On Bloch’s “Principle of topological continuity”

We discuss to what extent certain results about totally ramiﬁed values of entire and meromorphic functions remain valid if one relaxes the hypothesis that some value is totally ramiﬁed by assuming only that all islands over some Jordan domain are multiple. In particular, we prove a result suggested by Bloch which says that an entire function of order less than 1 has a simple island over at least one of two given Jordan domains with disjoint closures.


Introduction
About one hundred years ago, André Bloch [13] wrote a paper consisting mainly of heuristic speculations based on two philosophical principles.In this paper, he anticipated several important results of 20th century geometric function theory -and even today reading this paper can still be rewarding.
The first of these principles Bloch phrased as "Nihil est in infinito quod non prius fuerit in finito".This principle has often been interpreted as meaning that if all entire functions with a certain property are constant, then the family of functions which are holomorphic in some domain and have this property is normal.This is an important guideline in the theory of normal families; see [9,31,32] for a discussion.We will call it Bloch's normal family principle in the sequel.Another interpretation of "Nihil est in infinito quod non prius fuerit in finito" given by Bloch is that features of transcendental entire and meromorphic functions are in some form already present in polynomials and rational functions.
The second principle that Bloch discusses he calls the principle of topological continuity.It says that certain true statements remain true if one modifies the data from a metric point of view, but not from the topological point of view.
To illustrate this principle, he quotes the following theorem.
Theorem A. Let D be a domain and let a 1 , a 2 , a 3 ∈ C be distinct.Let F be the family of all functions holomorphic in D which do not have a simple a j -point, for all j ∈ {1, 2, 3}.Then F is normal.
He then argues [13, p. 87] that the principle of topological continuity should give the following result.
Theorem A ′ .Let D be a domain and let D 1 , D 2 and D 3 be three disks in C which have pairwise disjoint closures.Let F be the family of all functions holomorphic in D such that there does not exist a domain U with U ⊂ D such that U is mapped univalently onto one of the disks D j .Then F is normal.
Theorem A was known when Bloch wrote this.Theorem A ′ was not known then.It was first proved by Ahlfors six years later, see [1,2,3,4], his definite account being [5].Theorem A ′ , and its generalization Theorem D ′ below, are among the principal results of his theory of covering surfaces which earned him one of the two first Fields medals in 1936.
Ahlfors did not use Theorem A in his proof of Theorem A ′ , nor do the more recent proofs in [16,17,27].Ahlfors does, however, give an interesting discussion of Bloch's principle of topological continuity in [3, pp. 202-203].A deduction of Theorem A ′ from Theorem A -and thus in some sense a confirmation of Bloch's principle of topological continuity -was given in [8].This deduction was based on a rescaling principle of Zalcman [31].
Zalcman's lemma has become a major tool in the theory of normal families by giving a rigorous formulation of Bloch's normal family principle.For example, it shows that Theorems A and A ′ can be deduced from the following corresponding results about entire functions.
Theorem B. Let f be entire and let a 1 , a 2 , a 3 ∈ C be distinct.Suppose that all a j -points are multiple, for all j ∈ {1, 2, 3}.Then f is constant.Theorem B ′ .Let f be entire and let D 1 , D 2 and D 3 be three disks in C which have pairwise disjoint closures.Suppose that there does not exist j ∈ {1, 2, 3} and a bounded domain U in C which is mapped univalently onto D j .Then f is constant.
As a second example for his principle of topological continuity Bloch considers the following result.
Theorem C. Let f : C → C be a non-constant entire function of order less than 1 and let a 1 , a 2 ∈ C be distinct.Then there exists j ∈ {1, 2} such that f has a simple a j -point.
This theorem can be proved using Nevanlinna theory, and Bloch was aware of this proof [12].We will sketch a proof after Theorem D ′ below.
Bloch argues [13, p. 88] that Theorem C together with his principle of topological continuity should yield the following result.
Theorem C ′ .Let f : C → C be a non-constant entire function of order less than 1 and let D 1 and D 2 be two disks in C with disjoint closures.Then there exists j ∈ {1, 2} and a domain U in C which is mapped univalently onto D j .
So far as we know, there is no proof of Theorem C ′ in the literature, neither by Bloch nor by someone else.We will give a proof of (a generalization of) Theorem C ′ in section 3, using some ideas of Goldberg and Tairova [18].
Theorems B and B ′ are actually special cases of more general results.The generalization of Theorem B (which Bloch also knew) is the following result due to Nevanlinna [23,Chapitre IV,no. 51].
Here m j = ∞ means that f does not take the value a j at all, and we put 1/m j = 1/∞ = 0 in this case.
To state the corresponding generalization of Theorem B ′ we introduce some terminology.Let f : C → C be meromorphic and let D be a Jordan is bounded and simply-connected.Then f : U → D is a proper mapping.The degree of this proper mapping is called the multiplicity of the island U.With this terminology we can phrase the generalization of Theorem B ′ .It is due to Ahlfors, who called it "Scheibensatz" [5, p. 190].
Theorem D ′ .Let f : C → C be meromorphic and q ∈ N. Let D 1 , . . ., D q be Jordan domains in C with pairwise disjoint closures and let m 1 , . . ., m q ∈ N ∪ {∞}.If, for all j ∈ {1, . . ., q}, all islands over D j have multiplicity at least m j , then (1.1) holds or f is constant.
Theorems B and B ′ are obtained from Theorems D and D ′ by choosing There are also normal family analogues of Theorems D and D ′ according to Bloch's normal family principle.These can be obtained from Theorems D and D ′ via Zalcman's lemma.
Suppose now that we have equality in (1.1); that is, Then, apart from permutation of the m j , we have one of the following six cases: (iv Selberg [25, Satz II] determined all transcendental meromorphic functions f of finite order such that f has only finitely many a j -points of multiplicity less than m j , with the m j chosen such that (1.2) holds.He used this to determine the possible orders of these functions [25, Satz IV-VII].Let ρ(f ) denote the order of a meromorphic function f ; see [19, Chapter 2, Section 1] for the definition of the order and other concepts from the theory of meromorphic functions used in the sequel.We briefly summarize Selberg's reasoning and results.First, in case (i) we assume that (a 1 , a 2 ) = (0, ∞) and find that the function f has the form f = Qe P with a rational function Q and a polynomial P .Thus ρ(f ) ∈ N in this case.
In case (ii) we assume that has poles only at the simple ±1-points of f .Thus R has only finitely many poles.Since f has finite order, the lemma on the logarithmic derivative [19, Chapter 3, Section 1] yields that Nevanlinna's proximity functions m(r, R) satisfies m(r, R) = O(log r).We conclude that R is a rational function.
Hence f has the form This implies that ρ(f ) is an integer multiple of 1/2.(Equation (1.4) and its implication for ρ(f ) were actually already obtained by Valiron [28, p. 77], before Selberg's work and in fact before Nevanlinna developed his theory.)If all ±1-points are multiple, then R has no poles and is thus a polynomial.In this case we find that ρ(f ) ≥ 1.Note that this proves Theorem C.
In cases (iii)-(vi), assume that all a j are in C and that M is the least common multiple of the m j .We then find that is a rational function and f has the form where E is an elliptic function.
We only state the conclusions about the order of f that Selberg drew from this.
Theorem E. Let f : C → C be a transcendental meromorphic function of finite order, q ∈ N, a 1 , . . ., a q ∈ C distinct and m 1 , . . ., m q ∈ N ∪ {∞}.Suppose that (1.2) holds so that we have one of the cases (i)-(vi) listed above.
If, for all j ∈ {1, . . ., q}, all but finitely many a j -points of f have multiplicity at least m j , then ρ(f If all (and not only all but finitely many) a j -points have multiplicity at least m j , then in addition ρ(f ) ≥ 1 in cases (i) and (ii) while ρ(f ) ≥ 2 in cases (iii)-(vi).
The question that motivated this paper is whether there exists a Theorem E ′ which corresponds to Theorem E in the same way that Theorems A ′ , B ′ , C ′ and D ′ correspond to Theorems A, B, C and D. In other words, we ask to what extent the conclusion of Theorem E remains valid if instead of the hypotheses of Theorem D we assume the hypotheses of Theorem D ′ .Note that Theorem C ′ , which was envisaged by Bloch using his principle of topological continuity, says that this holds in case (ii) if f is entire.In contrast, we will see that the corresponding result does not hold in cases (i) and (iii).
The following result corresponds to case (i) of Theorem D. Conversely, for every ρ ∈ [1/2, ∞) and every Jordan domain D there exists an entire function f of order ρ such that f has no island over D.
Our next result corresponds to case (ii) of Theorem D. It has Theorem C ′ as a corollary.Here and in the following we will denote the multiplicity of an island U over some Jordan domain by µ(U).
Theorem 1.2.Let f be a transcendental entire function of finite order and let D 1 and D 2 be Jordan domains in C with disjoint closures.Suppose that f has only finitely many simple islands over D 1 and D 2 .Let N be the number (counting multiplicity) of critical points c of f such that f where the sum is taken over all islands U over D 1 or D 2 .Then 1 ≤ p < ∞ and there exists c > 0 such that as r → ∞.In particular, ρ(f ) = p/2.
Remark.In the proof of Theorem 1.2 the hypothesis that f is entire and of finite order will be used only to conclude that f −1 has finitely many transcendental singularities.So actually we prove a more general theorem: Let U be a simply connected domain in C and let f : U → C be holomorphic.Suppose that f −1 has only finitely many transcendental singularities over C and that f has only finitely many simple islands over two Jordan domains in C with disjoint closures.Then U = C, and f is either a polynomial or an entire function satisfying (1.6).
Under the hypothesis that there exist a 1 , a 2 ∈ C such that f has only finitely many simple a j -points, the conclusions of Theorem 1.2 and the above remark were obtained by Goldberg and Tairova [18].Their proof was based on topological arguments and it probably can be extended to a proof of Theorem 1.2.
Note that we have µ(U) ≥ 1 and hence µ(U) − 2 ≥ −1 for all islands U, with µ(U) − 2 = −1 only for the at most finitely many simple islands.Thus p < ∞ implies in particular that N < ∞ and that µ(U) > 2 for at most finitely many islands U.
Note also that if all islands over D 1 and D 2 are multiple, then µ(U) ≥ 2 for all islands U and thus p ≥ 2. Thus Theorem C ′ follows from Theorem 1.2.
Theorems 1.1 and 1.2 concern entire functions.The analogous results do not hold if instead of entire functions we consider meromorphic functions which have no island over a domain containing ∞.
Theorem 1.3.Let D 1 and D 2 be Jordan domains in C with disjoint closures.Then, given ρ ∈ [0, ∞), there exists a meromorphic function f of order ρ which has no island over D 1 and D 2 .
Theorem 1.4.Let D be a Jordan domain in C and let a 1 , a 2 ∈ C \ D be distinct.Then, given ρ ∈ [0, ∞), there exists a meromorphic function f of order ρ which has no island over D and for which all a 1 -points and a 2 -points are multiple.
The functions f constructed in the proofs of Theorems 1.3 and 1.4 have the property that for some c > 0 and all large r.Question 1.1.Do we have (1.7) for every meromorphic function f satisfying the conditions of Theorems 1.3 or 1.4?
The following result says that in case (iii) there is no analogue of Theorem E if instead of multiple a j -points one considers multiple islands over certain Jordan domains.In fact, it suffices to replace a j -points by multiple islands over some Jordan domain for one j.
Theorem 1.5.Let a 1 , a 2 , a 3 ∈ C be distinct and ρ ∈ (0, ∞).Then there exists a meromorphic function f of order ρ and a Jordan domain D whose closure is contained in C \ {a 1 , a 2 , a 3 } such that all a j -points of f are multiple for j ∈ {1, 2, 3} and such that f has no simple island over D. Question 1.2.Does the conclusion of Theorem 1.5 also hold for ρ = 0? Theorem 1.5 deals with case (iii).We do not know whether there are analogous results for the cases (iv)-(vi).Question 1.3.In cases (iv)-(vi), is there an analogue of Theorem E if instead of multiple a j -points we consider multiple islands over Jordan domains D j with disjoint closures?
The answer may depend on whether we replace multiple a j -points by multiple islands over the D j for all j ∈ {1, . . ., q} or only for some j.
There exist functions of order 0 for which all but finitely many islands over D j have multiplicity m j .In fact, this is the case already in the situation of Theorem E. Given a j and m j as there there exists a function of order 0 for which all but finitely many a j -points have multiplicity at least m j .Such a function f was considered already by Teichmüller [26, p. 734].It satisfies as r → ∞, for some c > 0. This leads to the following question analogous to Question 1.1.Question 1.4.Let f : C → C be meromorphic and q ∈ N. Let D 1 , . . ., D q be Jordan domains in C with pairwise disjoint closures and let m 1 , . . ., m q ∈ N ∪ {∞} satisfy (1.2).Suppose that, for all j ∈ {1, . . ., q}, all but finitely many islands over D j have multiplicity at least m j .Does there exists c > 0 such that (1.7) holds for all large r?We will use some results about quasiconformal and quasiregular mappings.We refer to [22] for the definition and basic properties, noting that quasiregular mappings are called quasiconformal functions there.Let D be a domain and f : D → C be quasiregular.We use the notation The Hölder continuity of quasiconformal mappings [22, § II.4.2] yields the following result.
A basic result [22, § V.1] in the theory of quasiconformal mappings says that there exist quasiconformal mappings with prescribed dilatation.One consequence of this is the following result.Lemma 2.6.Let φ : C → C be quasiconformal.Suppose that Then there exists c ∈ C \ {0} such that Remark.The condition (2.1) is satisfied in particular if the set A of all z satisfying |z| > 1 where φ is not conformal satisfies We denote the open and closed disk of radius r around a point a ∈ C by D(a, r) and D(a, r).We also put D := D(0, 1).
We will use the following sufficient condition for quasisymmetry.It is surely known, but we did not find a reference.Lemma 2.7.Let h : ∂D → ∂D be an orientation-preserving homeomorphism.Suppose that there exists a finite subset A of ∂D such that h is continuously differentiable with non-zero derivative in ∂D \ A. Suppose also that for all a ∈ A there exists γ a > 0 such that |h(ae it ) − h(a)| t γa has one-sided, non-zero limits as t → 0 ± .Then h is quasisymmetric.

Sketch of proof. The restriction of h to a closed arc which contains no point of
A is clearly quasisymmetric.To see that h is quasisymmetric on an arc which has a point of A as one of its endpoints it is convenient to consider quasisymmetric mappings on R rather than ∂D.Here this claim follows since t → t γ is quasisymmetric on [0, 1] and since the composition of quasisymmetric mappings is again quasisymmetric [20,Theorem 9].Finally, quasisymmetry in the union of these intervals (or arcs) follows from [20,Theorem 3].
Let B : D → D be a Blaschke product of degree d ≥ 2 fixing 0. Thus B has the form Suppose that r ∈ (0, 1) is such that all zeros of B are in D(0, r).By a result of Walsh [29, Theorem 1], all critical points of B in D are also contained in D(0, r).Since B D(0, r) ⊂ D(0, r) by Schwarz's lemma we see that B −1 D(0, r)) \ D(0, r) is an annulus.
Branner and Fagella [15, p. 163] showed that there exists a quasiregular mapping A : They state this only for the case d = 2, but their proof extends to the general case.Note that in order to prove this result one has to define the mapping in the annulus B −1 (D(0, r)) \ D(0, r)).The construction of this mapping is done in [15, Exercise 2.3.3] for arbitrary degree.
Branner and Fagella also note that we may choose A to depend continuously on B. This implies that the dilatation K(A) of A also depends continuously on B. Restricting to a compact set of Blaschke products B we find that the dilatation of the corresponding quasiregular maps A is uniformly bounded.
We summarize the above discussion in the following result.
Lemma 2.8.Let 0 < r 1 < r 2 < 1 and let B : D → D be a Blaschke product of degree d ≥ 2. Suppose that B(0) = 0 and that the zeros of B are contained in D(0, r 1 ).Then there exists a quasiregular mapping Moreover, there exists a constant C depending only on r 1 , r 2 and d such that A may be chosen to satisfy K(A) ≤ C.
We will also need the following result.Lemma 2.9.Let U be a simply connected, unbounded domain in C which is bounded by piecewise analytic curves.Suppose that each disk D(0, t) intersects only finitely many of these boundary curves.
Let f : U → C be a bounded, continuous function which is holomorphic in U. Suppose that there exist r, R > 0 such that |f (z)| > r for all z ∈ U while |f (z)| = r for all z ∈ ∂U satisfying |z| > R. Then there exists a curve γ tending to ∞ in U and a ∈ ∂D(0, r) such that f (z) → a as z → ∞ on γ.
Proof.Let φ : D → U be a conformal mapping.The hypotheses imply that the boundary of U in C is locally connected.Thus φ has a continuous extension φ : . Then E is a compact subset of ∂D.By a result of Beurling [24,Theorem 9.19], E has logarithmic capacity zero.Put g = f • φ.

There exists an open arc
Noting that |g(z)| > r and thus g(z) = 0 for z ∈ D, we deduce from the Schwarz reflection principle that g can be extended to a function holomorphic Since E has logarithmic capacity zero, it also has analytic capacity zero; see, e.g., [30,Proposition 3.5].Since g is bounded this yields that g has a holomorphic extension to Next we note that U is unbounded and that ∞ is accessible in U. Thus E = ∅.Taking a point ξ ∈ E we have φ(tξ) → ∞ as t → 1.The conclusion follows for the curve γ : [0, 1) → C, γ(t) = φ(tξ), and a = g(ξ).
We will also use the following result about the growth of composite meromorphic functions; see [10, Satz 2.3 and Satz 5.7] and [7,Corollary 4].
Lemma 2.10.Let f be a meromorphic function and g be an entire function.Then In [10,Satz 5.7] the left inequality in (2.4) is proved only under the additional hypothesis that lim sup This additional hypothesis is removed in [7, Corollary 4].It will be satisfied, however, in our applications.In fact, we will consider only the case where (log T (r, f ))/ log log r tends to a finite limit.In this case we deduce from (2.4) that 3 Proofs of Theorems 1.1-1.5 Proof of Theorem 1.1.Let f be a transcendental entire function which has only finitely many islands over some Jordan domain D. Then there exists a connected component U of f −1 (D) which is unbounded.This implies that the minimum modulus min |z|=r |f (z)| is bounded.Wiman's theorem (Lemma 2.1) now yields that ρ(f For the converse result, let D be a Jordan domain in C. Without loss of generality we may assume that 0 ∈ D. Thus there exists ε > 0 such that D(0, ε) ⊂ D. For ρ ∈ [1/2, 1) we consider the function g defined by Then g is a function of order ρ which has only positive zeros and satisfies g(x) → 0 as x → ∞.For ρ ∈ (1/2, 1) the last statement follows from, e.g., [14,Theorem 4.1.8]while for ρ = 1/2 it follows from the explicit representation We conclude that for sufficiently small δ > 0 the function f := δg has no island over D(0, ε) and hence no island over D.
To obtain functions of order ρ ∈ [1, ∞) we write ρ = pρ 0 with p ∈ N and ρ 0 ∈ [1/2, 1).Choosing g of order ρ 0 and δ > 0 as above we find that the function f given by f (z) = δg(z p ) has order ρ and that f has no island over D.
Proof of Theorem 1.2.Let r > 0 be such that the closures of D 1 and D 2 and all finite asymptotic values are contained in D(0, r).Note that by the Denjoy-Carleman-Ahlfors theorem (Lemma 2.3) there are only finitely many asymptotic values.We also assume that ∂D(0, r) contains no critical value.This can be achieved since the set of critical values is countable.
We consider the following graph Γ on the sphere C. It has two vertices, which we denote by × and • and which lie on ∂D(0, r), and three edges, two of which are arcs on the circle ∂D(0, r) connecting × and •, while the third edge is a crosscut of D(0, r) connecting × and • which separates D 1 and D 2 and which contains no critical or asymptotic value.The components of the complement of the set of vertices and edges are called faces.We then have three faces.The face C \ D(0, r) will be denoted by F ∞ and, for j ∈ {1, 2}, the face containing D j will be denoted by F j .
We consider the preimage Γ * = f −1 (Γ) of Γ.It yields a partition of the plane into faces, edges and vertices.It is similar to a line complex ; see [19,Chapter 7,Section 4].One difference is that a line complex is always connected, while Γ * need not be connected.We will see, however, that Γ * is connected if r is chosen sufficiently large.
For our purposes only the topology of Γ * is relevant.Thus we do not distinguish between the preimage Γ * and its image under a homeomorphism of the plane.In figures like Figure 1 we usually draw only a homeomorphic image of Γ * = f −1 (Γ), not the true preimage.In such figures we use the labels 1  ○, 2 ○ and ∞ ○ for the faces F 1 , F 2 and F ∞ as well as for their preimages in Γ * .The same remark applies to the vertices × and •.
Figure 1 shows Γ and Γ * for a function having only islands of multiplicity 2 over D 1 and D 2 , and no critical or finite asymptotic values outside D 1 and D 2 .An example is given by the sine or cosine function if  Clearly, Γ * is a bipartite, properly embedded graph.(Here "properly embedded" means that it not only lies in the plane, but also that its vertices and edges do not accumulate to a point in the plane.)We list some properties of this graph.
(a) Each vertex has degree 3 and lies on the boundaries of three faces with labels 1 ○, 2 ○ and ∞ ○.(e) If r is large enough, then Γ * is connected.
Property (a) is obvious from the definition of Γ * .To prove (b) we note that for a digon V labeled j ○ with j ∈ {1, 2} the mapping f : V → F j is bijective and hence V contains a simple island over D j .Thus (b) follows from the hypothesis that there are only finitely many simple islands over D 1 and D 2 .
To prove (c) we note that in a face labeled ∞ ○ the function f is unbounded.Thus such a face is unbounded and by the Denjoy-Carleman-Ahlfors theorem (Lemma 2.2) there are only finitely many such faces.
To prove (d), let V be an unbounded face labeled j ○ with j ∈ {1, 2}.Then ∂V contains an infinite chain Since every vertex lies on the boundary of a face labeled ∞ ○, and since there are only finitely many faces labeled ∞ ○, there exist a face V ′ labeled ∞ ○ such that that this chain contains infinitely many vertices which lie on both ∂V and ∂V ′ .Let v 1 and v 2 be two such vertices and let v 0 be a vertex between them.Then there exists a face V ′′ labeled ∞ ○ such that v 0 ∈ ∂V ′′ .Connecting v 1 and v 2 by a crosscut in V ′ we see that V ′′ must intersect this crosscut.Thus V ′′ = V ′ .We conclude that there are infinitely many triplets of adjacent vertices which are on the boundary of both V and V ′ .As the middle vertex of such a triplet has degree 3, it must be connected to one of the other two vertices of the triplet by a double edge.In other words, such a triplet leads to a digon.Since there are only finitely many digons by (b), this is a contradiction, completing the proof of (d).
As a preparation for the proof of (e), we note that there is a one-toone correspondence between the components of Γ * and the components of f −1 D(0, r) .In fact, given a component C of Γ * , the corresponding component of f −1 D(0, r) is obtained by "filling" those faces labeled 1  ○ or 2 ○ whose boundaries are contained in C. Note that these faces are all bounded by (d).Reversing this process, one obtains a component of Γ * from a component of f −1 D(0, r) .We conclude from this that the number of components of Γ * is a non-increasing function of r.
For the proof of (e), as well as some subsequent arguments, it will be convenient to consider a graph ∆ which in some sense is dual to Γ * : To each bounded face V we associate a point v ∈ V .(Recall that by (c) and (d) the bounded faces are those labeled 1  ○ or 2 ○.)These points v are the vertices of ∆.Two vertices are connected by an edge if the two faces of Γ * that contain these vertices share a common edge in Γ * .We take this edge in ∆ to be in the union of the closures of the two faces in Γ * , crossing the edge in Γ * which separates these faces once.So f is bounded on the set of edges of ∆. Figure 3 shows the graph ∆ corresponding to the graph Γ * in Figure 2. (Again we only consider a homeomorphic image.) Similarly as above we see that there is a one-to-one correspondence between the components of ∆ and the components of f −1 D(0, r) , and hence to the components of Γ * .To pass from a component of ∆ to a component of f −1 D(0, r) consider for a vertex v the closure V of the face V that contains v. Taking the union of the closures V over all v in a component of ∆ yields a component of f −1 D(0, r) .This process can be reversed, so indeed there is the one-to-one correspondence mentioned.Bounded components of Γ * correspond to bounded components of ∆ and unbounded components of Γ * correspond to unbounded components of ∆.
Next we show that every component of ∆ is a tree.Otherwise there exists a closed curve and in fact a Jordan curve in ∆.We thus have a Jordan curve γ contained in the union of the faces labeled 1  ○ or 2 ○ which crosses each edge of Γ * at most once.The interior of γ contains some vertex, and this vertex is on the boundary of a face V labeled ∞ ○.Since V is unbounded, it must intersect γ, which is a contradiction.Thus every component of ∆ is a tree.
For a face V labeled 1 ○ or 2 ○ the degree of the vertex v of ∆ such that v ∈ V is given by µ(V ).We will also write µ(v) instead of µ(V ).By hypothesis, there are only finitely many vertices of degree 1.Since each bounded tree contains a vertex of degree 1, we conclude that ∆ has only finitely many bounded components.This implies that ∆ also has an unbounded component.
We connect the finitely many bounded components of ∆ by some paths to an unbounded component.Then there exists C > 0 such that |f (z)| ≤ C for z on one of these paths.This implies that r < C and hence shows that if r is chosen large enough at the beginning, then all components of ∆ are unbounded.It follows that all components of Γ * are unbounded.
To complete the proof of (e), suppose that Γ * is disconnected.Then there exists a face V whose boundary contains two (unbounded) components of Γ * .We connect these two components by a crosscut γ in V .This crosscut separates V into two domains V 1 and V 2 .
Since Γ * has no bounded components, V and hence V 1 and V 2 are simply connected.By (c), the face V is labeled ∞ ○ and hence f is unbounded there.We claim that f is unbounded in each of the domains V 1 and V 2 .In fact, suppose that f is bounded in V j where j ∈ {1, 2}.Lemma 2.9 yields that f has an asymptotic value of modulus r with asymptotic path contained in V j .This contradicts our assumption made at the beginning that all asymptotic values are contained in D(0, r).Hence f is unbounded in both V 1 and V 2 .
Again there exists C > 0 such that |f (z)| ≤ C for z on the crosscut γ.Increasing r to some value greater than C thus increases the number of components of {z ∈ C : |f (z)| > r}.Since, by Lemma 2.2, the number of these components is at most max{1, 2ρ(f )} we conclude that if r is large enough, then Γ * has only one component; that is, Γ * is connected.This yields (e) and completes the proof of statements (a)-(e).
As mentioned, a line complex is always connected, while Γ * and ∆ need not be.But (e) says that this holds for large r.
Let p be the number of unbounded faces of Γ * .Thus p is the number of components of {z ∈ C : |f (z)| > r}.An argument similar to the one above about the one-to-one correspondence between the components of Γ * and the components of ∆ yields that p equals the number of complementary components of ∆.
Next we show that with the sum taken over all vertices v of ∆.The proof of (3.1) will only use that ∆ is an infinite properly embedded tree and that ∆ has only finitely many vertices of degree 1.To prove (3.1) suppose first that the number of vertices v for which µ(v) = 2 is finite.If there are no such vertices, then ∆ is an infinite chain If ∆ does not have a vertex of degree 1, then ∆ has a vertex of degree at least 3, and there exists such a vertex v 0 with µ(v This proves (3.1) if the number of vertices v for which µ(v) = 2 is finite.However, minor modifications of the argument show that if the number of vertices v for which µ(v) = 2 is infinite, then the number of complementary components of ∆ is infinite.Thus we also see that if p is finite, then the number of vertices v with µ(v) = 2 is finite.This means that the sum in (3.1) is finite.
To pass from (3.1) to (1.5), let v be a vertex of ∆ and let V be the face of Γ * containing v. Then V is labeled j ○ with j ∈ {1, 2} and it is an island over F j .Its boundary contains µ(V ) vertices × and µ(V ) vertices •.So it is an n-gon with n = 2 µ(V ).
Also, V contains at least one island over D j .Let U 1 , . . ., U m be the islands over D j that are contained in V .Then Let N(V ) be the number of critical points of f contained in V that are not mapped to D j .Since V contains µ(V )−1 critical points and each U k contains contains µ(U k ) − 1 critical points, we deduce from (3.2) that This allows to rewrite (3.2) in the form To obtain (1.5) we want to sum this over all V .Note that there may be (finitely many) vertices v in ∆ of degree 2 for which the corresponding face V contains one critical point and two simple islands over D j .For such a face V both the left and right hand side of (3.3) are equal to 0. Whether we include such a face or not will not affect the sum.Similarly, the sum in (1.5) is unchanged if we remove this critical point and the two simple islands.Summing (3.3) over all bounded faces V and combining this with (3.1) thus yields (1.5) with V N(V ) instead of N, where the sum is taken over all bounded faces V of Γ * .
To complete the proof of (1.5) we thus need to show that In other words, we have to show that the unbounded faces of Γ * contain no critical points of f .In order to do so, let V be an unbounded face of Γ * .Since V is simply connected and bounded by a single curve, unbounded in both directions, there exists a conformal mapping φ from D to V such that φ has a continuous extension to D \ {1}, mapping ∂D \ {1} to ∂V and satisfying φ(z) → ∞ as z → 1.Then defines a positive harmonic function in D which extends continuously to D \ {1}, with u(z) = 0 for z ∈ ∂D \ {1}.This yields that u is a constant multiple of the Poisson kernel; see, e.g., [6,Theorem 6.19].But this implies that log f • φ has no critical points in D. Hence f has no critical point in V .This completes the proof of (3.4) and hence of (1.5).For j ∈ {1, 2} we choose a j ∈ D j and a conformal mapping ψ j : F j → D satisfying ψ j (a j ) = 0.For a face V labeled j ○ we choose a conformal mapping ).We will show that there exists r 1 ∈ (0, 1), depending only on the choice of the F j but not on V , such τ −1 V (f −1 (D j )) and hence the zeros of B are contained in D(0, r 1 ).In order to do so we note that since p < ∞, for all but finitely many faces V labeled j ○ there is exactly one island U of multiplicity 2 over D j contained in V , but there are no further islands over D j contained in V , and there are no critical points in V \ U.For such a face V and island U we then have f −1 (D j ) ∩ V = U and the mapping f : V \ U → F j \ D j is a covering map of degree 2. Denoting by mod(Ω) the modulus of an annulus Ω we find that Thus the modulus of ) is bounded below.This implies that there exists r 1 ∈ (0, 1) such that τ −1 V (f −1 (D j )) ⊂ D(0, r 1 ) for all such faces V .Increasing r 1 if necessary we may assume that this inclusion also holds for the finitely many faces V where f : We now choose r 2 ∈ (r 1 , 1) and apply Lemma 2.8 to B. With the function A obtained from this lemma we define f Thus f V is a quasiregular mapping having one a j -point of multiplicity µ(V ) and no other a j -point in V , and there exists a neighborhood W of ∂V with f V (z) = f (z) for z ∈ V ∩ W . Since there are only finitely many faces V labeled j ○ with j ∈ {1, 2} for which the degree of the mapping f V : V → F j is greater than 2, Lemma 2.8 also yields that we may choose the mappings f V with uniformly bounded dilatation.
We now define a mapping g : C → C by putting g(z) = f V (z) if z ∈ V for such a face V , and g(z) = f (z) otherwise.Then g is quasiregular.By Lemma 2.5 there exists a quasiconformal homeomorphism φ : C → C such that the mapping h := g • φ is entire.It follows that all except possibly finitely many a j -points are multiple, for j ∈ {1, 2}.Moreover, Lemma 2.4 yields that ρ(h) ≤ K(φ)ρ(f ) < ∞.
As in (1.3) we now consider and deduce from the lemma on the logarithmic derivative that R is a rational function.Assuming without loss of generality that a 1,2 = ±1 we find as in (1.4) that h has the form We saw above that {z ∈ C : |f (z)| > r} has p components if r is sufficiently large.This implies that {z ∈ C : |h(z)| > r} has p components for large r.
It is easy to see that the infinite product converges and thus defines a function f meromorphic in C. For x ≥ a there exists n ∈ N and η ∈ [0, 1) such that x = a n+η .Hence This implies that f has no island over D(0, ε) and hence no island over D 1 .An analogous argument shows that f has no island over {z ∈ C : |z| > 1/ε}∪{∞} and hence no island over D 2 .Standard arguments show that the function f defined by (3.7) has order 0 and in fact that for some c > 0 as r → ∞.This completes the proof for the case that ρ = 0. To deal with the general case we note that if g is any entire function, then f • g has no island over D 1 and D 2 .Since lim r→∞ log T (r, f ) log log r = 2 (3.9) by (3.8) we deduce from (2.5) that ρ(f • g) = 2ρ(g).Thus we can achieve that f • g has the preassigned order ρ by choosing g with ρ(g) = ρ/2.
Proof of Theorem 1.4.Without loss of generality we may assume that a 1,2 = ±2i and that ∞ ∈ D. Let δ ∈ (0, 1) and β ∈ (0, 1/2) and let U be the domain which contains the imaginary axis and which is bounded by the curve and the curves −γ, γ and −γ obtained from γ by reflections.The idea is to glue the restriction (of a modification) of 2 sinh(arcsin z) to the domain U and the restriction of the function given by (3.7) to a half-plane.
In order to do so we note that arcsin γ is an injective curve in the first quadrant which connects arcsin δ with ∞.A computation shows that as t → ∞.
Next we consider G(z) = f (−z), with the function f from (3.7); that is, with some a > 1 to be determined later.Denoting by log G the branch of the logarithm with log G(0) = 1 we find that log as t → ∞ by (3.12).Let now V be the domain bounded by the curves γ and γ which contains the interval (δ, ∞).Thus Let b > 0 and define as t → ∞.This holds for any choice of a, b and β.We choose β = π/(2 log a) and b = (1/2−β)π.Note that this still leaves the possibility to choose a later.Then the right hand sides of (3.10) and (3.17) have the same asymptotics as t → ∞, apart from an additive constant.
Using interpolation it can now be shown that there exists an odd quasiconformal mapping ψ : U → C, symmetric with respect to R, which agrees with the arcsine in a neighborhood of the imaginary axis and which satisfies ψ(γ(t)) = log H(γ(t)) for all t ≥ 0. (3.18) Note that log H and hence ψ map the curve γ, and hence the boundary of V , to the line {z ∈ C : Im z = b}.
Next we consider the function Thus S(z) = ±e ±z if Im z = ±b.It is easy to see that S is quasiregular.Finally we define F : C → C by Note that by (3.18) and (3.19) we S(ψ(z)) = exp ψ(z) = H(z) for z ∈ ∂V .Thus F defines a quasiregular mapping.Lemma 2.5 yields that there exists a quasiconformal homeomorphism φ : C → C such that f := F • φ is meromorphic.
Noting that the hyperbolic sine has the totally ramified values ±i and recalling that we have assumed that a 1,2 = ±2i we see that all a j -points are multiple for j ∈ {1, 2}.A similar argument as in the proof of Theorem 1.3 shows that given ε > 0, we can choose a in (3.11) such that f has no island over {z ∈ C : |z| > 1/ε} ∪ {∞}.Choosing ε sufficiently small we conclude that f has no island over D.
Finally, we have as r → ∞, for j ∈ {1, 2}.Lemma 2.4 together with standard arguments now shows that f has order 0 an in fact that (3.9) holds.
This proves the theorem for ρ = 0.As at the end of the proof of Theorem 1.3 we can use this to obtain the result for any preassigned order ρ ∈ (0, ∞) by considering f • g instead of f for an entire function g satisfying ρ(g) = ρ/2.
Proof of Theorem 1.5.The idea behind the construction is due to Künzi [21].The details are somewhat different though.
An outline of the construction is as follows.We consider two elliptic function g 1 and g 2 , both having periods 2 and 2iτ , where τ > 0. We restrict g 1 and g 2 to the sectors We will modify the g j near ∂S j to obtain quasiregular mappings f j : With α := arctan τ this yields that defines a quasiregular mapping f 0 : C → C. By Lemma 2.5 there exists a quasiconformal mapping φ : C → C such that F := f 0 • φ is meromorphic.
The mapping f we want to construct is then given by f (z) = F (z p ) for some p ≥ 2. Using Lemma 2.6 we will see that ρ(f ) = 4pα/π so that we can achieve any preassigned positive order for f .Moreover, both f 1 and f 2 will have the critical values a 1 , a 2 and a 3 .The fourth critical value of f 1 will be different from that of f 2 .The domain D will be such that it contains these fourth critical values.
We now come to the details of the construction.Without loss of generality we may assume that {a 1 , a 2 , a and let g 1 be a conformal mapping from the interior of R onto the upper half-plane.The mapping g 1 extends continuously to the boundary of R and we may normalize it to satisfy g 1 (0) = 1, g 1 (1) = ∞ and g 1 (1 + iτ ) = −1.Then a := g 1 (iτ ) ∈ (−1, 1).The mapping g 1 can be extended by reflections to an elliptic function with periods 2 and 2iτ .
The mapping g 1 can be expressed in terms of the Weierstrass ℘-function with these periods.In fact, if L is the fractional linear transformation satisfying L(∞) = 1, L(e 1 ) = ∞ and L(e 3 ) = −1, then g 1 = L • ℘.
To define the quasiregular mapping f (This is the case if m ≥ 0 and n = m or n = −m − 1.) Note that ∆ m,n is a triangle for such m and n.We begin by defining f 1 on ∆ := ∆ 0,0 .In fact, we will first define f 1 on ∂∆.This will be done in such a way that it can be extended quasiconformally to the interior of ∆ using Lemma 2.7.
It remains to define f 1 in the still missing triangles in the lower half-plane, which are of the form ∆ m,−m−1 with m ∈ N 0 .This we do by reflection in the real axis; that is, we put f 1 (z) = f 1 (z) for z in such a triangle.
For m, n ∈ Z such that R m,n ⊂ S 2 and z ∈ R m,n we put f 2 (z) = g 2 (z).We also put f 2 (x) = g 2 (x) for −1 < x ≤ 0 and define f 2 on ∂S 2 by (3.20).This defines f 2 on the sector S 2 except for the interior of the triangles R −m,m−1 ∩S 2 and R −m,−m ∩ S 2 with m ∈ N. As before we can extend f 2 quasiconformally to these triangles.Here the triangle R −m,m−1 is mapped onto the half-disk Ω 2 := {z ∈ C : Im z > 0, |z| < 1} while R −m,−m is mapped onto Ω 2 .
Thus for j ∈ {1, 2} we have defined a quasiregular mapping f j : S j → C such that (3.20) holds.This implies that the mapping f 0 defined by (3.21) is quasiregular.
By construction, all (−1)-points and all poles of f 0 are multiple, and except for the origin all 1-points of f 0 are also multiple.Moreover, all apoints in the right half-plane are multiple and all (−a)-points in the left half-plane are multiple.It follows from (3.20) and (3.22) that f 0 maps the imaginary axis to ∂D.
Let D be a Jordan domain which contains a and −a and whose closure is contained in D. Then an island over D cannot intersect the imaginary axis.Thus all islands over D are contained in the right or left half-plane.Those contained in the right half-plane contain a multiple a-point while those contained in the left half-plane contain a multiple (−a)-point.We deduce that there are no simple islands over D.
As mentioned above, Lemma 2.5 yields that there exists a quasiconformal mapping φ : C → C such that F := f 0 • φ is meromorphic.It is easy to see that the set A of all z satisfying |z| > 1 where f 1 and f 2 are not meromorphic satisfies (2.3).Lemma 2.6 now yields that φ satisfies (2.2).Since an elliptic function has order 2 this implies that ρ(F ) = 4α/π.
As it is the case for f 0 , the function F has no simple island over D and all poles and all (±1)-points of F are multiple, except for the simple 1-point at the origin.We finally put f (z) = F (z p ) for some p ∈ N with p ≥ 2. Then the origin is a multiple 1-point of f and we conclude that f has no simple island over D and that all poles and all (±1)-points of f are multiple.Moreover, ρ(f ) = 4pα/π.

Theorem 1 . 1 .
Let f be a transcendental entire function and let D be a Jordan domain in C. If f has only finitely many islands over D, then ρ(f ) ≥ 1/2.

2Lemma 2 . 1 .Lemma 2 . 2 .Lemma 2 . 3 .
Results used in the proofsWe begin with a classical result of Wiman [19, Chapter 5, Theorem 1.3].Let f be a non-constant entire function.Suppose that there exists C > 0 such that min |z|=r |f (z)| < C for all r > 0. Then ρ(f ) ≥ 1/2.The next result [19, Chapter 5, Theorem 1.2] is a version of the Denjoy-Carleman-Ahlfors theorem.Let f be an entire function of finite order and let C > 0. Then {z ∈ C : |f (z)| > C} has at most max{1, 2ρ(f )} connected components.The following result [19, Chapter 5, Theorem 1.4] is a consequence of Lemma 2.2 and a result of Lindelöf.It is also called Denjoy-Carleman-Ahlfors theorem.Let f be a non-constant entire function of finite order.Then f has at most 2ρ(f ) finite asymptotic values.

Figure 2
Figure 2 shows Γ * for a function having one simple island and one island of multiplicity 4 over D 1 , one island of multiplicity 3 over D 2 , while all other islands over D 1 and D 2 have multiplicity 2.Clearly, Γ * is a bipartite, properly embedded graph.(Here "properly embedded" means that it not only lies in the plane, but also that its vertices and edges do not accumulate to a point in the plane.)We list some properties of this graph.

Figure 3 :
Figure 3: The graph ∆ corresponding to the graph Γ * in Figure 2.
0 and thus (3.1) holds.Let now n ∈ N and suppose that (3.1) has been proved if the number of vertices v with µ(v) = 2 is less than n.Let ∆ be a graph for which there are n such vertices.If ∆ has a vertex of degree 1, we replace this vertex by an infinite half-chain • − × − • − • • • (or × − • − × • • • ).The new graph ∆ ′ obtained has only n − 1 vertices v with µ(v) = 2 and both left and right side of (3.1) differ for ∆ and ∆ ′ by 1. Thus (3.1) holds for ∆ since it holds for ∆ ′ by induction hypothesis.
while all other vertices of ∆ are on the remaining part of ∆.Replacing the µ(v 0 ) − 1 half-chains by only one such half-chain yields a graph ∆ ′ which has only n−1 vertices v with µ(v) = 2.The left and right side of (3.1) differ for ∆ and ∆ ′ by µ(v 0 ) − 2. Again (3.1) holds for ∆ since it holds for ∆ ′ by induction hypothesis.