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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. Keywords Meromorphic function · Radially distributed value · Normal family Mathematics Subject Classiﬁcation 30D30 · 30D35 · 30D45 1 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 [5]. The restriction on the order can be omitted by a later result of Edrei [9]. Dedicated to the memory of Walter K. Hayman. Communicated by David Drasin. Alexandre Eremenko: Supported by NSF Grant DMS-1665115. B Walter Bergweiler bergweiler@math.uni-kiel.de Alexandre Eremenko eremenko@math.purdue.edu Mathematisches Seminar, Christian-Albrechts-Universität zu Kiel, Ludewig-Meyn-Str. 4, Kiel 24098, Germany Department of Mathematics, Purdue University, West Lafayette, IN 47907, USA 123 672 W. Bergweiler, A. Eremenko 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 [3,Thm.1.1]. Here D denotes the unit disk. Theorem B Let L and L be two distinct rays emanating from the origin and let F 0 1 be the family of all functions holomorphic in D for which all zeros lie on L and all 1-points lie on L . 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, 1-points 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) = zAi(z); see, e.g., [24, §9.2]. Example 1.1 Let 2πi /3 Ai e z πi /3 f (z) = e . (1.1) −2πi /3 Ai e z i π/3 −i π/3 Then all zeros of f are on the ray {re : r > 0}, all poles of f are on {re : r > 0} and all 1-points of f are on the negative real axis. We will verify at the beginning of Sect. 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 ,L and L be three distinct rays emanating from the origin. If 0 1 ∞ the rays are not equally spaced, then there is no transcendental meromorphic function for which all but ﬁnitely many zeros lie on L , all but ﬁnitely many 1-points lie on L 0 1 and all but ﬁnitely many poles lie on L . Theorem 1.2 Let L ,L and L be three distinct rays emanating from the origin 0 1 ∞ 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 ,all 1-points are on L and all 0 1 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, Kap. 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. 123 Meromorphic Functions with Three Radially Distributed Values 673 The differential equation satisﬁed by the Airy function implies that the function f given by (1.1) satisﬁes the differential equation S( f )(z) =−2z, where 2 2 f 1 f f 3 f S( f ) = − = − , f 2 f f 2 f 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. 1 2 3 Theorem 1.3 Let L ,L and L be three equally spaced rays and let f be a transcen- dental meromorphic function. Then there exist distinct values a ,a and a such that 1 2 3 all but ﬁnitely many a -points are on L if and only if 3θi 3 S( f )(z) = e zR z , (1.2) where θ is the argument of one of the rays L 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 , a and a are mapped to 0, 1 and ∞. 1 2 3 The rational functions Q for which the equation S( f ) = Q has a meromorphic solution f have been classiﬁed by Elfving [10, Kap. IV]; see also [7], [17,Thm.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 [22] raised the following interpolation problem: Given points c ,..., c 1 q on the Riemann sphere and q sequences (z ) ,...,(z ) in C, when does 1,k k∈N q,k k∈N there exists a meromorphic function f such that the c -values are precisely the points z ?For q = 2 such a function exists by the Weierstraß factorization theorem, so the j ,k 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. 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. Proof (Veriﬁcation of Example 1.1) The zeros of the Airy function are all negative i π/3 [24, §9.9]. This implies that all zeros of f are on {re : r > 0} and all poles of f 123 674 W. Bergweiler, A. Eremenko −i π/3 are on {re : r > 0}.By[24, Eq. 9.2.12] we have −2πi /3 −2πi /3 2πi /3 2πi /3 Ai(z) + e Ai e z + e Ai e z = 0. This implies that πi /3 2πi /3 −2πi /3 e Ai e z − Ai e z f (z) − 1 = −2πi /3 Ai e z 2πi /3 2πi /3 −2πi /3 −2πi /3 e Ai e z + e Ai e z =−πi /3 −2πi /3 Ai e z Ai(z) −πi /3 =−e . −2πi /3 Ai e z Hence the 1-points of f are all negative. Let D(a, r ) denote the closed disk of radius r around a point a. The following result was proved in [3, Thm. 1.3] and [4, Prop. 1.1]. Lemma 2.1 Let D be a domain and let L be a straight line which divides D into two + − subdomains D and D . Let F be a family of functions holomorphic in D which do not have zeros in D and for which all 1-points lie on L. Suppose that F is not normal at z ∈ D ∩ L and let ( f ) be a sequence in F which 0 k does not have a subsequence converging in any neighborhood of z . Suppose that ( f | + ) converges. Then either ( f | + ) → 0 and ( f | − ) →∞ or ( f | + ) →∞ k k k k D D D D and ( f | − ) → 0. Let z be as before and let r > 0 with D(z , r ) ⊂ D. Then for sufﬁciently large k 0 0 there exists a 1-point a of f such that a → z and if M is the line orthogonal to k k k 0 k L which intersects L at a , then | f (z)| = 1 for all z ∈ M ∩ D(z , r )\{a }. k k k 0 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 : D → R we have max u(z) ≤ C min u(z). z∈K z∈K 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 ) of functions holo- morphic in D which satisfy the hypotheses of the lemma and a sequence (ζ ) in K 123 Meromorphic Functions with Three Radially Distributed Values 675 such that log | f (ζ )| k k →∞. (2.1) log | f (ξ )| Since the f have no zeros and 1-points in D\R, the sequence ( f ) is normal in D\R. k k Suppose ﬁrst that the sequence ( f ) is normal in D.If | f (ξ )| →∞, then there k k exists a subsequence of ( f ) which tends to a limit function holomorphic in D.This contradicts (2.1). Thus | f (ξ )|→∞ and hence f | →∞. But then for large k the k k D functions u given by log | f (z)| u (z) := (2.2) log | f (ξ )| are positive harmonic functions in some connected neighborhood of K , and a contra- diction to (2.1) is now obtained from Lemma 2.2. We may thus assume that ( f ) 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 ) converges in D but that there exists a ∈ D ∩ R such that no subsequence of ( f ) converges in any neighborhood of a. It follows from Lemma 2.1 that either ( f | + ) → 0 and ( f | − ) →∞ or ( f | + ) →∞ and ( f | − ) → 0. The k D k D k D k D ﬁrst possibility is ruled out since we assumed that ξ ∈ D and | f (ξ )| > 2. Thus ( f | + ) →∞ and ( f | − ) → 0. (2.3) k D k D We may assume that ζ → ζ ∈ K . Lemma 2.2 implies that the functions u given k 0 k 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 ∈ R. Without loss of generality we assume that ζ = 0. Choose ε> 0 such that D(0, 10ε) ⊂ D.We may assume that D(0, 10ε) ⊂ K . Put + − K := {z ∈ K : Im z ≥ ε} and K := {z ∈ K : Im z ≤−ε}. ε ε Then by Lemma 2.2 the functions u are uniformly bounded on K . This means that there exists M > 1 such that log | f (z)|≤ M log | f (ξ )| and hence k k M + | f (z)|≤| f (ξ )| for z ∈ K . (2.4) k k By (2.3)wealsohave | f (z)| >1for z ∈ K . (2.5) 123 676 W. Bergweiler, A. Eremenko Fig. 1 The curves α and β and the domains G (top) and H (bottom) k k k and M − | f (z)| < 1 < | f (ξ )| for z ∈ K , (2.6) k k provided k is large enough. On the other hand, we have | f (ζ )| > | f (ξ )| for large k by (2.4). Since ζ → k k k k ζ = 0we also have |ζ | <ε for large k. By the maximum principle, there exists a 0 k curve α connecting ζ with the circle {z:|z|= 5ε} such that k k | f (z)|≥| f (ζ )| > | f (ξ )| >1for z ∈ α . (2.7) k k k k k It follows from (2.5), (2.6) and (2.7) that + − α ⊂ D(0, 5ε)\ K ∩ K ={z:|z|≤ 5ε, |Im z| <ε}. ε ε It is no loss of generality to assume that α connects ζ with a point on the right arc of k k the boundary of the latter set; that is, α connects ζ with the arc {z:|z|= 5ε, |Im z| < k k ε, Re z > 0}; see Fig. 1. 123 Meromorphic Functions with Three Radially Distributed Values 677 It follows from (2.3) that no subsequence of ( f ) is normal at any point of D ∩ R. We may thus apply Lemma 2.1 for any z ∈ D ∩ R. Since D ∩ R ⊃[−10ε, 10ε] we may, in particular, choose z = 3ε/2or z = 7ε/2. Doing so we ﬁnd that if k is 0 0 sufﬁciently large, then there exist 1-points a and b of f satisfying ε< a < 2ε and k k k k 3ε< b < 4ε such that if M and N are the lines orthogonal to R which intersect k k k R in a and b , respectively, then we have | f (z)| = 1for z ∈ M ∩ D(0, 10ε)\{a } k k k k k and z ∈ N ∩ D(0, 10ε)\{b }. This implies that k k | f (z)| >1for z ∈ (M ∪ N ) ∩ D(0, 10ε) ∩ D , (2.8) k k k and | f (z)| < 1for z ∈ (M ∪ N ) ∩ D(0, 10ε) ∩ D . k k k The curve α intersects both lines M and N .Let β be the subcurve of α which k k k k k begins at the last intersection point of α with M and ends at the ﬁrst intersection k k point of α with N . Note that by (2.7) the starting point u of β lies on M ∩{z : 0 < k k k k k Im z <ε} while the end point v lies on N ∩{z : 0 < Im z <ε}. k k We consider the domain G bounded by the arc {z:|z|= 5ε, Im z ≥ ε},the horizontal line segments √ √ x + i ε:− 24ε ≤ x ≤ a and x + i ε : b ≤ x ≤ 24ε , k k the vertical line segments {a + iy : Im u ≤ y ≤ ε} and {b + iy : Im v ≤ y ≤ ε}, k k k k and the curve β ; see Fig. 1. We want to show that the harmonic measure of β in the domain G at the point k k 2i ε is bounded away from 0. This will be done by comparing it with the harmonic measure of the line segment γ := {x − i ε : 2ε ≤ x ≤ 3ε} in the domain H which is bounded by the horizontal line segments γ , √ √ x + i ε:− 24ε ≤ x ≤ 2ε and x + i ε : 3ε ≤ x ≤ 24ε , 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 Fig. 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(2i ε, γ , H ) ≤ ω(2i ε, β , G ). (2.9) k k In order to do so we note that (see [26, Cor. 4.3.9]) ω(z,β , G ) ≥ ω(z,β ∩ H , G ) ≥ ω(z,β ∩ H , G ∩ H ) (2.10) k k k k k k 123 678 W. Bergweiler, A. Eremenko for all z ∈ G ∩ H . Another standard harmonic measure estimate (see [4, Lem. 2.6]) yields that ω(z,β ∩ H , G ∩ H ) ≥ ω(z,γ, H ) k k for all z ∈ G ∩ H . Combining this with (2.10) we obtain (2.9). By (2.5), (2.7) and (2.8)wehave | f (z)| >1for z ∈ ∂G (2.11) k k and large k. Since β ⊂ α we have k k | f (z)|≥| f (ζ )| for z ∈ β (2.12) k k k k by (2.7). Using the Two-Constant Theorem [26, Thm. 4.3.7] we deduce from (2.11) and (2.12) that log | f (z)|≥ ω(z,β , G ) log | f (ζ )| (2.13) k k k k k for all z ∈ G . Combining (2.13) with (2.4) and (2.9) we ﬁnd that M log | f (ξ )|≥ log | f (2εi )| k k ≥ ω(2εi,β , G ) log | f (ζ )| k k k k ≥ ω(2εi,γ, H ) log | f (ζ )|. k k This contradicts (2.1). Proof of Theorem 1.2 Suppose ﬁrst the the rays L , L and L are equally spaced. 0 1 ∞ Let f be the function of Example 1.1. Then there exists θ ∈ R such that either i θ i θ g(z) := f (e z) or g(z) := 1/ f (e z) deﬁnes a meromorphic function g for which all zeros are on L , all 1-points are on L and all poles are on L . Thus for each 0 1 ∞ k ∈ N the function g deﬁned by g (z) = g(kz) is contained in F. It is easy to see k k that {g : k ∈ N} is not normal at any point on any of the rays L , L and L . Thus k 0 1 ∞ F is not normal there. Suppose now that F is not normal. The rays L , L and L divide A := {z : r < 0 1 ∞ |z| < R} into three sectors which we denote by S , S and S .Here S is the sector 0 1 ∞ 0 “opposite” to L ; that is, the sector bounded by L and L . Similarly, S and S are 0 1 ∞ 1 ∞ the sectors opposite to L and L , respectively. 1 ∞ By Montel’s theorem, F is normal in S ∪ S ∪ S = A\(L ∪ L ∪ L ). Thus 0 1 ∞ 0 1 ∞ our assumption that F is not normal implies that F is not normal at some point in A ∩ (L ∪ L ∪ L ). Without loss of generality we may assume that F is not normal 0 1 ∞ at some point z ∈ A ∩ L .Let ( f ) be a sequence in F which does not have a 1 1 k subsequence converging in any neighborhood of z . Since F is normal in S and S , 1 0 ∞ we may assume that ( f ) converges in S and S . Lemma 2.1 implies that either k 0 ∞ ( f | ) → 0 and ( f | ) →∞ or ( f | ) →∞ and ( f | ) → 0. k S k S k S k S 0 ∞ 0 ∞ 123 Meromorphic Functions with Three Radially Distributed Values 679 This implies that ( f ) is not normal at some point of A ∩ (L ∪ L ). Assuming k 0 ∞ without loss of generality that ( f ) is not normal at some point z ∈ A ∩ L ,we k 0 0 deduce from Lemma 2.1, applied to 1 − f instead of f , that either ( f | ) → 1 and k k k S ( f | ) →∞ or ( f | ) →∞ and ( f | ) → 1. The latter possibility contradicts k S k S k S ∞ 1 ∞ our previous ﬁnding that either ( f | ) →∞ or ( f | ) → 0. Altogether we thus k S k S ∞ ∞ have ( f | ) → 0, ( f | ) → 1 and ( f | ) →∞; that is, k S k S k S 0 1 ∞ ⎪ 0for z ∈ S , f (z) → (2.14) 1for z ∈ S , k 1 ∞ for z ∈ S . Let now ξ ∈ S . Then | f (ξ )|→∞ as k →∞. Hence we may assume that ∞ k | f (ξ )| > 2 for all k ∈ N. Lemma 2.3 yields that the functions u deﬁned by k k log | f (z)| u (z) := , (2.15) log | f (ξ )| are locally bounded in T := S ∪ S ∪ (L ∩ A). Note that the u are also harmonic 1 0 ∞ 1 k in T . Passing to a subsequence if necessary we may thus assume that there exists a function u harmonic in T such that u (z) → u(z) for z ∈ T . (2.16) k 1 Similarly, put log | f (z) − 1| v (z) := . (2.17) log | f (ξ )| Then the v are harmonic in T := S ∪ S ∪ (L ∩ A). Since | f (ξ )|→∞ we have k 0 1 ∞ 0 k | f (ξ ) − 1|→∞ and log | f (ξ )|∼ log | f (ξ ) − 1| as k →∞. Using Lemma 2.3 k k k again we see that the functions log | f (z) − 1| z → , log | f (ξ ) − 1| and hence the v are locally bounded in T . Passing to a subsequence if necessary we k 0 thus ﬁnd that there exists a function v harmonic in T such that v (z) → v(z) for z ∈ T . (2.18) k 0 Moreover, u(z) = v(z) for z ∈ T ∩ T = S . (2.19) 0 1 ∞ 123 680 W. Bergweiler, A. Eremenko We now consider the functions w deﬁned by 1 f (z) w (z) := u (z) − v (z) = · log . (2.20) k k k log | f (ξ )| f (z) − 1 k k These functions w are harmonic in T := S ∪ S ∪ (L ∩ A).Wehave w → u k ∞ 0 1 ∞ k in S and w →−v in S . It follows that there is a function w harmonic in T such 0 k 1 ∞ that w (z) → w(z) for z ∈ T , (2.21) k ∞ and u(z) for z ∈ S , w(z) = (2.22) −v(z) for z ∈ S . Let now S and S be the two preimages of S under z → z . Analogously we ∞ ∞ deﬁne S , S , S and S . We may assume that these sectors are arranged in the cyclic 0 0 1 1 order S , S , S , S , S , S . ∞ 0 1 ∞ 0 1 We now deﬁne a function h : A → R as follows: 2 2 ⎪ v z = u z for z ∈ S , ⎪ ∞ 2 2 ⎪ u z = w z for z ∈ S , ⎪ 0 2 2 w z =−v z for z ∈ S , h(z) = (2.23) 2 2 −v z =−u z for z ∈ S , ⎪ 2 2 −u z =−w z for z ∈ S , 2 2 −w z = v z for z ∈ S . Here the two expressions used in the deﬁnition are equal by (2.19) and (2.22). It follows from (2.14) that u(z) ≥ 0for z ∈ S while u(z) ≤ 0for z ∈ S . Since ∞ 0 u(ξ ) = 1 we see that u is non-constant and thus u(z)> 0for z ∈ S while u(z)< 0 for z ∈ S . Analogously we see that v(z)> 0for z ∈ S while v(z)< 0for z ∈ S . 0 ∞ 1 This implies that >0for z ∈ S ∪ S ∪ S , ∞ 1 0 h(z) (2.24) <0for z ∈ S ∪ S ∪ S . 0 1 Let L be any ray separating two of the sectors S , S , S , S , S and S . Thus L ∞ 0 1 ∞ 0 1 is one of the preimages of L , L or L under z → z .Let σ be the reﬂection 0 1 ∞ L in L. The reﬂection principle for harmonic functions [26, Thm. 1.2.9] implies that h(σ (z)) =−h(z). This implies that all sectors S , S , S , S , S and S have the ∞ 0 1 ∞ 0 1 same opening angle. It follows that S , S or S all have opening angle 2π/3. Thus 0 1 ∞ the rays L , L or L are equally spaced. 0 1 ∞ 123 Meromorphic Functions with Three Radially Distributed Values 681 3 Proof of Theorem 1.1 As mentioned in the introduction, normal functions cannot be dealt with by Theo- rem 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 ,L and L be three distinct rays emanating from the origin. 0 1 ∞ Then there is no transcendental meromorphic function of order less than 1 for which all but ﬁnitely many zeros lie on L , all but ﬁnitely many 1-points lie on L and all 0 1 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 and bq lie in a sector opening angle δ. Trivially, there is a half-plane (that is, a sector of opening π) containing 1, ap and bq . The point is that δ is strictly less than π. To prove Lemma 3.1, we will use several other lemmas. Lemma 3.2 Let A, B ∈ ∂ D such that Re(A + B)> 0. Then 1, A and B lie on an arc of ∂ D of length at most arccos(Re(A + B) − 1). Proof The hypothesis implies that A =−1 and B =−1. We may assume that ImB ≥ 0, since otherwise we can pass to the complex conjugates of A and B.We i α i β may thus write A = e and B = e with α ∈ (−π, π ) and β ∈[0,π). Then α + β β − α Re(A + B) = cos α + cos β = 2 cos cos . (3.1) 2 2 Suppose ﬁrst that α< 0. Then −π< α + β< π and thus cos((α + β)/2)> 0. Hence cos((β − α)/2)> 0 so that 0 ≤|α + β|≤ β − α< π. Hence β − α Re(A + B) ≥ 2 cos = 1 + cos(β − α). 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 α + β Re(A + B) ≥ 2 cos = 1 + cos(α + β). Thus β ≤ α + β ≤ arccos(Re(A + B) − 1), and again the conclusion follows. 123 682 W. Bergweiler, A. Eremenko For a sequence (z ) in C, we deﬁne the average n n∈N av((z ) ) := lim z , n n∈N k n→∞ k=1 provided that the limit exists. For c ∈ C and ξ ∈ ∂ D\{1} we have n+1 1 1 ξ − ξ n n av (cξ ) = lim c ξ = lim c = 0. n∈N n→∞ n→∞ n n 1 − ξ k=1 i γ i τ Taking the real part yields for c = e and ξ = e that 0if τ ≡ 0 (mod 2π), av((cos(γ + nτ)) ) = (3.2) n∈N cos γ if τ ≡ 0 (mod 2π). We will use the following lemma. Lemma 3.3 Let (x ) be a bounded real sequence satisfying av((x )) = 0.If n n∈N n av((x )) exists, then lim sup x · lim sup |x |≥ av (x ) . n n n→∞ n→∞ Proof Let α, β > 0 and suppose that x ≤ α and |x |≤ β for all large n ∈ N,say for n n n ≥ N . Then (x − α) ≤−(α + β)(x − α) for all n ≥ N and thus n n n n n 1 1 n − N + 1 1 2 2 2 x − 2α x + α = (x − α) k k n n n n k=N k=N k=N n n 1 n − N + 1 1 ≤−(α + β) (x − α) = (α + β)α − (α + β) x . k k n n n k=N k=N 2 2 It follows that av((x )) + α ≤ (α + β)α and hence that av (x ) ≤ αβ, from which the conclusion follows. Proof of Lemma 3.1 Since the conclusion depends only on the arguments of a, b, p i α i β and q, we may assume that |a|=|b|=| p|=|q|= 1. We write a = e , b = e , i φ i ψ p = e and q = e , with α, β, φ, ψ ∈ (−π, π ]. Since 1, p and q are distinct we have φ = 0,ψ = 0 and φ = ψ. (3.3) 123 Meromorphic Functions with Three Radially Distributed Values 683 We will apply Lemma 3.3 with n n x := Re ap + bq = cos(α + nφ) + cos(β + nψ). (3.4) It follows from (3.2) that av((x )) = 0. We have 2 2 2 x = cos (α + nφ) + cos (β + nψ) + 2 cos(α + nφ) cos(β + nψ) 1 1 = 1 + cos(2α + 2nφ) + cos(2β + 2nψ) + cos(α + β + n(φ + ψ)) 2 2 + cos(α − β + n(φ − ψ )). Suppose ﬁrst that 2φ ≡ 0 (mod 2π) and 2ψ ≡ 0 (mod 2π). Equivalently, φ = π and ψ = π.Itfollows from(3.2) and (3.3) that 1if φ =−ψ, av((x )) = 1 + cos(α + β) if φ =−ψ. Thus av((x )) > 0 unless φ =−ψ and α + β ≡ π(mod 2π). Postponing this excep- tional case, and noting that lim sup |x |≤ 2by(3.4), we deduce from Lemma 3.3 n→∞ that there exist arbitrarily large n ∈ N such that x ≥ av((x ))/4. Lemma 3.2 n n implies that for such n the points 1, ap and bq are contained in an arc of length 2 2 arccos(av((x ))/4 − 1). In this case we may thus take δ = arccos(av((x ))/4 − 1). n n Suppose now that 2φ ≡ 0 (mod 2π). Then φ = π and thus φ =−ψ. Hence av (x ) = 1 + cos(2α) > 0, and the conclusion follows as before. The case that 2ψ ≡ 0 (mod 2π) and thus ψ = π is analogous. It remains to consider the case that φ =−ψ and α + β ≡ π(mod 2π). Then n n q = p and b =−a. Thus ap and bq are symmetric with respect to the imaginary n n axis so that Im(ap ) and Im(bq ) have the same sign. If δ with the properties claimed does not exist, we thus must have n n n n min |ap + 1|, |bq + 1| = min |ap + 1|, |ap − 1| → 0. Thus the only accumulation points of the sequence (ap ) are ±1. This implies that p =±1 and q =∓1, contradicting the hypothesis that 1, p and q are distinct. Lemma 3.4 Let F, G and H be transcendental entire functions for which the argu- ments of the Taylor coefﬁcients tend to 0. Let p, q, r ∈ ∂ D be distinct. Then F (pz), G(qz) and H (rz) are linearly independent. Proof Let ∞ ∞ ∞ n n n F (z) = α z , G(z) = β z and H (z) = γ z . n n n n=0 n=0 n=0 123 684 W. Bergweiler, A. Eremenko The hypothesis says that arg α → 0, arg β → 0 and arg γ → 0, (3.5) n n n 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 n n α ap + β bq + γ = 0, (3.6) n n n for all n ≥ 0. Lemma 3.1 implies that there exists arbitrarily large n such that the n n arguments of ap , bq and 1 lie in an interval of length at most δ. It thus follows n n from (3.5) that the arguments of α ap , β bq and γ lie in an interval of length less n n n than π. This contradicts (3.6). Lemma 3.5 Let F be an entire function of the form F (z) = P(z) 1 + , k=1 where (x ) 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. Proof Let ∞ ∞ d ∞ n n n 1 + = a z , P(z) = b z and F (z) = c z . n n n k=1 n=0 n=0 n=0 Then c = b a . (3.7) n k n−k k=0 It is well-known and easy to prove that a > 0 and a > a a for all n ∈ N.(More n n−1 n+1 generally, the sequence (a ) is totally positive; see [1]). Thus a /a is decreasing. n n+1 n Since F is entire, this implies that a /a → 0. n+1 n Dividing (3.7)by a we ﬁnd that n−d d−1 c a n n−k = b + b → b , d k d a a n−d n−d k=0 123 Meromorphic Functions with Three Radially Distributed Values 685 as n →∞. Since a > 0 and b > 0 we conclude that arg c → 0. n−d d n 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 , all but ﬁnitely many 1-points lie on L and all but ﬁnitely many poles lie on L . Without loss of generality we 1 ∞ may assume that f (0) ∈ C\{0}.Let , and be the canonical products of the 0 1 ∞ zeros, 1-points and poles. Then 0 1 f = f (0) and f − 1 = C , ∞ ∞ for some constant C. It follows that f (0) = C + . (3.8) 0 1 ∞ Let −p, −q and −r be the points where the rays L , L and L intersect ∂ D. Then 0 1 ∞ can be written in the form (z) = aF (pz) where F satisﬁes the hypothesis 0 0 of Lemma 3.5 and a ∈ C\{0}. Similarly, (z) = bG(pz) and (z) = cH (pz) 1 ∞ for entire functions G and H satisfying the hypothesis of Lemma 3.5, and b, c ∈ C\{0}. Equation (3.8) says that F (pz), G(qz) and H (rz) are linearly dependent. This contradicts Lemma 3.4. Proof of Theorem 1.1 Let f be a transcendental meromorphic function for which all but ﬁnitely many zeros lie on L , all but ﬁnitely many 1-points lie on L and all but 0 1 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. 4 Proof of Theorem 1.3 Let g:[r , ∞) → R be a positive increasing function and λ ≥ 0. A sequence (r ) 0 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 g(tr ) ≤ (1 + ε)t g(r ) for ε ≤ t ≤ , (4.1) k k for all large k. If instead of (4.1)wehave g(tr ) ≥ (1 − ε)t g(r ) for ε ≤ t ≤ , k k for all large k, then (r ) is called a sequence of Pólya peaks of the second kind (of order λ for g). Put g(tr ) ρ := sup p ∈ R : lim sup =∞ , (4.2) t g(r ) r ,t →∞ 123 686 W. Bergweiler, A. Eremenko and g(tr ) ρ := inf p ∈ R : lim inf = 0 . (4.3) r ,t →∞ t g(r ) Then log g(r ) log g(r ) 0 ≤ ρ ≤ lim inf ≤ lim sup ≤ ρ ≤∞. (4.4) r →∞ log r log r r →∞ 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 ). The following result is due to Drasin and Shea [8]. Lemma 4.1 Let g:[r , ∞) → R be a positive increasing function and λ ≥ 0. Then the following are equivalent: (a)ρ ≤ λ ≤ ρ . (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. Lemma 4.2 Let u be a positive harmonic function in the right half-plane which extends continuously to i R\{0}, with u(iy) = 0 for y ∈ R\{0}. Then u(z) = Re(az + b/z) with a, b ≥ 0. To prove this result, we note that [2, Thm. 7.26] yields that u has the form aRe z + P(z) where P is a Poisson integral for the right half-plane. Applying [2, Thm. 7.19] to u(z) − aRez 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 [23] 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. Lemma 4.3 Let Q be a rational function satisfying Q(z) ∼ az as z →∞, with d ∈ N and a ∈ C\{0}. Let f be a meromorphic function satisfying S( f ) = Q. (4.5) 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. 123 Meromorphic Functions with Three Radially Distributed Values 687 j i (2 j −1)π/3 j Lemma 4.4 For j ∈{1, 2, 3},let L ={re : r > 0}. The rays L divide C into three congruent sectors. Let V be the sector opposite to L . Let c > 0 and let Q be a rational function satisfying Q(z) ∼−cz as z →∞. Let f be a meromorphic function satisfying (4.5). Then there exist distinct values a , a , a ∈ C := C ∪{∞} such that f (z) → a as 1 2 3 j |z|→∞ in any closed subsector of V . These values a are logarithmic singularities, j j and f has no other asymptotic values. For each j ∈{1, 2, 3}, the function f has inﬁnitely any a -points, and given ε> 0, all but ﬁnitely many a -points are contained in the sector of opening angle ε bisected by L . Moreover, a meromorphic function F satisﬁes S(F ) = Q if and only if F is of the form F = L ◦ f with a linear fractional transformation L. Replacing f by L ◦ f with a linear fractional transformation L we can replace the values a , a and a by three other distinct values, in particular by the values 0, 1 1 2 3 and ∞. The following result is due to Gundersen [14, Thm. 3]. Here a meromorphic function f is called real if f (x ) ∈ R ∪{∞} for all x ∈ R. Otherwise it is called non-real. Lemma 4.5 Let A be a non-real polynomial of degree n, put A(z) − A(z) F (z) = , 2i and let p denote the number of distinct real zeros of F. Let w be a non-trivial solution of w + Aw = 0. (4.6) 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 and w are linearly independent solutions of (4.6) , then f := w /w satisﬁes 1 2 1 2 S( f ) = 2A. (4.7) 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. 123 688 W. Bergweiler, A. Eremenko Lemma 4.6 Let Q be a rational function and let f be a meromorphic functions sat- isfying 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]. Lemma 4.7 Let r > 0, γ> −2 and 0 <α <π with (2 + γ)α < π. Let u and A be holomorphic in S := {z:|z| > r , |arg z| <α} and suppose that u + Au = 0. Suppose also that both u and A are real on the real axis and that there exists c > 0 such that A(z) ∼ cz as z →∞, z ∈ S. (4.8) Then there exists x > r such that all zeros of u in {z:|arg(z − x )| <α} are real. 1 1 Proof A classical result of Kneser [16, Sect. 6] implies that u has arbitrarily large positive zeros. (Kneser’s result says that this is the case if there exists δ> 0 such that x A(x ) ≥ 1/4 + δ for all large x.) It follows from (4.8) that arg A(z) = γ arg z + o(1) as z →∞. Since arg A(x ) = γ arg x = 0for x > r this actually implies that argA(z) = (γ + o(1))arg z as z →∞, z ∈ S. (4.9) If x is large and |arg z| <α, then |x + z| is also large. Thus (4.9) yields that 1 1 arg z A(z + x ) = 2arg z + (γ + o(1))arg(z + x ), 1 1 as x →∞. Since |arg(z + x )| < |arg z|, it now follows from the hypothesesγ> −2 1 1 and (2 + γ)α < π that Im z A(z + x ) >0for z ∈ S with Im z > 0, (4.10) provided x is sufﬁciently large. Put v(z) := u(x + z) and B(z) := A(x + z), with a large zero x of u. Then 1 1 1 v + Bv = 0 and v(0) = 0. Let a, b ∈ S − x ={z − x : z ∈ S}. Then 1 1 2 2 a v (at )v(bt ) − b v(at )v (bt ) = a v (at )v(bt ) − b v(at )v (bt ) dt 2 2 =− a B(at ) − b B(bt ) v(at )v(bt ). (4.11) 123 Meromorphic Functions with Three Radially Distributed Values 689 Let now ξ ∈ S := {z:|arg(z − x )| <α} be a non-real zero of u. Then ξ is also a 1 1 zero of u. We may assume that Im ξ> 0. With a := ξ − x ∈ S and b := ξ − x ∈ S 1 1 we have v(a) = v(b) = 0. It follows from (4.11) that 2 2 0 = a B(at ) − b B(bt ) v(at )v(bt )dt 2 2 = 2i Im a B(at ) |v(at )| dt . (4.12) By (4.10)wehave 2 2 Im a B(at ) = Im (ta) A(at + x ) > 0. This contradicts (4.12). Thus all zeros in the sector S lie on the positive axis. Remark 4.1 Considering u(−z) instead of u(z) we see that Lemma 4.7 remains valid if we put S := {z:|z| > r , |arg z − π|≤ α} and assume that there exists c > 0 such that A(z) ∼−cz as z →∞ in S. Proof of Theorem 1.3 Suppose ﬁrst that there exist a , a and a such that all but 1 2 3 ﬁnitely many a -points are on the ray L . We may assume that a = 0, a = 1 and j 1 2 a =∞, 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 = L , 2 3 L = L and L = L . 1 ∞ Let n(r ) := n(r , 0) + n(r , 1) + n(r , ∞). and let ρ and ρ be deﬁned by (4.2) and (4.3), with g(r ) replaced by n(r ). Since f has at most two Borel exceptional values [13, Cha. 3, Thm. 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 ) tending to ∞, we consider the sequence ( f ) deﬁned by f (z) = k k k f (r z). We will prove the following: (a) If ( f ) is normal in C\{0}, then (r ) has a subsequence which is a sequence of k k Pólya peaks of order 0 for n(r ). (b)ρ ≤ 3/2. (c) If (r ) 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 ) is a sequence tending to ∞, then the sequence ( f ) cannot be normal in C\{0}. k k To prove (a),let ( f ) be normal in C\{0}. Passing to a subsequence if necessary, we may assume that ( f ) converges, say f (z) → φ(z) in C\{0}.Let 0 <ε < 1 and k k 123 690 W. Bergweiler, A. Eremenko let K be the number of zeros, 1-points and poles of φ in {z : ε/2 < |z| < 2/ε}.For large k we then have n(r /ε) − n(εr ) ≤ K . k k ε For ε ≤ t ≤ 1/ε and large k it follows that n(tr ) ≤ n(r /ε) ≤ n(εr ) + K ≤ n(r ) + K ≤ (1 + ε)n(r ), k k k ε k ε k as well as n(tr ) ≥ n(εr ) ≥ n(r /ε) − K ≥ n(r ) − K ≥ (1 − ε)n(r ). k k k ε k ε k Thus (r ) 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 ) such that ( f ) is not normal in C\{0}. k k We will proceed as in the proof of Theorem 1.2, but this time S will be the sector in C which is opposite to L , and not its intersection with the annulus A. Similarly, S 0 1 and S are sectors in C, and so are the sectors T , S and S with a ∈{0, 1, ∞}.For ∞ a a a example, T := S ∪ S ∪ L \{0}. As the rays L , L and L are equally spaced, 1 0 ∞ 1 0 1 ∞ the sectors S , S and S have opening angles 2π/3. 0 1 ∞ As in the proof of Theorem 1.2 we deﬁne u , v and w by (2.15), (2.17) and (2.20). k k k Passing to a subsequence of (r ) 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. i τ 3/2 3/2 Lemma 4.2 yields that u has the form u(z) = Re(e (az + b/z )) where a, b,τ ∈ R with a, b ≥ 0. Since log | f (ξ /r )| log | f (ξ )| k k u (ξ /r ) = = → 0, k k log | f (ξ /r )| log | f (r ξ)| k k k we deduce that b = 0. This implies that h has the form h(z) = Re cz , (4.13) for some c ∈ C\{0}. It follows from (2.16) and (4.13) there exists a sequence (c ) in C such that 3/2 log f (r z) ∼ c z for z ∈ T . (4.14) k k 1 Now f (r z) = 1 if and only if log f (r z) = 2πim for some m ∈ Z. This implies that k k if 0 <δ <ε < 1, then |c | 1 3/2 3/2 n(tr , 1) − n(δr , 1) ∼ t − δ for ε ≤ t ≤ . k k 2π ε 123 Meromorphic Functions with Three Radially Distributed Values 691 Putting 3/2 |c |δ |c | k k a = n(δr , 1) − and b = , k k k 2π 2π we ﬁnd that there exists a sequence (ε ) tending to 0 such that 3/2 n(tr , 1) − a ∼ b t for ε ≤ t ≤ . k k k k The same reasoning can be made for zeros and poles and this yields that 3/2 n(tr ) − A ∼ B t for ε ≤ t ≤ , (4.15) k k k k for suitable A , B ∈ R with B > 0. Noting that n(ε r ) ≥ 0 we deduce from (4.15) k k k k k that 3/2 A ≥−(1 + o(1))B ε . k k Together with (4.15) this implies that if 1 ≤ t ≤ 1/ε , then 3/2 3/2 3/2 2t n(r ) − n(tr ) = 2t (A + (1 + o(1))B ) − A − (1 + o(1))t B k k k k k k 3/2 3/2 = (2t − 1)A + (1 + o(1))t B k k 3/2 3/2 3/2 ≥ −(1 + o(1))(2t − 1)ε + (1 + o(1))t B ≥ 0 for large k. Thus n(tr ) 1 ≤2for 1 ≤ t ≤ , 3/2 t n(r ) ε k k for large k. This implies that ρ ≤ 3/2. To prove (c),let (r ) be a sequence of Pólya peaks (of the ﬁrst kind) for n(r ) of order λ> 0. It follows from (a) that ( f ) 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 ) ≤ (1 + ε)ε n(r ), k k for large k. Together with (4.15) this yields that 3/2 (1 − ε)B ε ≤ n(εr ) − A k k k ≤ (1 + ε)ε n(r ) − A k k ≤ (1 + ε)ε (A + (1 + ε)B ) − A . k k k 123 692 W. Bergweiler, A. Eremenko Hence λ 2 λ 3/2 1 − (1 + ε)ε A ≤ (1 + ε) ε − (1 − ε)ε B . k k Similarly, λ 2 λ 3/2 1 − (1 + ε)M A ≤ (1 + ε) M − (1 − ε)M B . k k The last two inequalities imply that 2 λ 3/2 3/2 2 λ (1 + ε) ε − (1 − ε)ε A (1 − ε)M − (1 + ε) M ≥ ≥ . λ λ 1 − (1 + ε)ε B (1 + ε)M − 1 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 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 ) tends to ∞, then ( f ) 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 , T and T contains a closed 0 1 ∞ subsector which contains inﬁnitely many critical points. Without loss of generality we may assume that this holds for T ;say (z ) is a sequence of critical points contained 1 k in a closed subsector T of T such that r := |z |→∞. As the sequence ( f ) is not 1 k k k normal, we may assume that (4.14) holds. Differentiating we obtain r f (r z) 3 k k 1/2 ∼ c z for z ∈ T . k 1 f (r z) 2 This contradicts the assumption that T contains a critical point of modulus r . 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 [13, Sect. 3.1] or [15, Sect. 2.2]) yields that m(r , S( f )) = O(log r ).Itfollows that T (r , S( f )) = N (r , S( f )) + m(r , S( f )) = O(log r ) 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. 123 Meromorphic Functions with Three Radially Distributed Values 693 1 2 3 Lemma 4.4 implies that the set {L , L , L } of rays considered there coincides with the set {L , L , L }.As L is the negative real axis, Lemma 4.6 implies that Q is 0 1 ∞ real. 2πi /3 2 Let ω = e and put f (z) := f (ωz). Then S( f )(z) = ω Q(ωz). Lemma 4.6 1 1 implies that ω Q(ωz) is also real. Writing Q(z) =−cz + c z j =−∞ we have 2 2+ j j ω Q(ωz) =−cz + c ω z . j =−∞ 2+ j It follows that both c and c ω are real for all j ≤ 0. This implies that c = 0 j j j if c = 1 (mod 3). Hence Q has the form Q(z) =−zR(z ) where R(∞) = c > 0. Thus f satisﬁes (1.2). 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 Eq.(1.2) also has a meromorphic solution f with the asymptotic values 0, 1 and ∞. 3θi Without loss of generality we may assume that e =−1. Putting Q(z) := −zR(z ) we thus have S( f ) = Q. In view of Lemma 4.4 we may assume with- out loss of generality that all but ﬁnitely many 1-points of f are contained in a small sector bisected by L = (−∞, 0]. The functions f (z) and 1/ f (z) have the same asymptotic values in the sectors V . 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, = f (z). (4.16) f (z) It follows from (4.16) that the 1-points of f are symmetric with respect to the real axis. We may write f = w /w where the w satisfy w + Aw = 0 with A = Q/2. We 1 2 j j have f = 1 if and only if w := w − w = 0. Thus the zeros of w are also symmetric 1 2 i γ with respect to the real axis. This implies that w(z) = cw(z) where c = e for iγ/2 some γ ∈ R. Thus u := e 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 2 1 3 axis L . The proof that the other two rays L and L contain all but ﬁnitely many zeros and poles follows with the same argument. 123 694 W. Bergweiler, A. Eremenko Fig. 2 Two line complexes Remark 4.2 The main objective of the papers of Nevanlinna [23] 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, Sect. 7.4] and [21, Sect. XI.2]. Two line complexes are sketched in Fig. 2. The left one was also considered by Elfving [10, Sect. 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, Sect. 47] considered how symmetry of the line complex is reﬂected in the function; see also [23, Sect. 42]. For the line complexes given in Figure 2, and the associated meromorphic functions f ,itfollows [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 Fig. 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 a b R(z) =−c + + . (4.17) z − p (z − p) We recall that Elfving [10, Kap. IV] determined for which rational functions Q the equation S( f ) = Q has a meromorphic solution f . It can be deduced from his result 123 Meromorphic Functions with Three Radially Distributed Values 695 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 + 36a + 45)/72 p. 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 Fig. 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. Remark 4.3 In the proof of Theorem 1.3, we have used Lemma 4.7 to prove that zeros, 1-points and poles are on the respective rays. Alternatively, we could have used the symmetry of the associated line complex, similarly to the reasoning in Remark 4.2. Acknowledgements 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. Funding Open Access funding enabled and organized by Projekt DEAL. 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Computational Methods and Function Theory – Springer Journals
Published: Dec 1, 2021
Keywords: Meromorphic function; Radially distributed value; Normal family; 30D30; 30D35; 30D45
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