Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 31, 2006, 523–540

Fixed points of composite entire and quasiregular maps

Walter Bergweiler

Christian-Albrechts-Universität zu Kiel, Mathematisches Seminar Ludewig-Meyn-Str. 4, DE-24098 Kiel, Germany; bergweiler@math.uni-kiel.de

Abstract. We give a new proof of the result that if f and g are entire transcendental functions, then f∘g has infinitely many fixed points. The method yields a number of generalizations of this result. In particular, it extends to quasiregular maps in R^{d}.

1. Introduction and main results

The following result was conjectured by Gross (see [11, p. 542] and [15, Problem 5]) and first proved in [4].

Theorem A. Let f and g be entire transcendental functions. Then f ∘ g has infinitely many fixed points.

The following generalization of Theorem A was proved in [5].

Theorem B. Let f and g be entire transcendental functions. Then f ∘ g has infinitely many repelling fixed points.

Here a fixed point ξ of a holomorphic function h is called repelling if |h′(ξ)| > 1. The repelling fixed points play an important role in iteration theory.

The purpose of this paper is twofold. Firstly, we give a new proof of Theorem A. Secondly, we obtain some generalizations of Theorem A (and B).

The main difference between the method employed here and the previous proofs of Theorem A and B is that the Wiman–Valiron method which was crucial in [4], [5] is not used here. Instead we use some ideas from normal families. This method is also applicable for quasiregular maps; see [23] for the definition and basic properties of quasiregular maps.

Theorem 1. Let d ≥ 2 and let f, g: R^{d} → R^{d} be quasiregular maps with an essential singularity at ∞. Then f ∘ g has infinitely many fixed points.

2000 Mathematics Subject Classification: Primary 30D05, 30C65, 30D35.

Part of this research was done while the author was visiting Purdue University. He thanks Alexandre Eremenko and the Purdue Department of Mathematics for the hospitality. He also thanks the referee for useful comments. Support by the Alexander von Humboldt Foundation, the National Science Foundation (grant of A. Eremenko), and the G.I.F., the German–Israeli Foundation for Scientific Research and Development, Grant G -809-234.6/2003, is gratefully acknowledged.

For functions in the plane we also obtain some extensions of the previously known results.

Theorem 2. Let f and g be entire transcendental functions. Then there exists a sequence (ξ_{n}) such that (f ∘ g)(ξ_{n}) = ξ_{n} and (f ∘ g)′(ξ_{n}) → ∞.

Theorem 3. Let f and g be entire transcendental functions. Then f ∘ g has infinitely many nonreal fixed points.

Theorem 3 answers a question of Clunie [10] who had shown that at least one of the two functions f ∘ g and g ∘ f has infinitely many nonreal fixed points. The special case f = g had been dealt with earlier in [8], answering a question of Baker [2]. Theorem 3 implies that for any straight line there are infinitely many fixed points not lying on this line.

Similar ideas to the ones employed in this paper were used—in the context of iteration rather than composition—in [3], [7], [12], [13] for holomorphic maps, and in [25], [26] for quasiregular maps.

Although each of the Theorems 1–3 contains Theorem A as a special case, we will first give a proof of Theorem A in Section 2, as this explains the underlying idea best. In Sections 3–5 we will then prove Theorems 1–3. These sections will make occasional reference to Section 2, but are independent of each other.

2. Proof of Theorem A

2.1. Preliminary lemmas. We shall need a result from the Ahlfors theory of covering surfaces; see [1], [17, Chapter 5] or [21, Chapter XIII] for an account of this theory. To state the result of the Ahlfors theory that we need, let D ⊂ C be a domain and let f: D → C be holomorphic. Given a Jordan domain V ⊂ C, we say that f has an island over V if f^{−1}(V) has a component whose closure is contained in D. Note that if U is such a component, then f|_{U}: U → V is a proper map.

Lemma 2.1. Let D ⊂ C be a domain and let D_{1}, D_{2} ⊂ C be Jordan domains with disjoint closures. Let ℱ be a family of functions holomorphic in D which is not normal. Then there exists a function f ∈ ℱ which has an island over D_{1} or D_{2} .

For example, Lemma 2.1 follows from Theorem 5.5 (applied with a domain D_{3} containing ∞) and Theorem 6.6 in [17].

For a different proof of Lemma 2.1 see [6, Section 5.1]. The proof given there is particularly simple in the case where the D_{j} are small disks. It turns out that this special case suffices for our purposes.

The following lemma is a simple consequence of the maximum principle.

Lemma 2.2. Let D ⊂ C be a domain and let (f_{n}) be sequence of functions holomorphic in D which is not normal. If (f_{n}) converges locally uniformly in D\E for some finite set E, then f_{n} → ∞ in D\E.

This lemma will be useful when dealing with quasinormal families. By definition, a family ℱ of functions holomorphic in a domain D is called quasinormal (cf. [9], [20], [24]) if for each sequence (f_{n}) in ℱ there exists a subsequence (f_{n_{k}}) and a finite set E ⊂ D such that (f_{n_{k}}) converges locally uniformly in D\E. If the cardinality of the exceptional set E can be bounded independently of the sequence (f_{n}), and if q is the smallest such bound, then we say that ℱ is quasinormal of order q.

We denote the maximum modulus of an entire function f by M(r, f).

Lemma 2.3. Let f be an entire transcendental function and A > 1. Then

lim_{r→∞} \frac{M(Ar, f)}{M(r, f)} = ∞.

This result follows easily from the convexity of log M(r, f) in log r and the transcendency of f. We omit the details. For an alternative proof of Lemma 2.3 see the proof of Lemma 3.3 in Section 3.1 below.

2.2. Proof of Theorem A. We first choose a sequence (c_{n}) tending to ∞ such that |f(c_{n})| ≤ 1. We may assume that |c_{n}| ≥ |g(0)| for all n and define r_{n} by M(r_{n}, g) = |c_{n}|.

We define

f_{n}(z) := \frac{f(c_{n}z)}{r_{n}} and g_{n}(z) := \frac{g(r_{n}z)}{c_{n}}.

It is easy to see that no subsequence of (f_{n}) is normal at 0. Since f_{n}(1) → 0 it follows from Lemma 2.2 that (f_{n}) is not normal in C\{0}. Passing to a subsequence if necessary we may thus assume that no subsequence of (f_{n}) is normal at a_{1} := 0 and some a_{2} ∈ C\{0}.

It follows from Lemma 2.3 that if n → ∞, then M(r, g_{n}) → 0 if r < 1 and M(r, g_{n}) → ∞ if r > 1. Lemma 2.2 implies that the sequence (g_{n}) is not quasinormal. Passing to a subsequence if necessary we may thus assume that there exist b_{1}, b_{2}, b_{3} ∈ C\{0} where no subsequence of (g_{n}) is normal. We choose 0 < ε < \frac{1}{2} such that the closed disks of radius ε around the b_{j} are pairwise disjoint and do not contain 0. In the following we denote by B(a, r) the open disk of radius r around a point a; that is, B(a, r) := {z ∈ C : |z − a| < r}.

It follows from Lemma 2.1 that if n is sufficiently large and j ∈ {1, 2}, then f_{n} has an island in B(a_{j}, ε) over at least two of the domains B(b_{k}, ε). This implies that there exists k ∈ {1, 2, 3} such that f_{n} has an island U_{1} ⊂ B(a_{1}, ε) and another island U_{2} ⊂ B(a_{2}, ε) over the same disk B(b_{k}, ε).

Moreover, it follows from Lemma 2.1 that if n is sufficiently large, then there exists j ∈ {1, 2} such that g_{n} has an island V ⊂ B(b_{k}, ε) over B(a_{j}, ε). Then V ∩ g_{n}^{−1}(U_{j}) contains a component W of (f_{n} ∘ g_{n})^{−1}(B(b_{k}, ε)) satisfying \overline{W} ⊂ B(b_{k}, ε).

For z ∈ ∂W we have

|((f_{n} ∘ g_{n})(z) − b_{k}) − ((f_{n} ∘ g_{n})(z) − z)| = |z − b_{k}| < ε = |(f_{n} ∘ g_{n})(z) − b_{k}|.

Rouché’s theorem implies that the number of fixed points of f_{n}∘g_{n} in W coincides with the number of zeros of f_{n} ∘ g_{n} − b_{k} in W. As f_{n} ∘ g_{n} is a proper map from W onto B(b_{k}, ε) the function f_{n} ∘ g_{n} − b_{k} has at least one zero in W and thus f_{n} ∘ g_{n} has a fixed point ξ ∈ W ⊂ B(b_{k}, ε). Then ξ_{n} := r_{n}ξ is a fixed point of f ∘ g. Since ξ_{n} ∈ B(r_{n}b_{k}, r_{n}ε) it follows that ξ_{n} → ∞ as n → ∞ so that f ∘ g has infinitely many fixed points.

3. Proof of Theorem 1

3.1. Preliminary lemmas. As a general reference for quasiregular maps we recommend [23]. We first state some lemmas analogous to those stated in Section 2.1, and begin with the analogue of Lemma 2.1.

We note that Lemma 2.1 is a generalization of Montel’s theorem, which in turn is the result that corresponds to Picard’s theorem in the context of normal families. The analogue of Picard’s theorem for quasiregular maps was given by Rickman [22] who proved that there exists q = q(d, K) ∈ N with the property that every K -quasiregular map f: R^{d} → R^{d} which omits q points is constant. We shall call this number q the Rickman constant. The corresponding normality result was proved by Miniowitz [19], using an extension of the Zalcman lemma [27] to quasiregular maps. We refer to [19] also for further information about normal families of quasiregular maps. Besides normal families we will also consider quasinormal families of quasiregular maps, which are defined in exactly the same way as for holomorphic functions.

Miniowitz’s extension of the Zalcman lemma has been used by Siebert [25],

[26] to deduce the following Lemma 3.1 from Rickman’s theorem. Here we call, as in Section 2.1, a domain U an island of the quasiregular map f: D → R^{d} over the simply connected domain V ⊂ R^{d}, if U is a component of f^{−1}(V) and if \overline{U} ⊂ D. And as in dimension 2 we denote by B(a, r) the open ball of radius r around a point a ∈ R^{d}; that is, B(a, r) := {x ∈ R^{d} : |x − a| < r}. Here |x| is the (Euclidean) norm of a point x ∈ R^{d}. With this notation Siebert’s result (see [25, Satz 2.2.2] or [26, Corollary 3.2.2]) can be stated as follows.

Lemma 3.1. Let d ≥ 2, K ≥ 1 and let q = q(d, K) be the Rickman constant. Let a_{1}, …, a_{q} ∈ R^{d} be distinct. Then there exists ε > 0 with the following property: if D ⊂ R^{d} is a domain and ℱ is a non-normal family of functions K -quasiregular in D, then there exists a function f ∈ ℱ which has an island over B(a_{j}, ε) for some j ∈ {1, …, q}.

The following lemma is literally the same as Lemma 2.2 in Section 2.1, and it is again a simple consequence of the maximum principle.

Lemma 3.2. Let D ⊂ R^{d} be a domain and let (f_{n}) be non-normal sequence of functions which are K -quasiregular in D. If (f_{n}) converges locally uniformly in D\E for some finite set E, then f_{n} → ∞ in D\E.

The next lemma is identical to Lemma 2.3 in Section 2.1. Again we denote by M(r, f) the maximum modulus; that is, M(r, f) := max_{|x|=r} |f(x)|.

Lemma 3.3. Let f: R^{d} → R^{d} be quasiregular with an essential singularity at ∞ and let A > 1. Then

lim_{r→∞} \frac{M(Ar, f)}{M(r, f)} = ∞.

Proof. Suppose that the conclusion does not hold. Then there exist C > 1 and a sequence (r_{n}) tending to ∞ such that M(Ar_{n}, f) ≤ CM(r_{n}, f). The sequence (f_{n}) defined by

f_{n}(x) := \frac{f(r_{n}x)}{M(r_{n}, f)}

is then bounded and thus normal in B(0, A). Passing to a subsequence we may assume that f_{n} → h for some quasiregular map h: B(0, A) → R^{d}. We have h(0) = 0 while M(1, h) = 1. Thus h is not constant.

Now there exists a ∈ R^{d} such that f has infinitely many a-points. Without loss of generality we may assume that a = 0 since otherwise we can consider f(x+a)−a instead of f(x). A contradiction will now be obtained from Hurwitz’s theorem (cf. [19, Lemma 2]).

More precisely, choose 0 < t < 1 such that h(x) ≠ 0 for |x| = t. For large n we then have μ(h, B(0, t), 0) = μ(f_{n}, B(0, t), 0) = μ(f, B(0, r_{n}t), 0). Here μ(h, B(0, t), 0) denotes the topological degree. Thus

μ(h, B(0, t), 0) = ∑_{x∈h^{−1}(0)∩B(0,t)} i(x, h)

where i(x, h) is the topological index. But μ(f, B(0, r_{n}t), 0) → ∞ as n → ∞ since f has infinitely many zeros. This is a contradiction.

The next lemma is a simple consequence of Lemma 3.3.

Lemma 3.4. Let f: R^{d} → R^{d} be quasiregular with an essential singularity at ∞. Then

lim_{r→∞} \frac{log M(r, f)}{log r} = ∞.

The following lemma (see [25, Lemma 1.3.14] or [26, Lemma 2.1.5]) replaces the argument where Rouché’s theorem was used in the proof of Theorem A.

Lemma 3.5. Let U ⊂ R^{d} be a domain and let a ∈ R^{d} and r > 0 be such that \overline{U} ⊂ B(a, r). Suppose that h: U → B(a, r) is proper and quasiregular. Then h has a fixed point in U.

Next we shall need the following result.

Lemma 3.6. Let K > 1, let D ⊂ R^{d} a domain and let C be a compact subset of D. Then there exist α, β > 0 with the following property: if f is K -quasiregular in D and satisfies |f(x)| ≥ 1 for all x ∈ D, then log |f(y)| ≤ α + β log |f(x)| for all x, y ∈ C.

Proof. It follows from [23, Corollary 3.9, p. 91] that there exist A, B > 0 such that if B(a, 2δ) ⊂ D, then log |f(x)| ≤ A + B log |f(a)| for all x ∈ B(a, δ). We may assume that C is connected. Since C is compact there exist a_{1}, …, a_{N} ∈ C and δ > 0 with

C ⊂ ⋃_{j=1}^{N} B(a_{j}, \frac{1}{2}δ) and ⋃_{j=1}^{N} B(a_{j}, 2δ) ⊂ D.

The conclusion follows with β := B^{N+1} and some α.

Finally we need the following observation which seems to have been made first in [14, Lemma 3].

Lemma 3.7. Let A, B be sets and let f: A → B and g: B → A be functions. Then the set of fixed points of f ∘ g and the set of fixed points of g ∘ f have the same cardinality.

To prove this lemma we only have to observe that g is a bijection from the set of fixed points of f ∘ g to the set of fixed points of g ∘ f.

3.2. Proof of Theorem 1. Let (c_{n}) be a sequence in R^{d} which tends to ∞ and define F_{n}(x) := f(|c_{n}|x)/|c_{n}| and G_{n}(x) := g(|c_{n}|x)/|c_{n}|. Lemma 3.4 yields that M(r, F_{n}) → ∞ and M(r, G_{n}) → ∞ as n → ∞ if r > 0, while F_{n}(0) → 0 and G_{n}(0) → 0. Thus no subsequence of (F_{n}) or (G_{n}) is normal at 0.

We distinguish between two cases.

Case 1. For every choice of (c_{n}) the sequences (F_{n}) and (G_{n}) are both quasinormal.

We may choose the sequence (c_{n}) such that |g(c_{n})| ≤ 1 for all n. Applying Lemma 3.2 and, passing to subsequences if necessary, we may assume that F_{n} → ∞ in R^{d}\E_{f} and that G_{n} → ∞ in R^{d}\E_{g} for two finite sets E_{f} and E_{g} containing 0. Moreover, we may assume that E_{g} contains at least one point b ∈ R^{d} with |b| = 1, with no subsequence of (G_{n}) converging in a neighborhood of b.

We choose ε > 0 such that 2ε < |a−b| for all a ∈ E_{g}\{b} and 2ε < |a| for all a ∈ E_{f}\{0}. For sufficiently large n we then have |G_{n}(x)| > 2 for x ∈ ∂B(b, ε) and |F_{n}(x)| > 2 for x ∈ ∂B(0, ε), while |G_{n}(y)| < 1 for some y ∈ B(b, ε) and |F_{n}(z)| < 1 for some z ∈ B(0, ε). Since B(0, ε)∪B(b, ε) ⊂ B(0, 2) this implies that G_{n} has an island V ⊂ B(b, ε) over B(0, ε) while F_{n} has an island U ⊂ B(0, ε) over B(b, ε). As in the proof of Theorem A we find that V ∩ G_{n}^{−1}(U) contains a component W of (F_{n} ∘ G_{n})^{−1}(B(b, ε)) satisfying \overline{W} ⊂ B(b, ε). Lemma 3.5 now implies that F_{n} ∘ G_{n} has a fixed point ξ ∈ B(b, ε) and thus f ∘ g has a fixed point ξ_{n} ∈ B(|c_{n}|b, |c_{n}|ε). It follows that f ∘ g has infinitely many fixed points.

Case 2. The sequence (c_{n}) can be chosen such that one of the sequences (F_{n}) and (G_{n}) is not quasinormal.

Because of Lemma 3.7 we may assume that the sequence (F_{n}) is not quasinormal. Passing to a subsequence if necessary, we may in fact assume that no subsequence of (F_{n}) is quasinormal.

As in the proof of Theorem A we may assume that |c_{n}| ≥ |g(0)| and define r_{n} by M(r_{n}, g) = |c_{n}|. As there we also define f_{n}(x) := f(|c_{n}|x)/r_{n} = F_{n}(x)|c_{n}|/r_{n} and g_{n}(x) := g(r_{n}x)/|c_{n}|. Again we find that no subsequence of (f_{n}) is normal at 0.

We now show that (f_{n}) is not quasinormal. To do this we assume that (f_{n}) is quasinormal. Passing to a subsequence we then may assume that f_{n} → ∞ in R^{d}\E for some finite set E. Let C ⊂ R^{d}\E be a compact set containing ∂B(0, r) for some r > 0. Then there exists a domain D ⊃ C such that |f_{n}(x)| ≥ 1 for x ∈ D if n is large. Lemma 3.6 yields that

(3.1)  log M(r, f_{n}) ≤ α + β log |f_{n}(x)|

for x ∈ C and large n. On the other hand, Lemma 3.4 implies that

log M(r, f_{n}) = log M(|c_{n}|r, f) − log r_{n} ≥ 4β log(|c_{n}|r) − log r_{n}

if n is large. Lemma 3.4 also yields β log |c_{n}| = β log M(r_{n}, g) ≥ log r_{n} for large n. Thus log M(r, f_{n}) ≥ 3β log |c_{n}|+4β log r ≥ 2β log |c_{n}| for large n. We deduce from this and (3.1) that

log |f_{n}(x)| ≥ \frac{log M(r, f_{n}) − α}{β} ≥ 2 log |c_{n}| − \frac{α}{β} ≥ log |c_{n}|

for x ∈ C and large n. It follows that

log |F_{n}(x)| = log |f_{n}(x)| − log |c_{n}| + log r_{n} ≥ log r_{n}

for x ∈ C and large n. Hence F_{n} → ∞ in R^{d}\E, contradicting the assumption that no subsequence of (F_{n}) is quasinormal. Thus (f_{n}) is not quasinormal. Passing to a subsequence if necessary we may assume that no subsequence of (f_{n}) is quasinormal.

From Lemma 3.3 we deduce that M(r, g_{n}) → 0 if r < 1 and M(r, g_{n}) → ∞ if r > 1. Lemma 3.2 now implies that no subsequence of (g_{n}) is quasinormal.

Let K be such that f and g are K -quasiregular and define p := 2q − 1 ≥ q where q = q(d, K) is the Rickman constant. Passing to subsequence if necessary we may assume that there exist a_{1}, …, a_{p} ∈ R^{d}\{0} where no subsequence of (f_{n}) is normal and that there exist b_{1}, …, b_{p} ∈ R^{d}\{0} where no subsequence of (g_{n}) is normal.

It follows from Lemma 3.1 that there exists ε > 0 such that if n is sufficiently large and j ∈ {1, …, p}, then f_{n} has an island in B(a_{j}, ε) over at least p − q + 1 of the p balls B(b_{k}, ε), and g_{n} has an island in B(b_{j}, ε) over at least p − q + 1 of the p balls B(a_{k}, ε).

This implies that there exists k ∈ {1, …, p} such that f_{n} has an island in B(a_{j}, ε) over B(b_{k}, ε) for at least p − q + 1 = q values of j. Lemma 3.1 implies that for at least one such value j the function g_{n} has an island in B(b_{k}, ε) over B(a_{j}, ε).

Thus we obtain j, k ∈ {1, …, p} such that f_{n} has an island U ⊂ B(a_{j}, ε) over B(b_{k}, ε) and g_{n} has an island V ⊂ B(b_{k}, ε) over B(a_{j}, ε). As before we find that V ∩ g_{n}^{−1}(U) contains a component W of (f_{n} ∘ g_{n})^{−1}(B(b_{k}, ε)) satisfying \overline{W} ⊂ B(b_{k}, ε). Lemma 3.5 now implies that f_{n} ∘ g_{n} has a fixed point in B(b_{k}, ε), and thus f ∘ g has a fixed point in B(|c_{n}|b_{k}, |c_{n}|ε). Thus f ∘ g has infinitely many fixed points.

4. Proof of Theorem 2

4.1. Preliminary lemmas. We shall require an additional result from the Ahlfors theory. An island U of a function f over a domain V is called simple if f: U → V is univalent. The following result can also be found in the references given in Section 2.1.

Lemma 4.1. Let D ⊂ C be a domain and let D_{1} , D_{2} and D_{3} be Jordan domains with pairwise disjoint closures. Let ℱ be a non-normal family of functions holomorphic in D. Then there exists a function f ∈ ℱ which has a simple island over D_{1} , D_{2} or D_{3} .

It follows from Lemma 4.1 that a non-constant entire function f has a simple island over one of three Jordan domains D_{1}, D_{2}, D_{3} with pairwise disjoint closures. We shall need the simple observation made in [7] that we need only two domains D_{1}, D_{2} if f is a polynomial or, more generally, a proper holomorphic map whose range contains D_{1} and D_{2} .

Although the formulation of the following Lemma 4.2 was slightly different in [7, Lemma 2.2], we omit the simple proof based on the Riemann–Hurwitz formula, but note that it is analogous to the proof of Lemma 4.3 which we will give below.

Lemma 4.2. Let f: U → V be a proper holomorphic map and let D_{1} and D_{2} be Jordan domains with disjoint closures contained in V . Then there exist two domains U_{1}, U_{2} ⊂ U which are simple islands over D_{1} or D_{2} .

Here U_{1} and U_{2} need not be islands over the same domain. We allow the possibility that U_{1} is an island over D_{1} and U_{2} is an island over D_{2} , or vice versa. For example, this will always be the case if f is univalent. For proper maps of higher degree, however, we have the following lemma.

Lemma 4.3. Let f: U → V be a proper holomorphic map of degree at least 2 and let D_{1} , D_{2} and D_{3} be Jordan domains with pairwise disjoint closures contained in V . Then there exists k ∈ {1, 2, 3} such that f has two simple islands over D_{k}.

Proof. Let U_{1}, …, U_{m} be the components of f^{−1}(⋃_{k=1}^{3} D_{k}). Thus the U_{j} are the islands over the domains D_{k}. Now f|_{U_{j}} is a proper map of some degree μ_{j} and ∑_{j=1}^{m} μ_{j} = 3d, where d is the degree of f. By the Riemann–Hurwitz formula the number of critical points contained in U_{j} is μ_{j} − 1, and f has d − 1 critical points in U. Thus

3d − m = ∑_{j=1}^{m}(μ_{j} − 1) ≤ d − 1

so that m ≥ 2d + 1. Since f has d − 1 critical points in U we conclude that the number n of domains U_{j} which do not contain a critical point satisfies

n ≥ m − (d − 1) ≥ (2d + 1) − (d − 1) = d + 2 ≥ 4.

Thus among the U_{j} there are at least 4 simple islands, and hence two of them must be over the same domain D_{k}.

We shall also use the following well-known result; see, e.g., [7, Lemma 2.3] for the simple proof.

Lemma 4.4. Let 0 < δ < \frac{1}{2}ε and let U ⊂ B(a, δ) be a simply-connected domain. Let f: U → B(a, ε) be holomorphic and bijective. Then f has a fixed point ξ in U which satisfies |f′(ξ)| ≥ ε/4δ.

We shall also need the following lemma concerning entire functions of small growth.

Lemma 4.5. Let g be an entire function of the form

g(z) = ∏_{k=1}^{∞}(1 − \frac{z}{z_{k}} )

where 0 < |z_{1}| ≤ |z_{2}| ≤ ⋯ and lim_{k→∞} |z_{k+1}/z_{k}| = ∞. Denote the zeros of g′ by z′_{k}, ordered such that 0 ≤ |z′_{1}| ≤ |z′_{2}| ≤ · · ·. Then lim_{k→∞} |z′_{k+1}/z′_{k}| = ∞.

Proof. For sufficiently large n there exists r satisfying |z_{n}| ≤ r and 8r ≤ |z_{n+1}|. We show first that for such r

(4.1)  min_{|z|=4r} |g(z)| > max_{|z|=r} |g(z)|,

provided n is large enough. Let |u| = r and |v| = 4r. We will show that |g(v)| > |g(u)|. To this end we write

log \frac{|g(v)|}{|g(u)|} = ∑_{k=1}^{∞} log \frac{|v − z_{k}|}{|u − z_{k}|} = ∑_{k=1}^{n} log \frac{|v − z_{k}|}{|u − z_{k}|} + ∑_{k=n+1}^{∞} log \frac{|v − z_{k}|}{|u − z_{k}|}
= S_{1} + S_{2}.

For k ≤ n we have |z_{k}| ≤ r < |v| so that

\frac{|v − z_{k}|}{|u − z_{k}|} ≥ \frac{|v| − |z_{k}|}{|u| + |z_{k}|} ≥ \frac{4r − r}{r + r} = \frac{3}{2}.

Thus S_{1} ≥ n log \frac{3}{2} .

For k ≥ n + 1 we have |z_{k}| ≥ 8r > |v| so that

log \frac{|v − z_{k}|}{|u − z_{k}|} ≥ log \frac{|z_{k}| − |v|}{|u| + |z_{k}|} = log(1 − \frac{4r}{|z_{k}|}) − log(1 + \frac{r}{|z_{k}|}) ≥ −\frac{8r}{|z_{k}|} − \frac{r}{|z_{k}|}.

Here we have used the inequalities log(1 + x) ≤ x and log(1 − x) ≥ −2x valid for 0 ≤ x ≤ \frac{1}{2} . For large n we also have |z_{k+1}/z_{k}| ≥ 2 if k ≥ n + 1. We find that

log \frac{|v − z_{k}|}{|u − z_{k}|} ≥ −\frac{9r}{|z_{k}|} ≥ −\frac{9r}{|z_{n+1}|}2^{n+1−k} ≥ −\frac{9}{8}2^{n+1−k}

for k ≥ n + 1. It follows that

S_{2} ≥ −\frac{9}{8} ∑_{k=n+1}^{∞} 2^{n+1−k} = −\frac{9}{4}.

Together with the estimate for S_{1} this implies that (4.1) holds for large n.

Let now U be the component of g^{−1}(B(0, M(r, g))) which contains B(0, r). It follows from (4.1) that U ⊂ B(0, 4r). By our choice of r the number of zeros of g in U is n. The Riemann–Hurwitz formula yields that g′ has n − 1 zeros in U. Thus g′ has at most n − 1 zeros in B(0, r) and at least n − 1 zeros in B(0, 4r). Thus r ≤ |z′_{n}| and |z′_{n−1}| ≤ 4r. Since this holds for any r satisfying |z_{n}| ≤ r and 8r ≤ |z_{n+1}| we conclude, choosing r = |z_{n}| or r = \frac{1}{8}|z_{n+1}|, that |z′_{n−1}| ≤ 4|z_{n}| and \frac{1}{8}|z_{n+1}| ≤ |z′_{n}| for large n. Thus \frac{1}{8}|z_{n+1}| ≤ |z′_{n}| ≤ 4|z_{n+1}| for large n. The conclusion follows.

The following result is a variant of Lemma 3.7.

Lemma 4.6. Let f and g be entire transcendental functions. Suppose that there exists a sequence (ξ_{n}) such that (f ∘ g)(ξ_{n}) = ξ_{n} and (f ∘ g)′(ξ_{n}) → ∞. Then η_{n} := g(ξ_{n}) satisfies (g ∘ f)(η_{n}) = η_{n} and (g ∘ f)′(η_{n}) = (f ∘ g)′(ξ_{n}) → ∞.

The proof is straightforward and thus omitted.

4.2. Proof of Theorem 2. We proceed as in the proof of Theorem A and define the sequences (c_{n}), (r_{n}), (f_{n}) and (g_{n}) as there. Again we find that (g_{n}) is not quasinormal in B(0, 2). In the proof of Theorem A we noted that by passing to a subsequence we can achieve that no subsequence of (g_{n}) is normal at any of three points b_{1}, b_{2}, b_{3} ∈ C\{0}. The same argument yields this for any number of points b_{k}. Moreover, we can achieve that these points are in B(0, 2)\{0}.

We will have to distinguish several cases now.

Case 1. It is possible to choose the sequence (c_{n}) such that (f_{n}) is not quasinormal of order 2.

Passing to a subsequence if necessary we may assume that there exist a_{1}, a_{2}, a_{3} ∈ C where no subsequence of (f_{n}) is normal and that there exist b_{1}, …, b_{7} ∈ C\{0} where no subsequence of (g_{n}) is normal.

We choose 0 < ε < \frac{1}{2} such that the closures of the disks B(a_{j}, ε) are pairwise disjoint. Moreover, we require that the closures of the disks B(b_{j}, ε) are pairwise disjoint and do not contain 0. We also choose 0 < δ < \frac{1}{2}ε.

It follows from Lemma 4.1 that if n is sufficiently large and j ∈ {1, 2, 3}, then f_{n} has a simple island in B(a_{j}, ε) over at least five of the seven disks B(b_{k}, ε). Overall we obtain at least 15 domains in the union of the three disks B(a_{j}, ε) which are simple islands over one of the seven disks B(b_{k}, ε). This implies that there exists k ∈ {1, …, 7} such that f_{n} has three simple islands U_{j} ⊂ B(a_{j}, ε), j ∈ {1, 2, 3}, over the same disk B(b_{k}, ε). It also follows from Lemma 2.1 that if n is sufficiently large, then there exists j ∈ {1, 2, 3} such that g_{n} has a simple island V ⊂ B(b_{k}, δ) over B(a_{j}, ε). Then W := V ∩ g_{n}^{−1}(U_{j}) is a simple island of f_{n} ∘ g_{n} over B(b_{k}, ε), and W ⊂ B(b_{k}, δ). Lemma 4.4 implies that f_{n} ∘ g_{n} has a fixed point ξ ∈ W with |(f_{n} ∘ g_{n})′(ξ)| ≥ ε/4δ. Then ξ_{n} := r_{n}ξ is a fixed point of f ∘ g with |(f ∘ g)′(ξ_{n})| ≥ ε/4δ. Since δ can be chosen arbitrarily small the conclusion follows.

Case 2. For every choice of the sequence (c_{n}) the sequence (f_{n}) is quasinormal of order 2.

Then f_{n} → ∞ in C\{0, 1} by Lemma 2.2. Let K > 2. For sufficiently large n we then have |f_{n}(z)| > 2 if \frac{1}{4} ≤ |z| ≤ K and |z − 1| ≥ \frac{1}{4} .

We now distinguish two subcases.

Case 2.1. There are infinitely many n such that B(1, \frac{1}{4}) contains at least two zeros of f_{n}.

Passing to a subsequence if necessary we may assume that this holds for all n. We note that if n is sufficiently large, then B(0, \frac{1}{4}) also contains at least two zeros of f_{n}. With a_{1} := 0 and a_{2} := 1 thus both disks B(a_{j}, \frac{1}{4}) contain at least two zeros of f_{n}.

We may assume that no subsequence of (g_{n}) is normal at five points b_{1}, …, b_{5} ∈ B(0, 2)\\{0\}. Again we choose 0 < ε < \frac{1}{2} such that the closures of the disks B(b_{k}, ε) are pairwise disjoint and do not contain 0, and we choose 0 < δ < \frac{1}{2}ε. We can now deduce from Lemma 2.1 that if n is sufficiently large and k ∈ {1, …, 5}, then g_{n} has an island over one of the disks B(a_{j}, \frac{1}{4}) in B(b_{k}, δ). It follows that for at least three values of k the function g_{n} has an island over the same disk B(a_{j}, \frac{1}{4}) in B(b_{k}, δ). We may assume that the b_{k} are numbered such that this holds for k ∈ {1, 2, 3}.

We claim that there exists k ∈ {1, 2, 3} such that f_{n} has two simple islands U_{1}, U_{2} in B(a_{j}, \frac{1}{4}) over B(b_{k}, ε). To this end we assume first that there are two components X_{1}, X_{2} of f_{n}^{−1}(B(0, 2)) contained in B(a_{j}, \frac{1}{4}) in which f_{n} is univalent. Then we can simply take U_{l} := f_{n}^{−1}(B(b_{k}, ε)) ∩ X_{l}, for l ∈ {1, 2} and arbitrary k ∈ {1, 2, 3}. Suppose now that such components X_{1}, X_{2} do not exist. Since |f_{n}(z)| > 2 if |z − a_{j}| = \frac{1}{4} and since f_{n} has at least two zeros in B(a_{j}, \frac{1}{4}), there now exists a component X of f_{n}^{−1}(B(0, 2)) contained in B(a_{j}, \frac{1}{4}) such that f: X → B(0, 2) is a proper map of degree at least 2. Now Lemma 4.3 yields our claim that there exists k ∈ {1, 2, 3} such that f_{n} has two simple islands U_{1}, U_{2} ⊂ X ⊂ B(a_{j}, \frac{1}{4}) over B(b_{k}, ε).

Recall that in turn g_{n} has an island V ⊂ B(b_{k}, δ) over B(a_{j}, \frac{1}{4}). Lemma 4.3, applied to the proper map g_{n}: V → B(a_{j}, \frac{1}{4}), now implies that V contains a domain W which is a simple island of g_{n} over U_{1} or U_{2} . Then W is a simple island of f_{n} ∘ g_{n} over B(b_{k}, ε), and W ⊂ V ⊂ B(b_{k}, δ). As before Lemma 4.4 implies that f_{n} ∘ g_{n} has a fixed point ξ ∈ W with |(f_{n} ∘ g_{n})′(ξ)| ≥ ε/4δ. Again ξ_{n} := r_{n}ξ is a fixed point of f ∘ g with |(f ∘ g)′(ξ_{n})| ≥ ε/4δ, and the conclusion follows since δ can be chosen arbitrarily small.

Case 2.2. For all sufficiently large n the disk B(1, \frac{1}{4}) contains at most one zero of f_{n}.

Since |f_{n}(z)| > 2 for \frac{1}{4} ≤ |z| ≤ K and |z − 1| ≥ \frac{1}{4} , provided n is sufficiently large, we conclude that the annulus R := \{z ∈ C : \frac{1}{4} ≤ |z| ≤ K\} contains at most one zero of f_{n}. On the other hand, since f_{n}(1) → 0 we conclude that B(1, \frac{1}{4}) contains a component Y of f_{n}^{−1}(B(0, 2)) and thus in particular a zero of f_{n} for large n. Thus f_{n} has exactly one zero in R if n is sufficiently large. In fact, f_{n} takes every value in B(0, 2) exactly once in Y , and since there are no other components of f_{n}^{−1}(B(0, 2)) intersecting R, we see that f_{n} takes every value in B(0, 2) exactly once in R. We find that if z_{1} ∈ Y and z_{2} ∈ C with |z_{2}| ≥ |z_{1}| and f_{n}(z_{2}) = f_{n}(z_{1}) ∈ B(0, 2), then |z_{2}| ≥ K while |z_{1}| ≤ \frac{5}{4} so that |z_{2}| ≥ \frac{4}{5}K|z_{1}|.

In terms of f we see that f has infinitely many c-points for every c ∈ C\{0}, and the sequence (w_{n}) of c-points of f, arranged such that |w_{n+1}| ≥ |w_{n}|, satisfies |w_{n+1}| ≥ \frac{4}{5}K|w_{n}|. Since K can be chosen arbitrarily large we deduce that lim_{n→∞} |w_{n+1}/w_{n}| = ∞. It follows that the number n(r, 1/(f − c)) of c-points of f in \overline{B(0, r)} satisfies n(r, 1/(f − c)) = o(log r) as r → ∞, for any c ∈ C\\{0\}. Standard estimates from value distribution theory ([17], [18], [21]) now imply that log M(r, f) = o((log r)^{2}) as r → ∞. In particular, f has order 0.

Lemma 4.6 says that the conclusion is symmetric with respect to f and g. We may thus assume that interchanging the roles of f and g leads again to Case 2.2. We find that g is also of order 0 and that the sequence (z_{n}) of zeros of g satisfies lim inf_{n→∞} |z_{n+1}/z_{n}| = ∞.

Assuming without loss of generality that g(0) ≠ 0 we deduce from Lemma 4.5 that the sequence (z′_{n}) of zeros of g′ satisfies

(4.2)  lim_{n→∞} \frac{|z′_{n+1}|}{|z′_{n}|} = ∞.

Proceeding as before we may assume, passing to a subsequence if necessary, that no subsequence of (g_{n}) is normal at four points b_{1}, …, b_{4} ∈ B(0, 2)\{0}. As before we choose 0 < ε < \frac{1}{2} such that the closures of the disks B(b_{j}, ε) are pairwise disjoint and do not contain 0, and we choose 0 < δ < \frac{1}{2}ε.

With a_{1} := 0 and a_{2} := 1 we can deduce as before from Lemma 2.1 that if n is sufficiently large and k ∈ {1, …, 4}, then g_{n} has an island V_{k} ⊂ B(b_{k}, δ) over one of the disks B(a_{j}, \frac{1}{4}). It follows from (4.2) that at most one of these four islands contains a zero of g′_{n}, provided n is large enough. Thus V_{k} is a simple island for at least three values of k. Hence there exist two values of k such that V_{k} is a simple island over the same disk B(a_{j}, \frac{1}{4}). We may assume that the b_{k} are numbered such that this holds for k ∈ {1, 2}. By Lemma 4.2 the function f_{n} has a simple island U ⊂ B(a_{j}, \frac{1}{4}) over B(b_{1}, ε) or B(b_{2}, ε). Without loss of generality we can assume that U is a simple island over B(b_{1}, ε). Then W := V_{1} ∩g_{n}^{−1}(U) is a simple island of f_{n} ∘ g_{n} over B(b_{1}, ε), and W ⊂ B(b_{1}, δ). As before we deduce that f_{n} ∘ g_{n} has a fixed point ξ ∈ W with |(f_{n} ∘ g_{n})′(ξ)| ≥ ε/4δ. Again ξ_{n} := r_{n}ξ is a fixed point of f ∘ g with |(f ∘ g)′(ξ_{n})| ≥ ε/4δ so that the conclusion follows also in this case.

5. Proof of Theorem 3

5.1. Preliminary lemmas. We shall need the following quantitative version of Lemma 2.1.

Lemma 5.1. Let D_{1}, D_{2} ⊂ C be Jordan domains with disjoint closures and let f: B(a, r) → C be a holomorphic function which has no island over D_{1} or D_{2} . Then

\frac{|f′(a)|}{2μ(log μ + A)} ≤ \frac{1}{r}

where μ = max{1, |f(a)|} and A is a constant depending only on the domains D_{1} and D_{2} .

While Lemma 2.1 says that the family of holomorphic functions having no islands over any of two given domains is normal, Lemma 5.1 is based on the fact that this family is in fact a normal invariant family. Lemma 5.1 follows from Lemma 2.1 together with results of Hayman [16] on normal invariant families. It is a direct consequence of Theorems 6.8, 6.6, and 5.5 of his book [17].

We also need the following version of a classical growth lemma due to Borel.

Lemma 5.2. Let r_{0} > 0 and let T: [r_{0}, ∞) → [e, ∞) be increasing and continuous. Define

F := \{r ≥ r_{0} : T (r(1 + \frac{1}{(log T (r))^{2}} )) > eT (r)\}.

Then

∫_{F} \frac{dt}{t} ≤ \frac{π^{2}}{6} .

Proof. We may assume that F ≠ ∅ and define

r_{1} := inf F, r′_{1} := r_{1}(1 + \frac{1}{(log T (r_{1}))^{2}} )

and then inductively

r_{k} := inf(F ∩ [r′_{k−1}, ∞)), r′_{k} := r_{k}(1 + \frac{1}{(log T (r_{k}))^{2}} ).

If F ∩[r′_{k}, ∞) = ∅ for some k so that the process terminates, then we put N := k. Otherwise we put N := ∞.

We have T (r_{k}) ≥ T (r′_{k−1}) ≥ eT (r_{k−1}). Hence T (r_{k}) ≥ e^{k−1}T (r_{1}) ≥ e^{k} so that (log T (r_{k}))^{2} ≥ k^{2} . If N = ∞ we thus have r_{k} → ∞ as k → ∞. In any case we find that

F ⊂ ⋃_{k=1}^{N}[r_{k}, r′_{k}]

so that

∫_{F} \frac{dt}{t} ≤ ∑_{k=1}^{N} log \frac{r′_{k}}{r_{k}} = ∑_{k=1}^{N} log(1 + \frac{1}{(log T (r_{k}))^{2}} ).

Since log(1 + x) < x for x > 0 this yields

∫_{F} \frac{dt}{t} ≤ ∑_{k=1}^{N} \frac{1}{(log T (r_{k}))^{2}} ≤ ∑_{k=1}^{∞} \frac{1}{k^{2}} = \frac{π^{2}}{6} .

5.2. Proof of Theorem 3. We proceed as in the proof of Theorem A and define the sequences (c_{n}), (r_{n}), (f_{n}) and (g_{n}) and the points a_{1}, a_{2} and b_{1}, b_{2}, b_{3} as there. If we could choose all three points b_{1}, b_{2}, b_{3} nonreal, then the argument given there would imply that f ∘ g has infinitely many nonreal fixed points. We may thus assume that (g_{n}) is quasinormal of order 2 in C\R and thus g_{n} → 0 locally uniformly in C\R by Lemma 2.2.

If there exist three annuli Ω_{j} := {z ∈ C : S_{j} < |z| < T_{j}} with disjoint closures such that g_{n} has an island over B(a_{1}, ε) or B(a_{2}, ε) in Ω_{j}\R, for all j ∈ {1, 2, 3} and infinitely many n, then the argument used in the proof of Theorem A shows again that f_{n} ∘ g_{n} has a fixed point in one of these islands, and thus f ∘ g has infinitely many nonreal fixed points.

We may thus assume that such annuli Ω_{j} do not exist. Passing to a subsequence if necessary we thus find an annulus Ω := {z ∈ C : S − 1 < |z| < T + 1} with S > 1 and T > e^{2}S such that no g_{n} has an island over B(a_{1}, ε) or B(a_{2}, ε) in Ω\R. Since g_{n} → 0 in C\R we may assume that |g_{n}(z)| ≤ 1 for S ≤ |z| ≤ T and | Im z| ≥ 1.

To save indices, we now write h := g_{n}. For S ≤ |z| ≤ T and | Im z| ≤ 1 we conclude from Lemma 5.1 that

\frac{|h′(z)|}{2μ(log μ + A)} ≤ \frac{1}{| Im z|}

where μ = max{1, |h(z)|} and A is a constant. We may assume that A > 1. If S ≤ r ≤ T and log |h(re^{it})| ≥ A, then |h(re^{it})| ≥ e^{A} > 1 so that | Im(re^{it})| = r| sin t| ≤ 1 and thus

(5.1)  \frac{|h′(re^{it})|}{|h(re^{it})| log |h(re^{it})|} ≤ \frac{2|h′(re^{it})|}{|h(re^{it})|(log |h(re^{it})| + A)} ≤ \frac{4}{r| sin t|}.

We denote by T (r, h) the Nevanlinna characteristic of h and recall the inequality

(5.2)  log M(r, h) ≤ \frac{R + r}{R − r} T (R, h)

valid for 0 < r < R. Since M(r, g_{n}) → ∞ as n → ∞ if r > 1 we conclude that T (r, h) = T (r, g_{n}) → ∞ for r > 1. In particular we may assume that T (r, h) ≥ 3A > 3 for r ≥ S.

Choosing

R := r(1 + \frac{1}{(log T (r))^{2}} )

in (5.2) and noting that log(T/S) > 2 > π^{2}/6 we deduce from Lemma 5.2 that there exists r ∈ [S, T] such that

log M(r, h) ≤ (2 + \frac{1}{(log T (r, h))^{2}} )(log T (r, h))^{2}eT (r, h).

Now (log T (r, h))^{2} ≥ (log 3A)^{2} ≥ 1 and hence

(5.3)  log M(r, h) ≤ 3eT (r, h)(log T (r, h))^{2}.

For a value of r satisfying (5.3) we consider the set

E(r) := \{t ∈ [0, 2π] : log |h(re^{it})| ≥ \frac{1}{2}T (r, h)\}

and define λ(r) := meas E(r), where meas E denotes the measure of a set E. Then for at least one of the four sets E_{l}(r) := E(r) ∩ [l \frac{1}{2}π, (l + 1)\frac{1}{2}π], where l ∈ \{0, 1, 2, 3\}, we have meas E_{l}(r) ≥ \frac{1}{4}λ(r). We assume now that this is the case for l = 0. The modifications that have to be made for the other cases will be obvious. We define

α(r) := max E_{0}(r)

and

β(r) := min\{t ∈ [α(r), \frac{1}{2}π] : log |h(re^{it})| = A\}.

Then α(r) ≥ meas E_{0}(r) ≥ \frac{1}{4}λ(r). For α(r) ≤ t ≤ β(r) we have 1 < A ≤ log |h(re^{it})| ≤ \frac{1}{2}T (r, h). Thus log log h may be defined on the arc \{re^{it} : α(r) ≤ t ≤ β(r)}, and we may choose the branch of the logarithm such that |log h(re^{iβ(r)})| ≤ log|h(re^{iβ(r)})| + π = A + π and hence log|log h(re^{iβ(r)})| ≤ log(A + π). Thus

log log|h(re^{iα(r)})| − log(A + π) ≤ log|log h(re^{iα(r)})| − log|log h(re^{iβ(r)})|

≤ |log log h(re^{iα(r)}) − log log h(re^{iβ(r)})|

= |∫_{α(r)}^{β(r)} \frac{h′(re^{it})}{h(re^{it}) log h(re^{it})}ire^{it} dt|

≤ r ∫_{α(r)}^{β(r)} \frac{|h′(re^{it})|}{|h(re^{it})| log |h(re^{it})|} dt

≤ 4 ∫_{α(r)}^{β(r)} \frac{dt}{| sin t|}.

Here the last inequality follows from (5.1). Now sin t ≥ 2t/π for 0 ≤ t ≤ \frac{1}{2}π. Thus

log log|h(re^{iα(r)})| − log(A + π) ≤ 2π ∫_{α(r)}^{β(r)} \frac{dt}{t} = 2π log \frac{β(r)}{α(r)} ≤ 2π log \frac{π}{2α(r)}.

Since

log|h(re^{iα(r)})| = \frac{1}{2}T (r, h)

this yields

(5.4)  log T (r, h) ≤ 2π log \frac{π}{2α(r)} + log(A + π) + log 2.

Now

T (r, h) = \frac{1}{2π} ∫_{0}^{2π} log^{+} |h(re^{it})| dt

= \frac{1}{2π} ∫_{E(r)} log^{+} |h(re^{it})| dt + \frac{1}{2π} ∫_{[0,2π]\E(r)} log^{+} |h(re^{it})| dt

≤ \frac{1}{2π} λ(r) log M(r, h) + \frac{1}{2}T (r, h)

so that

\frac{λ(r)}{π} ≥ \frac{T (r, h)}{log M(r, h)}.

Hence

\frac{π}{2α(r)} ≤ \frac{2π}{λ(r)} ≤ \frac{2 log M(r, h)}{T (r, h)}.

Using (5.3) we find that

\frac{π}{2α(r)} ≤ 6e(log T (r, h))^{2}.

Together with (5.4) this yields

log T (r, h) ≤ 2π log(6e(log T (r, h))^{2}) + log(A + π) + log 2.

This implies that

log T (r, h) ≤ 4π log log T (r, h) + C

for some constant C, which is a contradiction since T (r, h) = T (r, g_{n}) → ∞ as n → ∞.

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Received 25 October 2005