(z) ± λ| < 2ε^{2}λ^{−1} + 4ε/π. Thus we have f(z) ∈ T ⊂ H (if ε was chosen small enough).

If f′(z) = 0 and |z − π| ≤ t = \frac{1}{4}π, then

\frac{ε}{(λe^{π/4})} ≤ |(z − π)^{2}| = |\frac{ε}{(λ cos z)}| ≤ \frac{\sqrt{2} ε}{λ},

since

| cos z| ≥ | cos x cos hy| ≥ 1/\sqrt{2}  and  | cos z| < e^{|y|}.

Thus |z − π| < 2^{1/4}\sqrt{ε} \sqrt{1/λ}, | sin z| < 2\sqrt{ε} \sqrt{1/λ}, ε|z − π|^{−1} < \sqrt{ε} \sqrt{λ} e, and |f(z)| < 2\sqrt{ε} \sqrt{λ} + \sqrt{ε} \sqrt{λ} e, so that f(z) ∈ T ⊂ H, provided ε was chosen small enough.

Thus the set E_{1}(f) of the singularities of f^{−1}(z) consists of a countable subset of T whose closure is compact in T, together with ∞. The same is true of the sets E(f) and E′(f) defined in Section 8 and E(f) ∪ E′(f) = ⋃_{0}^{∞} f^{j}(E_{1}(f) \ {∞}) ∪ {β, ∞}. It follows from the results of Section 8 that there are no Siegel discs or Herman rings. The only cyclic component of F (f) is H.

All singularities (except ∞) of f^{−1} are contained in H. Take a point z_{0} in H and a branch g of f^{−1} such that g(z_{0}) ∈ H. For any z_{1} in H and any branch h of f^{−1} at z_{1} we can reach h(z_{1}) by analytic continuation of g along a path γ from z_{0} to z_{1} . Now γ is homotopic to a path from γ in C \ E_{1}(f) to a path γ_{1} from z_{0} to z_{1} , and the continuation of g along γ_{1} is h at z_{1} . But g(γ_{1}) ⊂ F (f) and hence g(γ_{1}) ⊂ H. Thus H is completely invariant.

The only possible constant limits of sequences f^{n_{k}} in Fatou components are β and ∞. The only possible components G of F (f) other than H will be wandering components in which f^{n} → ∞ as n → ∞. Now we shall show that no such domains G exist.

If there is such G we may suppose that f^{n}(G) does not meet D(π, ϱ) for any N, since f^{n}(G) → ∞. We must have Im f^{n} → ∞ in G and hence also f′(f^{n}) → ∞ and (f^{n})′ → ∞ in G. It follows from Bloch’s theorem that f^{n+1}(G) contains some discs D(a, 4π), where | Im a| can be taken arbitrarily large, but then on the horizontal diameter of D(a, π), f(z) ∼ \frac{1}{2}e^{−ix+y} so that f^{n+2}(G) contains some real points, which must be in H. This is impossible, so there are no such domains G.

We have now proved our claim that F (f) = H which is multiply connected, since π is not in F (f). Then H must in fact have infinite connectivity and by Theorem A the singleton components of J(f) are everywhere dense in J(f). It follows from the corollary to Theorem F that the Julia set J(f) has some components which are not singletons. We show that this is true of the component which contains ∞ that is, there exist unbounded components of J(f). It follows from Theorem F that if R is a sufficiently large positive number, then there is no curve γ in F (f) ∩ \overline{D}(0, R)^{c} which winds round zero. This implies that the component K of the closed set J(f) ∪\overline{D}(0, R), which contains ∞, is not a singleton.

If K ⊂ J(f), then we are finished. If not we can take a point w in K ∩\overline{D}(0, R) and for each ε = 1/n we may form an ε-chain (in the spherical metric) in the connected compact set K which starts at ∞ and ends at w. Denote by A_{n} the set {z_{1}^{n} = {∞}, z_{2}^{n}, …, z_{j(n)}^{n}} where z_{j(n)+1}^{n} is the first point of the chain which lies in \overline{D}(0, R). Thus A_{n} ⊂ J(f). Since ∞ ∈ lim inf A_{n} ≠ ∅, we deduce that L = lim sup A_{n} = {z : a neighbourhood of z meets infinitely many A_{n}} is connected (see e.g. [43]). But L ⊂ J(f), ∞ ∈ L and L meets C(0, R). Thus {∞} is not a singleton component of J(f), that is, J(f) has an unbounded component.

For g(z) = λ sin z the Julia set contains segments ±{z : z = iy, 0 < y_{0} < y < ∞} of the imaginary axis. For f(z) one can show by direct discussion that J(f) contains continua which are ‘close to’ the above segments. We omit the discussion.

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Received 29 October 1996