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It is shown that two key results on transcendental singularities for meromorphic func- tions of ﬁnite lower order have reﬁnements which hold under the weaker hypothesis that the logarithmic derivative has ﬁnite lower order. Keywords Meromorphic function · Direct and indirect transcendental singularities · Logarithmic derivative Mathematics Subject Classiﬁcation 30D35 1 Introduction and Results Suppose that f is a transcendental meromorphic function on C such that, as z tends to inﬁnity along a path γ in the plane, f (z) tends to some α ∈ C. Then, for each t > 0, an unbounded subpath of γ lies in a component C (t ) of the set {z ∈ C : | f (z) − α| < t }. Here, C (t ) ⊆ C (s) if 0 < t < s, and the intersection C (t ) t >0 is empty [2]. The path γ then determines a transcendental singularity of the inverse −1 function f over the asymptotic value α and each C (t ) is called a neighbourhood of the singularity [2,18]. Two transcendental singularities over α are distinct if they have disjoint neighbourhoods for some t > 0. Following [2,18], a transcendental −1 singularity of f over α ∈ C is said to be direct if C (t ),for some t > 0, contains ﬁnitely many points z with f (z) = α, in which case there exists t > 0 such that C (t ) contains no α-points of f for 0 < t < t . A direct singularity over α ∈ C is logarithmic if there exists t > 0 such that log t /( f (z) − α) maps C (t ) conformally onto the right half plane. If, on the other hand, C (t ) contains inﬁnitely many α-points Communicated by Walter Bergweiler. B J. K. Langley james.langley@nottingham.ac.uk School of Mathematical Sciences, University of Nottingham, Nottingham NG7 2RD, UK 123 118 J. K. Langley of f , for every t > 0, then the singularity is called indirect: a well-known example is −1 + given by f (z) = z sin z, with α = 0 and γ the positive real axis R . Transcendental −1 singularities of f over ∞ and their corresponding neighbourhoods may be deﬁned and classiﬁed using 1/ f , and the asymptotic and critical values of f together comprise −1 the singular values of f . If f has ﬁnite (lower) order of growth, as deﬁned in terms of the Nevanlinna char- acteristic function T (r , f ) [8,18], then the number of direct singularities is controlled by the celebrated Denjoy–Carleman–Ahlfors theorem [9,18]. Theorem 1.1 (Denjoy–Carleman–Ahlfors theorem) Let f be a transcendental mero- morphic function in the plane of ﬁnite lower order μ. Then the number of direct −1 transcendental singularities of f is at most max{1, 2μ}. A key consequence of Theorem 1.1 is that a transcendental entire function of ﬁnite lower order μ has at most 2μ ﬁnite asymptotic values [9]. A result of Bergweiler and Eremenko [2] shows that the critical values of a meromorphic function of ﬁnite (lower) order have a decisive inﬂuence on indirect transcendental singularities. Theorem 1.2 [2] Let f be a transcendental meromorphic function in the plane of ﬁnite lower order. −1 (a) If f has an indirect transcendental singularity over α ∈ C = C ∪{∞}, then each neighbourhood of the singularity contains inﬁnitely many zeros of f which are not α-points of f ; in particular, α is a limit point of critical values of f . −1 (b) If f has ﬁnitely many critical values, then f has ﬁnitely many transcendental singularities, and all transcendental singularities are logarithmic. Theorem 1.2 was proved in [2]for f of ﬁnite order and was extended to ﬁnite lower order, using essentially the same method, by Hinchliffe [11]. Part (b) follows from part (a) combined with Theorem 1.1 and a well-known classiﬁcation theorem from [18, p. 287], which shows in particular that any transcendental singularity of the inverse function over an isolated singular value is logarithmic. Theorem 1.2 was employed in [2] to prove a long-standing conjecture of Hayman [7] concerning zeros of ff − 1, and has found many subsequent applications, including zeros of derivatives [12]. The reader is referred to [3,19] for further striking results on singularities of the inverse, both restricted to entire functions but independent of the order of growth. The starting question of the present paper concerns the extent to which Theorems 1.1 (k) and 1.2 hold under the weaker hypothesis that f / f has ﬁnite lower order for some k ∈ N ={1, 2,...}. The obvious example f (z) = exp(exp(z)) shows that f / f can −1 have ﬁnite order despite f having inﬁnite lower order; here, f has inﬁnitely many direct (indeed logarithmic) singularities over 0 and ∞, and one over 1. Furthermore, if k ∈ N and A is a transcendental entire function, then the lemma of the logarithmic derivative [8] shows that every non-trivial solution of (k) w − A (z)w = 0 (1.1) has inﬁnite lower order, even if A has ﬁnite order. Clearly, each of exp(exp(z)) and −1 exp(z sin z) satisﬁes an equation of form (1.1) with coefﬁcient of ﬁnite order. Note 123 Transcendental Singularities for a Meromorphic Function 119 further that if f is a transcendental meromorphic function in the plane and f / f has (k) ﬁnite lower order, then it is easy to prove by induction that so has A = f / f for every k ≥ 1, using the formula A = A A + A , whereas the example k+1 k 1 f (z) 1 f (z) 1 −z/2 z z z 2z f (z) = e sin(e ), =− + e cot(e ), = − e f (z) 2 f (z) 4 shows that f / f can have ﬁnite order despite f / f having inﬁnite lower order. Theorem 1.3 Let f be a transcendental meromorphic function in the plane such that −1 f has n ≥ 1 distinct direct transcendental singularities over ﬁnite non-zero values. (k) Let k ∈ N and let μ be the lower order of A = f / f . Then the following statements hold. (i) There exists a set F ⊆[1, ∞) of ﬁnite logarithmic measure such that log (min{|A (z)|:|z|= r }) lim =−∞. (1.2) r →+∞,r ∈ / F log r (ii) If n ≥ 2, then n ≤ 2μ. (iii) If n = 1 and there exist κ> 0 and a path γ tending to inﬁnity in the complement of the neighbourhood C (κ) of the singularity, then μ ≥ 1/2. Theorem 1.3 will be deduced from a version of the Wiman–Valiron theory for mero- morphic functions with direct tracts developed in [4], and part (ii) is sharp, by Example 1 in Sect. 2. Furthermore, if g is a transcendental entire function of lower order less than 1/2, then the inverse function of f = 1 − 1/g has a direct singularity over 1; in this case, A obviously has lower order less than 1/2, but the cos πλ theorem [9,Ch. 6] implies that every neighbourhood of the singularity contains circles |z|= r with r arbitrarily large, so that a path γ as in (iii) cannot exist. Theorem 1.4 Let f be a transcendental meromorphic function in the plane such that (k) −1 f / f has ﬁnite lower order for some k ∈ N. Assume that f has an indirect transcendental singularity over α ∈ C. Then each neighbourhood of the singularity (k) contains inﬁnitely many zeros of f f , which are not α-points of f . Theorem 1.4 will be proved using a modiﬁcation of methods from [2,11]. Corollary 1.1 Let f be a transcendental meromorphic function in the plane, with −1 ﬁnitely many critical values, such that f / f has ﬁnite lower order. Then f has ﬁnitely many transcendental singularities over ﬁnite non-zero values, and f has ﬁnitely −1 many asymptotic values. Moreover, all transcendental singularities of f are loga- rithmic. Corollary 1.1 follows from Theorems 1.3 and 1.4, coupled with [18, p. 287]. Corollary 1.2 Let f be a transcendental meromorphic function in the plane such that f / f has lower order μ< ∞ and f / f and f / f have ﬁnitely many zeros. Then f / f is a rational function and f has ﬁnite order and ﬁnitely many poles. 123 120 J. K. Langley To prove Corollary 1.2, observe that all but ﬁnitely many zeros of f f are zeros of −1 f . Thus, f has no indirect singularities, by Theorem 1.4, and hence f has ﬁnitely many asymptotic values, in view of Theorem 1.3. Since f evidently has ﬁnitely many critical values, the result follows via [12, Theorem 2]. The condition μ< ∞ holds if f / f has ﬁnite lower order, and is not redundant, because of an example in [12]. The last result of this paper is related to the following theorem from [14]. Theorem 1.5 [14] Let M be a positive integer and let f be a transcendental mero- morphic function in the plane with transcendental Schwarzian derivative f (z) 3 f (z) S (z) = − , (1.3) f (z) 2 f (z) such that: (i) f has ﬁnitely many critical values and all multiple points of f have multiplicity at most M; (ii) the inverse function of f has ﬁnitely many transcendental singularities. Then the following three conclusions hold: (a) f has inﬁnitely many multiple points; (b) the inverse function of S does not have a direct transcendental singularity over ∞; (c) the value ∞ is not Borel exceptional for S . Conclusion (a) is a result of Elfving [6] and Nevanlinna [17,18], but was proved in [14] by a completely different method. The following example shows that under the hypotheses of Theorem 1.5 the inverse of the Schwarzian can have a direct transcen- dental singularity over a ﬁnite value: write 3 tanh z g(z) = sinh z, S (z) = 1 − , −1 so that S has two logarithmic singularities over − 1/2. However, assumptions (i) and (ii) of Theorem 1.5 imply that f belongs to the Speiser class S [1,2] consisting of all meromorphic functions in the plane for which the inverse function has ﬁnitely many singular values. For f ∈ S, the following result excludes direct singularities of the inverse of S over 0. Theorem 1.6 Let f be a transcendental meromorphic function in the plane belonging to the Speiser class S, with transcendental Schwarzian derivative S . Then the inverse function of S does not have a direct transcendental singularity over 0. The example f (z) = tan z from [5] shows that for f ∈ S it is possible for 0 to be an asymptotic value of S . Here direct computation shows that f (z)/ f (z) tends to 0as z →∞ in the left half plane, and so does S (z). The author thanks the referees for their helpful comments. 2 Examples Illustrating Theorems 1.3 and 1.4 Example 1 A function extremal for Theorem 1.3(ii), but not for Theorem 1.1,isgiven by 123 Transcendental Singularities for a Meromorphic Function 121 f (z) π z f (0) = 1, = . f (z) sin π z Here f is meromorphic in the plane, having at each non-zero integer n a zero or pole of multiplicity |n|, depending on the sign and parity of n. Hence N (r , f ) and N (r , 1/ f ) have order 2. Because +∞ +∞ π y π y 0 <α = dy = dy sinh π y π y + (π y) /6 + ··· 0 0 1 ∞ < 1dy + dy <π 2 2 π y 0 1 ±i α and f / f is even, f has distinct asymptotic values e , approached as z tends to inﬁnity along the imaginary axis. As f / f has ﬁnite order and f has no ﬁnite non- −1 zero critical values, both of these singularities of f are direct by Theorem 1.4. Example 2 Deﬁne g by g (z) 1 g(0) = 1, = A (z) = √ . g(z) π cos z √ √ The zeros of cos z occur where z = b = (2n +1)π/2, with n ∈ Z, and the residue 2 + of A at b is ±(2n + 1). Thus g is meromorphic in C, with zeros and poles in R and no ﬁnite non-zero critical values. Integration along the negative real axis shows −1 that g has a non-zero real asymptotic value α, and g has a logarithmic singularity over α by Corollary 1.1. This gives δ> 0 and a simply connected component C of {z ∈ C :|g(z) − α| <δ} with (−∞, R) ⊆ C for some R < 0. Moreover, C is symmetric with respect to R, since g is real meromorphic, so that C ∩ R is bounded, and g is extremal for Theorem 1.3(iii). Example 3 Let F (z) = exp(−z/2 − (1/4) sin 2z) cos z, so that F /F is entire of ﬁnite + −1 order. Then F (z) tends to 0 along R and this singularity of F is evidently indirect. Example 4 Deﬁne entire functions A and v by v (z) 1 − cos z 1 v(0) = 1, = A (z) = = + ··· . v(z) z 2 Then there exists α ∈ R such that v(x ) → exp(±α) as x →±∞ on R and, since −1 A does not satisfy (1.2), Theorem 1.3 implies that v has no direct singularities over ﬁnite non-zero values. Because all critical points of v are real, all but ﬁnitely many of −1 them belong to neighbourhoods of the indirect singularities over exp(±α), and so v has no other indirect singularities, by Theorem 1.4. Thus applying [18, p. 287] again, −1 in conjunction with Iversen’s theorem, shows that v has logarithmic singularities over the omitted values 0 and ∞. 123 122 J. K. Langley Example 5 Let h(z) = exp(sin z − z), so that A = h /h is entire of ﬁnite order but does not satisfy (1.2). Since h(z) tends to 0 along R , and to ∞ on the negative real −1 axis, with h (2πn) = 0 for all n ∈ Z, these singularities of h are direct but not logarithmic. 3 Preliminaries The following well-known estimate may be found in Theorem 8.9 of [9]. Lemma 3.1 [9] Let D ,..., D be n ≥ 2 pairwise disjoint plane domains. If 1 n u ,..., u are non-constant subharmonic functions on C such that u vanishes outside 1 n j D , then h(r ) lim inf > 0, h(r ) = max B(r , u ), B(r , u ) = sup{u (z) :|z|= r }. j j j n/2 r →∞ r 1≤ j ≤n (3.1) For a ∈ C and R > 0 denote by D(a, R) the open disc of centre a and radius R, and by S(a, R) its boundary circle. Lemma 3.2 To each k ∈ N corresponds d ∈ (0, ∞) with the the following property. Suppose that 0 < R < ∞ and w = h(z) maps the domain U ⊆ C conformally onto D(a, R), with inverse function F : D(a, R) → U. Then there exists an analytic function V : D(a, R) → C with (k) k h (z)F (w) = V (w), |V (w)|≤ as |w − a|→ R − . (3.2) k k k−1 (R −|w − a|) Proof Assume that a = 0 and initially that R = 1. It is clear that (3.2) holds for k = 1, with V (w) = 1. If (3.2) holds for k, then it follows that F (w) (k+1) k+1 (k) k−1 h (z)F (w) = V (w) − kh (z)F (w) F (w) = V (w) − kV (w) . k k F (w) −1 Since F (w)/F (w) = O(1 −|w|) as |w|→ 1− by [10, p. 5, (1.6)], applying Cauchy’s estimate for derivatives to V proves the lemma by induction when R = 1. In the general case write w = h(z) = RH (z) = Rv and z = F (w) = G(v) so that, as |w|→ R−, 1−k d R d k k (k) k 1−k (k) k |h (z)F (w) |= R |H (z)G (v) |≤ = . k−1 k−1 (1 −|v|) (R −|w|) 123 Transcendental Singularities for a Meromorphic Function 123 Lemma 3.3 Let M ∈ N and s > 2 and let E ,..., E be N ≥ 24M pairwise 1 N disjoint domains in C, and for t > 0 let φ (t ) be the angular measure of S(0, t ) ∩ E . j j Then at least N − 12Mof the E satisfy π dt π dt > M log s and > M log s. (3.3) 1/2 t φ (t ) 2 t φ (t ) [4s ,s/4] j [4s,s /4] j Proof This is a standard application as in [9, Ch. 8] or [2] of the Cauchy–Schwarz inequality, which gives ⎛ ⎞ ⎛ ⎞ L L L L 1 1 π ⎝ ⎠ ⎝ ⎠ ≤ φ (t ) ≤ 2 (3.4) t t φ (t ) t φ (t ) j j j =1 j =1 j =1 for M ≤ L ≤ N and t > 0. If s > 2 and either inequality of (3.3) fails for L ≥ 6M of the E , without loss of generality for j = 1,..., L, then integrating (3.4) yields a contradiction via √ √ s s 2LM log s < 6LM log ≤ L log ≤ 2LM log s. 16 16 Lemma 3.4 [1] Let h be a transcendental meromorphic function in the plane belonging to the Speiser class S. Then there exist positive constants C, R and M such that zh (z) h(z) ≥ C log for |z|≥ R. (3.5) h(z) M 4 Proof of Theorem 1.3 −1 Let f be a transcendental meromorphic function in the plane such that f has n ≥ 1 direct singularities over (not necessarily distinct) ﬁnite non-zero values a ,..., a . 1 n (k) Let k ∈ N; then A = f / f does not vanish identically. There exist a small positive δ and non-empty components D of {z ∈ C :| f (z) − a | <δ},for j = 1,... n, j j such that f (z) = a on D , so that D immediately qualiﬁes as a direct tract for j j j g = δ/( f − a ) in the sense of [4, Section 2]. Here δ may be chosen so small j j that if n ≥ 2 then these D are pairwise disjoint. For each j, deﬁne a non-constant subharmonic function u on C by u (z) = log |g (z)|= log (z ∈ D ), u (z) = 0 (z ∈ / D ). j j j j j f (z) − a Then [4, Theorem 2.1] implies that, with B(r , u ) as in (3.1), B(r , u ) lim =+∞, lim a(r , u ) =+∞, a(r , u ) = rB (r , u ). j j j r →+∞ r →+∞ log r (4.1) 123 124 J. K. Langley Lemma 4.1 There exists a set F ⊆[1, ∞), of ﬁnite logarithmic measure, such that for each s ∈[1, ∞)\F and each j there exists z with 0 j (k) f (z ) |z |= s, A (z ) = = O exp(−B(s, u )/2) . (4.2) j k j j f (z ) Proof Fix τ with 1/2 <τ < 1 and apply the version of Wiman–Valiron theory developed in [4] for meromorphic functions with direct tracts. By [4, Theorem 2.2 and Lemma 6.10], there exists a set F ⊆[1, ∞) of ﬁnite logarithmic measure such that every s ∈[1, ∞)\F has the following two properties: ﬁrst, a(s, u ) is large, by 0 j (4.1), but satisﬁes a(s, u ) ≤ B(s, u ) ; (4.3) j j second, for each j there exists z with |z |= s and u(z ) = B(s, u ) such that j j j j −a(s,u ) f (z) − a z s ∼ for |z − z | < . (4.4) f (z ) − a z a(s, u ) j j j j A standard application of Cauchy’s estimate for derivatives in (4.4)now gives ( p) p+1 f a(s, u ) (z) = O for p = 0,..., k − 1 f − a s and |z − z | < . 2a(s, u ) It follows via [8, Lemma 3.5] that (k) (k) k f (z ) f (z ) f (z ) − a a(s, u ) exp(−B(s, u )) j j j j j j = · = O , f (z ) f (z ) − a f (z ) s j j j j which, by (4.3), yields (4.2) for large enough s ∈ / F . Combining (4.1) with (4.2)for j = 1 leads to (1.2). To prove the remaining asser- tions it may be assumed that A has ﬁnite lower order μ. Choose a positive sequence (r ) tending to inﬁnity such that μ+o(1) T (8r , A )< r . (4.5) m k Let m be large and let w ,...,w be the zeros and poles of A in r /4 ≤|z|≤ 4r , 1 q k m m repeated according to multiplicity: then (4.5) and standard estimates yield μ+o(1) q ≤ n(4r , A ) + n(4r , 1/A ) ≤ T (8r , A ) + O(1) ≤ r . (4.6) m m k m k m k log 2 123 Transcendental Singularities for a Meromorphic Function 125 −μ Let U be the union of the discs D(w , r ). Since the sum of the radii of the discs m j m of U is o(r ) by (4.6), there exists a set E ⊆[r /2, 2r ], of linear measure at least m m m m m r , and so logarithmic measure l ≥ 1/2, such that for r ∈ E the circle |z|= r does m m m not meet U . A standard application of the Poisson–Jensen formula [8] on the disc |ζ|≤ 4r then yields μ+o(1) |log |A (z)|| ≤ r for |z|∈ E . (4.7) k m Since m is large and l ≥ 1/2, there exists s ∈ E \F . m m m 0 Suppose now that n = 1 and there exist κ> 0 and a path γ tending to inﬁnity in the complement of the neighbourhood C (κ) of the singularity, or that n ≥ 2. Then (3.1) holds, by [9, Theorem 6.4] when n = 1, and by Lemma 3.1 when n ≥ 2. Combining (3.1) and (4.2), with s = s ≥ r /2, yields points z with |z |= s and, for at least m m j j m one j, n/2−o(1) A (z ) = O exp(−B(s , u )/2) = O exp −s . k j m j m On combination with (4.7), this forces 2μ ≥ n. 5 Indirect Singularities Proposition 5.1 Let f be a transcendental meromorphic function in the plane such (k) −1 that f / f has ﬁnite lower order μ for some k ∈ N. Assume that f has an indirect transcendental singularity over α ∈ C\{0}. Then for each δ> 0, the neighbourhood (k) C (δ) of the singularity contains inﬁnitely many zeros of f f . The proof of Proposition 5.1 will take up the whole of this section. The method is adapted from those in [2,11], but some complications arise, in particular when k ≥ 2. Assume throughout that f and α are as in the hypotheses, but C (ε), for some small (k) ε> 0, contains ﬁnitely many zeros of f f . It may be assumed that α = 1, and (k) that C (ε) contains no zeros of f f . Choose positive integers N , N ,..., N with 1 2 9 5μ + 12 < N and N /N large for each j. 1 j +1 j Lemma 5.1 For each j ∈{1,..., N } there exist z ∈ C (ε) and a ∈ C with 0 < r = 9 j j j |1 − a | <ε/2, as well as a simply connected domain D ⊆ C (ε), with the following j j properties. The a are pairwise distinct and the D pairwise disjoint. Furthermore, j j the function f maps D univalently onto D(1, r ), with z ∈ D and f (z ) = 1. j j j j j Moreover, 0 ∈ / D but D contains a path σ tending to inﬁnity, which is mapped by j j j f onto the half-open line segment [1, a ), with f (z) → a as z →∞ on σ . j j j This is proved exactly as in [2]. If 0 < T <ε/2 and z ∈ C (T ) is such that j j j −1 f (z ) = 1, let r be the supremum of t > 0 such that the branch of f mapping 1 to j j z admits unrestricted analytic continuation in D(1, t ). Then r < T because f is not j j j −1 univalent on C (T ), and there is a singularity a of f with |1 − a |= r ; moreover, j j j j a must be an asymptotic value of f .The z and T are then chosen inductively: for j j j the details see [2](or [13, Lemma 10.3]). 123 126 J. K. Langley Lemma 5.2 Let the z , a ,σ and D be as in Lemma 5.1.For t > 0,let t θ (t ) be the j j j j j length of the longest open arc of S(0, t ) which lies in D . Then f satisﬁes, as z tends to inﬁnity on σ , |z| r dt log ≥ . (5.1) | f (z) − a | 4t θ (t ) j j |z | −1 Proof Let z = H (w) be the branch of f mapping D(1, r ) onto D .For z ∈ σ , j j j the distance from z to ∂ D is at most |z|θ (|z|). Thus Koebe’s quarter theorem [10, j j Ch.1]impliesthat |(w − a )H (w)|≤ 4|z|θ (|z|) for z = H (w), w ∈[1, a ). j j j Hence, for large z ∈ σ and w = f (z), writing u = H (v) for v ∈[1,w] gives (5.1) via r |dv| |du| log = = | f (z) − a | |a − v| |(a − v)H (v)| j [1,w] j H ([1,w]) j |du| ≥ . 4|u|θ (|u|) H ([1,w]) j Since N > 5μ, there exists a positive sequence (s ) tending to inﬁnity such that 1 n 5 (k) 5 (k) N T (s , f / f ) + T (s , f / f ) ≤ s . (5.2) n n n Set (k) f (z) G(z) = z , N = N . (5.3) f (z) Applying [15, Lemma 4.1] to 1/G (with ψ(t ) = t in the notation of [15]) gives a small positive η such that G has no critical values w with |w|= η and such that the length L(r,η, G) of the level curves |G(z)|= η lying in D(0, r ) satisﬁes 6+N /2 4 6 8 4 1/2 1 N L(s ,η, G) = O(s T (e s , G) ) = O(s ) ≤ s as n →∞, (5.4) n n n n using (5.2) and the fact that N > 12. Assume henceforth that n is large. Lemma 5.3 At least N of the domain D and paths σ , without loss of generality 8 j j D ,..., D and σ ,...,σ , are such that 1 N 1 N 8 8 −N | f (z) − a |≤ s for z ∈ σ with |z|≥ s /4. (5.5) j j n 123 Transcendental Singularities for a Meromorphic Function 127 Proof By Lemma 3.3, it may be assumed that, for j = 1,..., N , π dt > N log s , 8 n 1/2 t θ (t ) [4s ,s /4] j n n which, on combination with Lemma 5.2, leads to (5.5). 1/4 Lemma 5.4 Let w ,...,w be the zeros and poles of G in s ≤|z|≤ s , repeated 1 q n n n according to multiplicity. Then 4 4 N q ≤ n(s , 1/G) + n(s , G) = o s (5.6) n n n and there exist t , T satisfying n n 1/2 1/2 2 2 s − 1 ≤ t ≤ s , s ≤ T ≤ s + 1, (5.7) n n n n n n such that N +1 max{| log |G(z)|| : z ∈ S(0, t ) ∪ S(0, T )}≤ s . (5.8) n n −N −1 Proof (5.6) follows from (5.2). Let U be the union of the discs D(w , s ): these n q n −1 discs have sum of radii at most s and so since n is large there exist t , T satisfying n n (5.7) such that the circles S(0, t ), S(0, T ) do not meet U . Hence the Poisson–Jensen n n n formula gives (5.8). Lemma 5.5 Deﬁne sets E, K and L by E ={z ∈ C :|G(z)| <η} and n n K ={z ∈ C : t < |z| < T }, L ={z ∈ C : s /4 < |z| < 4s }. n n n n n n Then the number of components E of E ∩ K which meet L is at most s . q n n n Proof If the closure F of E lies in K , then E must contain a zero of G, whereas q q n q if F K then ∂ E ∩ K has arc length at least s /8. Thus the lemma follows from q n q n n (5.4) and (5.6). Lemma 5.6 Let u lie on σ with s /4 ≤|u|≤ 4s . Then, with d as in Lemma 3.2, j n n k there exists v on σ such that: −N |u|≤|v|≤|u|+ s ;| f (v) − a |≤| f (u) − a |; j j (k) k kN | f (v)|≤ k d s | f (u) − a |. (5.9) k j Proof Starting at u, follow σ in the direction in which | f (z) − a | decreases. Then j j −N σ describes an arc γ joining the circles S(0, |u|) and S(0, |u|+ s ), such that the j n ﬁrst two inequalities of (5.9) hold for all v ∈ γ . Since f maps D univalently onto D(1, r ), the inverse function H of f maps a proper sub-segment I of the half-open 123 128 J. K. Langley line segment J =[ f (u), a ) onto γ . Assume that the last inequality of (5.9) fails for all v ∈ γ . Then Lemma 3.2 yields, on I , −1 −N −1/k 1/k−1 |H (w)|≤ k s | f (u) − a | (r −|w − 1|) . j j Since 1, f (u) and a are collinear, a contradiction arises via −N −1 −N −1/k 1/k−1 3 3 s ≤ H (w)dw ≤ k s | f (u) − a | (r −|w − 1|) |dw| j j n n I I −1 −N −1/k 1/k−1 < k s | f (u) − a | (r −|w − 1|) |dw| j j −1 −N −1/k 1/k−1 = k s | f (u) − a | (r − t ) dt j j | f (u)−1| −N −1/k 1/k −N 3 3 = s | f (u) − a | (r −| f (u) − 1|) = s . j j n n Lemma 5.7 Let E be a component of E ∩ K which meets L , and suppose that p n n there exists j = j (p) such that E contains k points ζ ,...,ζ ∈ D , each with p 1 k j −N −N 7 3 | f (ζ ) − a |≤ s . Assume further that |ζ − ζ |≥ s for q = q . Then q j n q q n −N | f (z) − a |≤ s for all z ∈ E , and E ⊆ C (ε). j n p p Proof Let M = sup{| f (z)|: z ∈ E }; then M < +∞ since poles of f in C\{0} 0 p 0 are poles of G,by(5.3), and |G(z)|≤ η on the closure of E . Choose u ∈ E with p 0 p | f (u )|≥ M /2. There exists a polynomial P,ofdegreeatmost k − 1, such that 0 0 z k−1 (z − t ) (k) f (z) = P(z) + f (t ) dt on E . (k − 1)! The length of the boundary of E is at most 2s by (5.4). Hence each z ∈ E can be p n p joined to u by a path in the closure of E , of length at most 4s , and so 0 p n −N k−1 N −N 5 1 4 | f (z) − P(z)|≤ M ηt (2T ) 4s ≤ M s , (5.10) 0 n 0 n n n −N by (5.3) and (5.7). In particular, this gives |P(ζ ) − a |≤ (1 + M )s for each q. q j 0 n For z in E , Lagrange’s interpolation formula leads to z − ζ |P(z) − a |= (P(ζ ) − a ) j q j ζ − ζ q ν q=1 ν=q −N k−1 (k−1)N −N 4 3 3 ≤ k(1 + M )s (2T ) s ≤ (1 + M )s . (5.11) 0 n 0 n n n Setting z = u in (5.11) then delivers M ≤ 2|P(u )|≤ 2|a |+ o(1 + M ) and so 0 0 0 j 0 −N M ≤ 5. Now combining (5.10) with (5.11) yields | f (z) − a |≤ s and hence 0 j n | f (z) − 1| <ε on E . Since E meets D ⊆ C (ε), this gives E ⊆ C (ε). p p j p 123 Transcendental Singularities for a Meromorphic Function 129 For each j ∈{1,..., N } choose λ = s points u ,..., u on σ , each with 8 n j ,1 j ,λ j −N s /2 ≤|u |≤ s and such that |u |≥|u |+2s . Applying Lemma 5.6 with n j ,κ n j ,κ+1 j ,κ n −N u = u gives points v ∈ σ with s /2 ≤|u |≤|v |≤|u |+ s ≤ 2s j ,κ j ,κ j n j ,κ j ,κ j ,κ n n and, using (5.3), (5.5) and (5.9), −N N (k) −N 7 5 6 | f (v ) − a |≤ s , |G(v )|≤ 2|v | | f (v )|≤ s <η. (5.12) j ,κ j j ,κ j ,κ j ,κ n n −N These points v satisfy |v |≥|v |+ s , and each lies in a component of j ,κ j ,κ+1 j ,κ n N N 2 1 E ∩ K which meets L . Since there are s of these v for each j, but at most s n n n j ,κ n available components E by Lemma 5.5, it must be the case that for each j there are at least k points v lying in the same component E . Lemma 5.7 then implies that j ,κ p E ⊆ C (ε) and f (z) = a + o(1) on E . p j p Thus for j = 1,..., N the following exist: a component C = E ⊆ C (ε) of 8 j p E ∩ K which meets L and on which f (z) = a + o(1); a point v ∈ C such that, n n j j j by (5.12), −N s /2 ≤|v |≤ 2s , |G(v )|≤ s . (5.13) n j n j Since C ⊆ C (ε), the function log |1/G(z)| is subharmonic on C . Moreover, because j j j = j gives f (z) → a = a as z →∞ on σ ,the C are pairwise disjoint and j j j j none of them contains a circle S(0, t ) with t ∈[t , T ].For t > 0let φ (t ) be the n n j angular measure of C ∩ S(0, t ). Then (5.7) and [20, p. 116] give a harmonic measure estimate T /2 dt ω(v , C , S(0, T ) ∪ S(0, t )) ≤ c exp −π j j n n 1 t φ (t ) 2|v | |v |/2 dt +c exp −π , t φ (t ) 2t for j = 1,..., N , in which c is a positive absolute constant. By Lemma 3.3 and 8 1 −N (5.7), there exists at least one j for which ω(v , C , S(0, T ) ∪ S(0, t )) ≤ 2c s . j j n n 1 n For this choice of j the two constants theorem [18] delivers, using (5.8), (5.13) and the fact that |G(z)|= η on ∂C ∩ K , j n 1 1 −N +N +1 7 1 N log s ≤ log ≤ log + 2c s , 6 n 1 |G(v )| η a contradiction since n is large. 6 Proof of Theorem 1.4 This is almost identical to the corresponding proof in [2], but with Theorem 1.3 standing in for the Denjoy–Carleman–Ahlfors theorem. Suppose that f , k and α are 123 130 J. K. Langley as in the hypotheses, but there exists ε> 0 such that in the neighbourhood C (ε) of (k) the singularity the function f f has ﬁnitely many zeros which are not α-points of f : it may be assumed that there are no such zeros. On the other hand, because the (k) singularity is indirect, f must have inﬁnitely many α-points in C (ε). Since f / f has −1 ﬁnite lower order, f cannot have inﬁnitely many direct transcendental singularities over ﬁnite non-zero values, by Theorem 1.3. Set A(ε) ={w ∈ C : 0 < |w − α| <ε} if α ∈ C, with A(ε) ={w ∈ C :|w| > 1/ε} if α =∞. In either case, it may be −1 assumed that ε is so small that A(ε) ⊆ C\{0} and there is no w in A(ε) such that f has a direct transcendental singularity over w. −1 Take z ∈ C (ε), with f (z ) = w = α, and let g be that branch of f mapping 0 0 0 w to z .If g admits unrestricted analytic continuation in A(ε) then, exactly as in 0 0 [2], the classiﬁcation theorem from [18, p. 287] shows that z lies in a component C of the set {z ∈ C : f (z) ∈ A(ε) ∪{α}} which contains at most one point z with f (z) = α, so that C (ε) C . But any z ∈ C (ε) can be joined to z by a path λ on 0 1 0 which f (z) ∈ A(ε) ∪{α}, which gives λ ⊆ C and hence C (ε) ⊆ C , a contradiction. 0 0 Hence there exists a path γ :[0, 1]→ A(ε), starting at w , such that analytic continuation of g along γ is not possible. This gives rise to S ∈[0, 1] such that, as t → S−, the image z = g(γ (t )) either tends to inﬁnity or to a zero z ∈ C (ε) of f with f (z ) = γ(S) ∈ A(ε), the latter impossible by assumption. It follows that setting z = σ(t ) = g(γ (t )),for 0 ≤ t < S, deﬁnes a path σ tending to inﬁnity in C (ε), on which f (z) → w ∈ A(ε) as z →∞. But then there exists δ> 0 such that an unbounded subpath of σ lies in a component C ⊆ C (ε) of the set (k) {z ∈ C :| f (z) − w | <δ}, with δ so small that f f has no zeros on C . Further, the singularity over w must be indirect, since direct singularities over values in A(ε) have been excluded, and this contradicts Proposition 5.1. 7 A Result Needed for Theorem 1.6 Theorem 7.1 [16, Theorem 1] Let u be a subharmonic function in the plane such that −1 B(r ) = sup{u(z) :|z|= r } satisﬁes lim (log r ) B(r ) =+∞. Then there exist r →∞ δ > 0 and a simple path γ :[0, ∞) → C with γ(t ) →∞ as t →+∞ and the following properties: u(z) (i) lim =+∞; (ii) if λ> 0 then exp(−λu(z)) |dz| < ∞; z→∞,z∈γ log |z| (iii) if z = γ(t ) then u(γ (s)) ≥ δ u(z) for all s ≥ t . (7.1) Conclusion (iii) and the fact that γ may be chosen to be simple are not stated in [16, Theorem 1], but both are implicit in the proof. Here γ = γ ∪ γ ∪ ... is constructed 1 2 in [16, Section 3] so that, for some ﬁxed δ ∈ (0, 1), each γ :[k − 1, k]→ C 1 k is a simple path from a ∈ D to a ∈ ∂ D , where D is the component of k k k+1 k k −1 {z ∈ C : u(z)<(1 − δ ) u(a )} containing a .By[16, (3.2) and (3.3)], the γ 1 k 1 k −1 are such that 0 <δ u(a ) ≤ u(z)<(1 − δ ) u(a ) on λ = γ \{a } and 1 k 1 k k k k+1 −1 u(a ) ≥ (1 − δ ) u(a )> u(a ). Hence if z = γ(t ) ∈ λ , then u(γ (s)) ≥ k+1 1 k k k δ u(a ) ≥ δ (1 − δ )u(γ (t )) for all s ≥ t. If the whole path γ is not simple, take the 1 k 1 1 123 Transcendental Singularities for a Meromorphic Function 131 least k ≥ 2 such that = γ ∪ ... ∪ γ is not simple. Then there exists a maximal k 1 k t ∈[k − 1, k] such that u = γ (t ) lies in the compact set , and t < k since k k k−1 γ (k) = a ∈ ∂ D . Replacing by the part of from a to u , followed by k k+1 k k k−1 1 k the part of γ from u to a , does not affect conclusions (i), (ii) and (iii), and the k k k+1 argument may then be repeated. Theorem 7.1 leads to the following result. Proposition 7.1 Let N ∈ N and let A be a transcendental meromorphic function in the plane such that the inverse function of A has a direct transcendental singularity over 0. Then there exists a path γ tending to inﬁnity in C and linearly independent solutions U, V of w + A(z)w =0(7.2) on a simply connected domain containing γ , such that U and V satisfy, as z →∞ on γ , O(1) O(1) O(1) O(1) U (z) = z + , U (z) = 1 + , V (z) = 1 + , V (z) = . N N N N z z z z (7.3) To prove Proposition 7.1, observe ﬁrst that, as in the proof of Theorem 1.3, there exist a small positive δ and a non-empty component D of {z ∈ C :|A(z)| <δ} such that A(z) = 0on D, as well as a non-constant subharmonic function u on C given by u(z) = log (z ∈ D), u(z) = 0 (z ∈ / D). A(z) Then u satisﬁes the hypotheses of Theorem 7.1,by[4, Theorem 2.1], and so there exists a path γ :[0, ∞) → D as in conclusions (i), (ii) and (iii). In particular, (iii) implies that 1−δ δ 0 0 if z = γ(t ) then |A(γ (s))|≤ δ |A(z)| for all s ≥ t . (7.4) Choose a simply connected domain on which A has no poles, such that γ ⊆ .By −1/4 2 (7.1) it may be assumed that |A(t )| ≥|t | ≥ 4on γ , and that 2 2 1/2 1/4 |t | |A(t )||dt|≤ |t | |A(t )| |dt|≤ |A(t )| |dt | < . (7.5) γ γ γ Lemma 7.1 Let v be a solution of (7.2)on . Then v(z) = O(|z|) as z →∞ on γ . Proof This is a standard argument along the lines of Gronwall’s lemma. Let y be the starting point of γ . Differentiating twice shows that there exist constants a , b such 1 1 that, on , v(z) = a z + b − (z − t )A(t )v(t ) dt . 1 1 123 132 J. K. Langley If φ(z) = v(z)/z is unbounded on γ, there exist ζ →∞ on γ such that φ(ζ ) →∞ n n and |φ(t )|≤|φ(ζ )| on the part of γ joining y to ζ .If n is large then (7.5) delivers n 0 n a contradiction via |φ(ζ )| |φ(ζ )|≤|a |+|b |+|φ(ζ )| (1 +|t |)|tA(t )||dt|≤|a |+|b |+ . n 1 1 n 1 1 Lemma 7.2 (a) Let N ∈ N. Then on γ every solution v of (7.2) has v (z) = α z + β + (z − t )A(t )v (t ) dt,α ,β ∈ C, (7.6) j j j j j j the integration being from z to inﬁnity along γ . Moreover, v satisﬁes, as z →∞ on γ , O(1) O(1) v (z) − α z − β = ,v (z) − α = . (7.7) j j j j N N z z (b) If v ,v are linearly independent solutions of (7.2) on then |α |+|α | > 0 in 1 2 1 2 (7.6), and if α = 0 then β = 0. 2 2 Proof First, (7.6) follows from (7.5) and Lemma 7.1.Next, (7.1), (7.4)–(7.6) and Lemma 7.1 imply that, as z →∞ on γ , |v (z) − α z − β |≤|z| (1 +|t |)|A(t )|O(|t |) |dt | j j j O(1) (1−δ )/2 δ /2 1/2 0 0 ≤|z| δ |A(z)| (1+|t |)|A(t )| O(|t |) |dt|= , |v (z) − α |= A(t )v (t ) dt j j O(1) (1−δ )/2 δ /2 1/2 0 0 ≤ δ |A(z)| |A(t )| O(|t |) |dt|= . Finally, suppose that v ,v are linearly independent solutions of (7.2)on , but the 1 2 conclusion of (b) fails. Then v (z)v (z) − v (z)v (z) → 0as z →∞ on γ,by(7.7), 1 2 2 1 contradicting the fact that W (v ,v ) is a non-zero constant by Abel’s identity. 1 2 Now ﬁx linearly independent solutions v ,v of (7.2)on . Then α ,α cannot 1 2 1 2 both vanish in (7.6). On the other hand, it is possible to ensure that one of α ,α is 0, 1 2 by otherwise considering α v − α v . Hence it may be assumed that α = 1, while 2 1 1 2 1 α = 0 and β = 1. Now write U = v and V = v , so that Lemma 7.2 gives (7.3). 2 2 1 2 123 Transcendental Singularities for a Meromorphic Function 133 8 Proof of Theorem 1.6 Assume that f and S are as in the hypotheses, but that the inverse function of S has f f a direct transcendental singularity over 0. Then evidently so has that of A = S /2, and it is well known that (1.3) implies that f is locally the quotient of linearly independent solutions of (7.2). Now Proposition 7.1 gives linearly independent solutions U , V of (7.2) satisfying (7.3) on a path γ tending to inﬁnity. Moreover, h = U /V has the form h = T ◦ f , for some Möbius transformation T , and so h ∈ S, whereas h(z) ∼ z and zh (z)/h(z) = O(1) on γ , contradicting (3.5). 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Computational Methods and Function Theory – Springer Journals
Published: Sep 10, 2018
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