Meromorphic functions with three radially distributed values

We consider transcendental meromorphic functions for which the zeros, 1-points and poles are distributed on three distinct rays. We show that such functions exist if and only if the rays are equally spaced. We also obtain a normal family analogue of this result.


Introduction and results
Our starting point is the following result.
Theorem A. There is no transcendental entire function for which all zeros lie on one ray and all 1-points lie on a different ray.
This was proved by Biernacki [6, p. 533] and Milloux [20] for functions of finite order; see also [3]. The restriction on the order can be omitted by a later result of Edrei [9]. This result yields that if all zeros and 1-points of an entire function f lie on finitely many rays, then f has finite order.
The following normal family analogue of Theorem A was proved in [4, Theorem 1.1]. Here D denotes the unit disk.
Theorem B. Let L 0 and L 1 be two distinct rays emanating from the origin and let F be the family of all functions holomorphic in D for which all zeros lie on L 0 and all 1-points lie on L 1 . Then F is normal in D\{0}.
The purpose of this paper is to consider analogues of these results for meromorphic functions, with poles being distributed on some further ray. First we note that there exist meromorphic functions for which zeros, 1points and poles lie on three distinct rays. Such a function is given by the following example. Recall here that the Airy function Ai is an entire function which satisfies the differential equation Ai ′′ (z) = z Ai(z); see, e.g., [24, §9.2]. We will verify at the beginning of Section 2 that the function f defined in Example 1.1 has the properties stated there.
We note that in Example 1.1 the rays are equally spaced. If the rays are not equally spaced, then we have analogues of Theorems A and B. Theorem 1.1. Let L 0 , L 1 and L ∞ be three distinct rays emanating from the origin. If the rays are not equally spaced, then there is no transcendental meromorphic function for which all but finitely many zeros lie on L 0 , all but finitely many 1-points lie on L 1 and all but finitely many poles lie on L ∞ . Theorem 1.2. Let L 0 , L 1 and L ∞ be three distinct rays emanating from the origin and let 0 ≤ r < R ≤ ∞. Let F be the family of all functions meromorphic in {z ∈ C : r < |z| < R} for which all zeros are on L 0 , all 1-points are on L 1 and all poles are on L ∞ .
Then F is normal if and only if the rays are not equally spaced.
One can deduce Theorem A from Theorem B by considering the family {f (rz) : r > 0}. Given any transcendental entire function f , this family is not normal at some point in C \ {0}. This approach does not suffice to deduce Theorem 1.1 from Theorem 1.2, since there are meromorphic functions f for which the family {f (rz) : r > 0} is normal in C \ {0}. Such functions were called Julia exceptional functions by Ostrowski [25, Kapitel II] who studied them in detail. They are also called normal functions. Lehto and Virtanen [19] introduced this terminology for functions meromorphic in a domain G, but in the case that G = C \ {0} it reduces to the property stated above; see also [11,12] for a discussion of normal functions.
The differential equation satisfied by the Airy function implies that the function f given by (1.1) satisfies the differential equation where denotes the Schwarzian derivative.
The following result says that -in some sense -all meromorphic functions for which zeros, 1-points and poles are distributed on three rays are related to the function of Example 1.1. Theorem 1.3. Let L 1 , L 2 and L 3 be three equally spaced rays and let f be a transcendental meromorphic function. Then there exist distinct values a 1 , a 2 and a 3 such that all but finitely many a j -points are on L j if and only if S(f )(z) = e 3θi zR(z 3 ), (1.2) where θ is the argument of one of the rays L j and R is a real rational function satisfying 0 < R(∞) < ∞.
If L is a linear fractional transformation, then S(L • f ) = S(f ). One may choose L such that a 1 , a 2 and a 3 are mapped to 0, 1 and ∞.
The rational functions Q for which the equation S(f ) = Q has a meromorphic solution f have been classified by Elfving [10,Kapitel IV]; see also [7], [17,Theorem 6.7] and [18].
An example of a rational function R with poles such that (1.2) has a solution f for which all (and not only all but finitely many) zeros, 1-points and poles are on the rays will be given in Remark 4.2.
Nevanlinna [23] raised the following interpolation problem: Given points c 1 , . . . , c q on the Riemann sphere and q sequences (z 1,k ) k∈N , . . . , (z q,k ) k∈N in C, when does there exists a meromorphic function f such that the c j -values are precisely the points z j,k ? For q = 2 such a function exists by the Weierstraß factorization theorem, so the problem addresses the case that q ≥ 3. Theorems 1.1 and 1.3 may also be considered as a contribution to this problem of Nevanlinna.
Acknowledgment. We thank Fedor Nazarov for suggesting the proof of Proposition 3.1. We also thank four referees and the editor, David Drasin, for many helpful suggestions.
2 Proof of Theorem 1.2 As we will use Example 1.1 in one direction of the proof, we begin by verifying the properties of this example.
This implies that Hence the 1-points of f are all negative.
Let D(a, r) denote the closed disk of radius r around a point a. Suppose that F is not normal at z 0 ∈ D ∩ L and let (f k ) be a sequence in F which does not have a subsequence converging in any neighborhood of z 0 .
Let z 0 be as before and let r > 0 with D(z 0 , r) ⊂ D. Then for sufficiently large k there exists a 1-point a k of f k such that a k → z 0 and if M k is the line orthogonal to L which intersects L at a k , then |f k (z)| = 1 for all z ∈ M k ∩ D(z 0 , r) \ {a k }.
The following lemma is a simple consequence of Harnack's inequality for the disk; see [2, Harnack's Inequality, 3.6] Lemma 2.2. Let D ⊂ C be a domain and let K ⊂ D be compact. Then there exists C > 1 such that for any positive harmonic function u : Lemma 2.3. Let D be a domain and let L be a straight line. Let ξ ∈ D \ L and let K be a compact subset of D. Then there exists C > 0 such that if f : D → C is a holomorphic function satisfying |f (ξ)| > 2 which has no zeros in D and for which all 1-points lie on L then log |f (z)| ≤ C log |f (ξ)| for all z ∈ K.
Proof. Without loss of generality we may assume that L = R and Im ξ > 0. Suppose that the conclusion is not true. Then there exists a sequence (f k ) of functions holomorphic in D which satisfy the hypotheses of the lemma and a sequence (ζ k ) in K such that Since the f k have no zeros and 1-points in D \ R, the sequence (f k ) is normal in D \ R. Suppose first that the sequence (f k ) is normal in D. If |f k (ξ)| → ∞, then there exists a subsequence of (f k ) which tends to a limit function holomorphic in D. This contradicts (2.1). Thus |f k (ξ)| → ∞ and hence f k | D → ∞. But then for large k the functions u k given by are positive harmonic functions in some connected neighborhood of K, and a contradiction to (2.1) is now obtained from Lemma 2.2. We may thus assume that (f k ) is not normal in D. In fact, with D + := {z ∈ D : Im z > 0} and D − := {z ∈ D : Im z < 0} we may assume (f k ) converges in D + but that there exists a ∈ D ∩ R such that no subsequence of (f k ) converges in any neighborhood of a. It follows The first possibility is ruled out since we assumed that ξ ∈ D + and |f k (ξ)| > 2. Thus We may assume that ζ k → ζ 0 ∈ K. Lemma 2.2 implies that the functions u k given by (2.2) are bounded on any compact subset of D + , with a bound depending on this compact subset, but not on k. Together with (2.1) this yields that ζ 0 ∈ R. Without loss of generality we assume that ζ 0 = 0. Choose ε > 0 such that D(0, 10ε) ⊂ D. We may assume that D(0, 10ε) ⊂ K. Put K + ε := {z ∈ K : Im z ≥ ε} and K − ε := {z ∈ K : Im z ≤ −ε}.
Then by Lemma 2.2 the functions u k are uniformly bounded on K + ε . This means that there exists M > 1 such that log |f k (z)| ≤ M log |f k (ξ)| and hence provided k is large enough.
On the other hand, we have |f k (ζ k )| > |f k (ξ)| M for large k by (2.1). Since ζ k → ζ 0 = 0 we also have |ζ k | < ε for large k. By the maximum principle, there exists a curve α k connecting ζ k with the circle {z : |z| = 5ε} such that It follows from (2.4), (2.6) and (2.7) that It is no loss of generality to assume that α k connects ζ k with a point on the right arc of the boundary of the latter set; that is, α k connects ζ k with the arc {z : |z| = 5ε, | Im z| < ε, Re z > 0}; see Figure 1. It follows from (2.3) that no subsequence of (f k ) is normal at any point of D ∩ R. We may thus apply Lemma 2.1 for any z 0 ∈ D ∩ R. Since D ∩ R ⊃ [−10ε, 10ε] we may, in particular, choose z 0 = 3ε/2 or z 0 = 7ε/2. Doing so we find that if k is sufficiently large, then there exist 1-points a k and b k of f k satisfying ε < a k < 2ε and 3ε < b k < 4ε such that if M k and N k are the lines orthogonal to R which intersect R in a k and b k , respectively, then we The curve α k intersects both lines M k and N k . Let β k be the subcurve of α k which begins at the last intersection point of α k with M k and ends at the first intersection point of α k with N k . Note that by (2.7) the starting point u k of β k lies on M k ∩ {z : 0 < Im z < ε} while the end point v k lies on N k ∩ {z : 0 < Im z < ε}.
We consider the domain G k bounded by the arc {z : |z| = 5ε, Im z ≥ ε}, the horizontal line segments the vertical line segments {a k + iy : Im u k ≤ y ≤ ε} and {b k + iy : Im v k ≤ y ≤ ε}, and the curve β k ; see Figure 1.
We want to show that the harmonic measure of β k in the domain G k at the point 2iε is bounded away from 0. This will be done by comparing it with the harmonic measure of the line segment γ : the vertical line segments {2ε + iy : − ε ≤ y ≤ ε} and {3ε + iy : − ε ≤ y ≤ ε}, and the arc {z : |z| = 5ε, Im z ≥ ε}; see Figure 1. For a bounded domain G and a compact subset A of ∂G, let ω(z, A, G) be the harmonic measure of A at a point z ∈ G. We want to show that ω(2iε, γ, H) ≤ ω(2iε, β k , G k ). (2.9) In order to do so we note that (see [26,Corollary 4.3.9]) for all z ∈ G k ∩ H. Combining this with (2.10) we obtain (2.9).
Proof of Theorem 1.2. Suppose first the the rays L 0 , L 1 and L ∞ are equally spaced. Let f be the function of Example 1.1. Then there exists θ ∈ R such that either g(z) := f (e iθ z) or g(z) := 1/f (e iθ z) defines a meromorphic function g for which all zeros are on L 0 , all 1-points are on L 1 and all poles are on L ∞ . Thus for each k ∈ N the function g k defined by g k (z) = g(kz) is contained in F . It is easy to see that {g k : k ∈ N} is not normal at any point on any of the rays L 0 , L 1 and L ∞ . Thus F is not normal there. Suppose now that F is not normal. The rays L 0 , L 1 and L ∞ divide A := {z : r < |z| < R} into three sectors which we denote by S 0 , S 1 and S ∞ . Here S 0 is the sector "opposite" to L 0 ; that is, the sector bounded by L 1 and L ∞ . Similarly, S 1 and S ∞ are the sectors opposite to L 1 and L ∞ , respectively.
Without loss of generality we may assume that F is not normal at some point z 1 ∈ A ∩ L 1 . Let (f k ) be a sequence in F which does not have a subsequence converging in any neighborhood of z 1 . Since F is normal in S 0 and S ∞ , we may assume that (f k ) converges in S 0 This implies that (f k ) is not normal at some point of A ∩ (L 0 ∪ L ∞ ). Assuming without loss of generality that (f k ) is not normal at some point The latter possibility contradicts our previous finding that either ( (2.14) Let now ξ ∈ S ∞ . Then |f k (ξ)| → ∞ as k → ∞. Hence we may assume that |f k (ξ)| > 2 for all k ∈ N. Lemma 2.3 yields that the functions u k defined by Note that the u k are also harmonic in T 1 . Passing to a subsequence if necessary we may thus assume that there exists a function u harmonic in T 1 such that and hence the v k are locally bounded in T 0 . Passing to a subsequence if necessary we thus find that there exists a function v harmonic in We now consider the functions w k defined by We may assume that these sectors are arranged in the cyclic order S ′ ∞ , S ′ 0 , S ′ 1 , S ′′ ∞ , S ′′ 0 , S ′′ 1 . We now define a function h : A → R as follows: for z ∈ S ′′ 1 . (2.23) Here the two expressions used in the definition are equal by (2.19) and (2.22). (2.24) Let L be any ray separating two of the sectors have the same opening angle. It follows that S 0 , S 1 or S ∞ all have opening angle 2π/3. Thus the rays L 0 , L 1 or L ∞ are equally spaced.
3 Proof of Theorem 1.1 As mentioned in the introduction, normal functions cannot be dealt with by Theorem 1.2. The results of Ostrowski [25] already mentioned imply in particular that normal functions have order 0. The following result actually covers functions of order less than 1.
Proposition 3.1. Let L 0 , L 1 and L ∞ be three distinct rays emanating from the origin. Then there is no transcendental meromorphic function of order less than 1 for which all but finitely many zeros lie on L 0 , all but finitely many 1-points lie on L 1 and all but finitely many poles lie on L ∞ .
To prove this proposition, we will use the following lemma. This lemma may be known, but since we are not aware of any reference, we include a detailed proof.
Lemma 3.1. Let a, b, p, q ∈ C \ {0} and suppose that 1, p and q are distinct. Then there exists δ ∈ (0, π) such that for some arbitrarily large n ∈ N the points 1, ap n and bq n lie in a sector opening angle δ.
Trivially, there is a half-plane (that is, a sector of opening π/2) containing 1, ap n and bq n . The point is that δ is strictly less than π/2.
As the arc on ∂D which connects A with B and contains 1 has length β − α, the conclusion follows. Suppose now that α ≥ 0. We may suppose that α ≤ β. Then there is an arc on ∂D of length β which contains 1, A and B. Now (3.1) yields that Re(A + B) ≥ 2 cos 2 α + β 2 = 1 + cos(α + β).
Thus the only accumulation points of the sequence (ap n ) are ±1. This implies that p = ±1 and q = ∓1, contradicting the hypothesis that 1, p and q are distinct. The hypothesis says that arg α n → 0, arg β n → 0 and arg γ n → 0 (3.5) as n → ∞. Suppose now that aF (pz) + bG(qz) + cH(rz) = 0, with a, b, c ∈ C. If c = 0, then we easily obtain a = b = 0. Thus suppose that c = 0. Without loss of generality we may assume that c = 1. We may also assume that r = 1. It follows that α n ap n + β n bq n + γ n = 0 (3.6) for all n ≥ 0. Lemma 3.1 implies that there exists arbitrarily large n such that the arguments of ap n , bq n and 1 lie in an interval of length at most δ. It thus follows from (3.5) that the arguments of α n ap n , β n bq n and γ n lie in an interval of length less than π. This contradicts (3.6).
Lemma 3.5. Let F be an entire function of the form where (x k ) is a sequence of positive numbers tending to ∞ and where P is a polynomial with positive leading coefficient. Then the arguments of the Taylor coefficients of F tend to 0. It is well-known and easy to prove that a n > 0 and a 2 n > a n−1 a n+1 for all n ∈ N. (More generally, the sequence (a n ) is totally positive; see [1].) Thus a n+1 /a n is decreasing. Since F is entire, this implies that a n+1 /a n → 0.

Proof. Let
Dividing (3.7) by a n−d we find that c n a n−d as n → ∞. Since a n−d > 0 and b d > 0 we conclude that arg c n → 0.
Proof of Proposition 3.1. Let f be a transcendental meromorphic function of order less than 1 for which all but finitely many zeros lie on L 0 , all but finitely many 1-points lie on L 1 and all but finitely many poles lie on L ∞ . Without loss of generality we may assume that f (0) ∈ C \ {0}. Let Π 0 , Π 1 and Π ∞ be the canonical products of the zeros, 1-points and poles. Then for some constant C. It follows that Proof of Theorem 1.1. Let f be a transcendental meromorphic function for which all but finitely many zeros lie on L 0 , all but finitely many 1-points lie on L 1 and all but finitely many poles lie on L ∞ . Proposition 3.1 implies that f has order at least 1. The results of Ostrowski [25] already quoted yield that the family {f (rz) : r > 0} is not normal in C \ {0}. The conclusion now follows from Theorem 1.2.

Proof of Theorem 1.3
Let g : [r 0 , ∞) → R be a positive increasing function and λ ≥ 0. A sequence (r k ) tending to ∞ is called a sequence of Pólya peaks (of the first kind) of order λ for g if for every ε > 0, we have for all large k. If instead of (4.1) we have for all large k, then (r k ) is called a sequence of Pólya peaks of the second kind (of order λ for g).  The upper and lower limits in (4.4) are called the order and lower order of g.
For a meromorphic function f the order and lower order are obtained by taking for g(r) the Nevanlinna characteristic T (r, f ).
(b) g has Pólya peaks of the first kind of order λ.
(c) g has Pólya peaks of the second kind of order λ.
We will also use the following standard result about positive harmonic functions. To prove this result, we note that [2,Theorem 7.26] yields that u has the form a Re z + P (z) where P is a Poisson integral for the right half-plane. Applying [2,Theorem 7.19] to u(z) − a Re z shows that P has the form P (z) = Re(b/z). Lemmas 4.1 and 4.2 will be used to prove that the Schwarzian S(f ) is rational. In order to prove that S(f ) is not only rational, but has the form (1.2), we need results of Elfving [10] and Nevanlinna [21] concerning meromorphic functions with rational Schwarzian derivative. These results were proved by Nevanlinna for the case of a polynomial Schwarzian derivative and extended to rational Schwarzian derivatives by Elfving.
The first result we need is the following.  Then f has order (d + 2)/2.
We will see that in our case the order of f is 3/2 so that d = 1. Thus Q(z) ∼ az as z → ∞. It is no loss of generality to assume that a < 0. The asymptotics of f are then described by the following result.
Let c > 0 and let Q be a rational function satisfying Let f be a meromorphic function satisfying (4.5).
Then there exist distinct values a 1 , a 2 , a 3 ∈ C := C ∪ {∞} such that f (z) → a j as |z| → ∞ in any closed subsector of V j . These values a j are logarithmic singularities, and f has no other asymptotic values.
For each j ∈ {1, 2, 3}, the function f has infinitely any a j -points, and given ε > 0, all but finitely many a j -points are contained in the sector of opening angle ε bisected by L j .
Moreover  Then the number k of real zeros of w is finite and we have k ≤ p + 1. In particular, k ≤ n + 1.
If A is a polynomial, then every solution w of (4.6) is entire. It follows from Lemma 4.5 that if there is a solution of (4.6) which has infinitely many real zeros, then A is real.
If w 1 and w 2 are linearly independent solutions of (4.6), then f : Conversely, every solution f of (4.7) is a quotient of two linearly independent solutions of (4.6). Thus we find that if f satisfies (4.7) for some polynomial A and if f has infinitely many real zeros, then A is real. Since S(L • f ) = S(f ) for every linear fractional transformation L we see that if a meromorphic function f satisfying (4.7) has infinitely many real a-points for some a ∈ C, then A is real. It turns out that this remains valid for rational functions A.
Lemma 4.6. Let Q be a rational function and let f be a meromorphic functions satisfying S(f ) = Q. If f has infinitely many real a-points for some a ∈ C, then Q is real.
As explained above, this result follows from Lemma 4.5 if Q is a polynomial. However, the proof extends to the case that Q is rational. We note that in order to prove Lemma 4.6 for rational Q it does not suffice to extend Lemma 4.5 to the case that A is rational and w is meromorphic, since for rational A the solutions of (4.6) may be multi-valued, but the quotient of two multi-valued solutions may be single-valued. However, the proof of Lemma 4.5 given in [14] also extends to multi-valued functions.
The proof of the following lemma uses Lommel's method to prove that the zeros of Bessel functions are real; see [27, p. 482].
Put v(z) := u(x 1 + z) and B(z) := A(x 1 + z), with a large zero x 1 of u.  By (4.10) we have This contradicts (4.12). Thus all zeros in the sector S 1 lie on the positive axis. Proof of Theorem 1.3. Suppose first that there exist a 1 , a 2 and a 3 such that all but finitely many a j -points are on the ray L j . We may assume that a 1 = 0, a 2 = 1 and a 3 = ∞, since otherwise we can replace f by L • f with a suitable linear fractional transformation L. We switch to the notation previously used by putting L 0 = L 1 , L 1 = L 2 and L ∞ = L 3 . Let n(r) := n(r, 0) + n(r, 1) + n(r, ∞).
Since f has at most two Borel exceptional values [13, Chapter 3, Theorem 2.2], the order of n(r) is equal to that of f . Since f has order at least 1 by Proposition 3.1, Lemma 4.1 yields that ρ * ≥ 1.
For a sequence (r k ) tending to ∞, we consider the sequence (f k ) defined by f k (z) = f (r k z). We will prove the following: (a) If (f k ) is normal in C \ {0}, then (r k ) has a subsequence which is a sequence of Pólya peaks of order 0 for n(r).
(c) If (r k ) is a sequence of Pólya peaks for n(r) of finite non-zero order λ, then λ = 3/2.
Since ρ * ≥ 1 we can deduce from (b), (c) and Lemma 4.1 that ρ * = ρ * = 3/2. This implies that n(r) and hence f have order 3/2. Moreover, it follows from (a) that if (r k ) is a sequence tending to ∞, then the sequence (f k ) cannot be normal in C \ {0}.
Thus (r k ) is a sequence of Pólya peaks for n(r) of order 0 of both the first and second kind.
To prove (b), we note that f has order at least 1 and thus is not normal by Ostrowski's result [25]. Hence there exists a sequence (r k ) such that (f k ) is not normal in C \ {0}. We will proceed as in the proof of Theorem 1.2, but this time S 0 will be the sector in C which is opposite to L 0 , and not its intersection with the annulus A. Similarly, S 1 and S ∞ are sectors in C, and so are the sectors T a , S ′ a and S ′′ a with a ∈ {0, 1, ∞}. For example, As the rays L 0 , L 1 and L ∞ are equally spaced, the sectors S 0 , S 1 and S ∞ have opening angles 2π/3.
As in the proof of Theorem 1.2 we define u k , v k and w k by (2.15), (2.17) and (2.20). Passing to a subsequence of (r k ) if necessary we find as in the proof of Theorem 1.2 that these sequences converge in the appropriate sectors; that is, we have (2.16), (2.18) and (2.21). With h defined by (2.23) we find again that (2.24) holds.
Lemma 4.2 yields that u has the form u(z) = Re(e iτ (az 3/2 +b/z 3/2 )) where a, b, τ ∈ R with a, b ≥ 0. Since we deduce that b = 0. This implies that h has the form h(z) = Re(cz 3 ) (4.13) for some c ∈ C \ {0}. It follows from (2.16) and (4.13) there exists a sequence (c k ) in C such that log f (r k z) ∼ c k z 3/2 for z ∈ T 1 . (4.14) Now f (r k z) = 1 if and only if log f (r k z) = 2πim for some m ∈ Z. This implies that if 0 < δ < ε < 1, then Putting a k = n(δr k , 1) − |c k |δ 3/2 2π and b k = |c k | 2π we find that there exists a sequence (ε k ) tending to 0 such that The same reasoning can be made for zeros and poles and this yields that k . Together with (4.15) this implies that if 1 ≤ t ≤ 1/ε k , then for large k. This implies that ρ * ≤ 3/2. To prove (c), let (r k ) be a sequence of Pólya peaks (of the first kind) for n(r) of order λ > 0. It follows from (a) that (f k ) is not normal. Thus we may assume that (4.15) holds.
Let M > 1 > ε > 0. By the definition of Pólya peaks we have n(εr k ) ≤ (1 + ε)ε λ n(r k ), for large k. Together with (4.15) this yields that Similarly, The last two inequalities imply that Suppose now that λ < 3/2. Then for small ε the left hand side is less than 1, while for large M the right hand side is greater than 1. This is a contradiction. This implies that there are no Pólya peaks of the first kind of order less than 3/2. The same arguments arguments can be made for Pólya peaks of the second kind. This yields there are no Pólya peaks of the second kind of order greater than 3/2. Lemma 4.1 yields that ρ * = ρ * = 3/2, meaning that all Pólya peaks of the first or second kind have order 3/2. This completes the proof of (c).
As explained above, it follows from (a), (b) and (c) that if (r k ) tends to ∞, then (f k ) is not normal in C \ {0}. Moreover, f and n(r) have order 3/2.
Next we show that f has only finitely many critical points; that is, f ′ has only finitely many zeros and f has only finitely many multiple poles. Suppose that f has infinitely many critical points. Then one of the sectors T 0 , T 1 and T ∞ contains a closed subsector which contains infinitely many critical points. Without loss of generality we may assume that this holds for T 1 ; say (z k ) is a sequence of critical points contained in a closed subsector T ′ 1 of T 1 such that r k := |z k | → ∞. As the sequence (f k ) is not normal, we may assume that (4.14) holds. Differentiating we obtain This contradicts the assumption that T ′ 1 contains a critical point of modulus r k . Hence f has only finitely many critical points. This implies that the Schwarzian S(f ) has only finitely many poles so that N(r, S(f )) = O(log r).
Since f has finite order, the lemma on the logarithmic derivative (see [ so that S(f ) is rational. Let Q := S(f ). Since f has order 3/2, Lemma 4.3 yields that there exists a ∈ C \ {0} such that Q(z) ∼ az as z → ∞. With out loss of generality we may assume that a is negative, say a = −c with c > 0. Lemma 4.4 implies that the set {L 1 , L 2 , L 3 } of rays considered there coincides with the set {L 0 , L 1 , L ∞ }. As L 2 is the negative real axis, Lemma 4.6 implies that Q is real.
Let ω = e 2πi/3 and put f 1 (z) := f (ωz). Then S(f 1 )(z) = ω 2 Q(ωz). Lemma 4.6 implies that ω 2 Q(ωz) is also real. Writing It follows that both c j and c j ω 2+j are real for all j ≤ 0. This implies that It remains to prove the converse direction. Thus suppose that R is a real rational functions satisfying 0 < R(∞) < ∞ and that (1.2) has a meromorphic solution. Then, as remarked after Lemma 4.4, the equation (1.2) also has a meromorphic solution f with the asymptotic values 0, 1 and ∞.
Without loss of generality we may assume that e 3θi = −1. Putting Q(z) := −zR(z 3 ) we thus have S(f ) = Q. In view of Lemma 4.4 we may assume without loss of generality that all but finitely many 1-points of f are contained in a small sector bisected by L 2 = (−∞, 0]. The functions f (z) and 1/f (z) have the same asymptotic values in the sectors V j . Since both functions have Schwarzian derivative Q, and thus by Lemma 4.4 differ only by a linear fractional transformation, this yields that they are actually equal; that is, It follows from (4.16) that the 1-points of f are symmetric with respect to the real axis.
We may write f = w 1 /w 2 where the w j satisfy w ′′ j + Aw j = 0 with A = Q/2. We have f = 1 if and only if w := w 1 − w 2 = 0. Thus the zeros of w are also symmetric with respect to the real axis. This implies that w(z) = cw(z) where c = e iγ for some γ ∈ R. Thus u := e iγ/2 w is real on the real axis. Choosing α < π/3 we deduce from Lemma 4.7 and Remark 4.1 all but finitely many zeros of u are negative.
It follows that all but finitely many 1-points are contained in the negative real axis L 2 . The proof that the other two rays L 1 and L 3 contain all but finitely many zeros and poles follows with the same argument.
Remark 4.2. The main objective of the papers of Nevanlinna [21] and Elfving [10] cited above was to study Riemann surfaces with finitely many branch points. They showed that such surfaces correspond to meromorphic functions with rational Schwarzian derivative.
Elfving described such surfaces (and functions) in terms of line complexes (also called Speiser graphs). We do not give the definition of a line complex here, but refer to [13,Section 7.4 Section 2, Abb. 3]. The function corresponding to this line complex has three logarithmic singularities and three critical points, and the critical values corresponding to these three critical points coincide with the three logarithmic singularities.
Elfving [10, Section 47] considered how symmetry of the line complex is reflected in the function; see also [21,Section 42]. For the line complexes given in Figure 2, and the associated meromorphic functions f , it follows [10, p. 59] that S(f ) has the form (1.2) with rational functions R satisfying R(∞) ∈ C \ {0}. In addition, the mirror symmetry of the line complexes implies that R is real.
For the left line complex in Figure 2, the function f has only three (simple) critical points. Hence S(f ) has three (double) poles. Thus R has only one (double) pole p and hence the form (4.17) We recall that Elfving [10, Kapitel IV] determined for which rational functions Q the equation S(f ) = Q has a meromorphic solution f . It can be deduced from his result that if R is given by (4.17), then (1.2) has a meromorphic solution if and only if b = −27p/2 and c = (4a 2 + 36a + 45)/72p. We may assume that −c = R(∞) < 0 and that f has logarithmic singularities over 0, 1 and ∞, with the 1-points close to the negative real axis, corresponding to the branch of the line complex which extends to the left. The simple 1-points then correspond to the double edges of the line complex on this branch, and there is one double 1-point corresponding to the diamond at the end of this branch. Since 1-points are symmetric with respect to the real axis, it follows that all 1-points must lie on the negative real axis.
Thus there are rational functions R with poles such that (1.2) has a solution f for which all (and not only all but finitely many) zeros, 1-points and poles lie on three rays.
For the right line complex in Figure 2 the situation is different. Assume again that the 1-points are distributed along the negative real axis, corresponding to the branch of the line complex which extends to the left. The center of the hexagon on this branch corresponds to a negative 1-point. However, there are also further 1-points corresponding to double edges of the hexagons on the other branches. So it may happen that not all but only all but finitely many zeros, 1-points and poles lie on the rays.
Putting more than one hexagon on the branches stretching to ∞, or replacing the hexagons by (4n + 2)-gons for some n > 1, we find that the rational function R in (1.2) may have arbitrarily high degree.