Transcendental singularities for a meromorphic function with logarithmic derivative of finite lower order

In this note it is shown that two key results on transcendental singularities for meromorphic functions of finite lower order have refinements which hold under the weaker hypothesis that the logarithmic derivative has finite lower order.


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 t>0 C(t) is empty [2]. The path γ then determines a transcendental singularity of the inverse function f −1 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 singularity of f −1 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 1 > 0 such that C(t) contains no α-points of f for 0 < t < t 1 . 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 of f , for every t > 0, then the singularity is called indirect: a well-known example is given by f (z) = z −1 sin z, with α = 0 and γ the positive real axis R + . Transcendental singularities of f −1 over ∞ and their corresponding neighbourhoods may be defined and classified using 1/ f , and the asymptotic and critical values of f together comprise the singular values of f −1 .
If f has finite (lower) order of growth, as defined in terms of the Nevanlinna characteristic function T (r , f ) [8,18], then the number of direct singularities is controlled by the celebrated Denjoy-Carleman-Ahlfors theorem [9,18]. 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 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 f f − 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 and 1.2 hold under the weaker hypothesis that f (k) / f has finite lower order for some k ∈ N = {1, 2, . . .}. The obvious example f (z) = exp(exp(z)) shows that f / f can have finite order despite f having infinite lower order; here, f −1 has infinitely many direct (indeed logarithmic) singularities over 0 and ∞, and one over 1. Furthermore, if k ∈ N and A k is a transcendental entire function, then the lemma of the logarithmic derivative [8] shows that every non-trivial solution of has infinite lower order, even if A k has finite order. Clearly, each of exp(exp(z)) and exp(z −1 sin z) satisfies an equation of form (1.1) with coefficient of finite order. Note further that if f is a transcendental meromorphic function in the plane and f / f has finite lower order, then it is easy to prove by induction that so has A k = f (k) / f for every k ≥ 1, using the formula A k+1 = A k A 1 + A k , whereas the example shows that f / f can have finite order despite f / f having infinite lower order.
(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 meromorphic 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 k 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.  [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 meromorphic function in the plane with transcendental Schwarzian derivative  [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 transcendental singularity over a finite value: write

Conclusion (a) is a result of Elfving
so that S −1 g 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 f 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 f . Then the inverse function of S f does not have a direct transcendental singularity over 0.
The example f (z) = tan 2 √ z from [5] shows that for f ∈ S it is possible for 0 to be an asymptotic value of S f . Here direct computation shows that f (z)/ f (z) tends to 0 as z → ∞ in the left half plane, and so does S f (z).
The author thanks the referees for their helpful comments.
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 and f / f is even, f has distinct asymptotic values e ±iα , approached as z tends to infinity along the imaginary axis. As f / f has finite order and f has no finite nonzero critical values, both of these singularities of f −1 are direct by Theorem 1.4.

Example 2 Define g by
The zeros of cos √ z occur where √ z = b n = (2n +1)π/2, with n ∈ Z, and the residue of A 1 at b 2 n 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 that g has a non-zero real asymptotic value α, and g −1 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).
Then F(z) tends to 0 along R + and this singularity of F −1 is evidently indirect.

Example 4 Define entire functions
Then there exists α ∈ R + such that v(x) → exp(±α) as x → ±∞ on R and, since A 1 does not satisfy (1.2), Theorem 1.3 implies that v −1 has no direct singularities over finite non-zero values. Because all critical points of v are real, all but finitely many of them belong to neighbourhoods of the indirect singularities over exp(±α), and so v −1 has no other indirect singularities, by Theorem 1.4. Thus applying [18, p. 287] again, in conjunction with Iversen's theorem, shows that v −1 has logarithmic singularities over the omitted values 0 and ∞.

Example 5
Let h(z) = exp(sin z − z), so that A 1 = 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 axis, with h (2π n) = 0 for all n ∈ Z, these singularities of h −1 are direct but not logarithmic.

Preliminaries
The following well-known estimate may be found in Theorem 8.9 of [9].
Lemma 3.1 [9] Let D 1 , . . . , D n be n ≥ 2 pairwise disjoint plane domains. If u 1 , . . . , u n are non-constant subharmonic functions on C such that u j vanishes outside D j , then 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.
Proof Assume that a = 0 and initially that R = 1. It is clear that (3.2) holds for k = 1, with V 1 (w) = 1. If (3.2) holds for k, then it follows that .
, applying Cauchy's estimate for derivatives to V k proves the lemma by induction when R = 1.
In the general case write w = h(z) = R H(z) = Rv and z = F(w) = G(v) so that, as |w| → R−, Proof This is a standard application as in [9,Ch. 8] or [2] of the Cauchy-Schwarz inequality, which gives Let f be a transcendental meromorphic function in the plane such that f −1 has n ≥ 1 direct singularities over (not necessarily distinct) finite non-zero values a 1 , . . . , a n . Let k ∈ N; then A k = f (k) / f does not vanish identically. There exist a small positive δ and non-empty components D j of {z ∈ C : | f (z) − a j | < δ}, for j = 1, . . . n, such that f (z) = a j on D j , so that D j immediately qualifies as a direct tract for g j = δ/( f − a j ) in the sense of [4, Section 2]. Here δ may be chosen so small that if n ≥ 2 then these D j are pairwise disjoint. For each j, define a non-constant subharmonic function u j on C by  second, for each j there exists z j with |z j | = s and u(z j ) = B(s, u j ) such that
Combining (4.1) with (4.2) for j = 1 leads to (1.2). To prove the remaining assertions it may be assumed that A k has finite lower order μ. Choose a positive sequence (r m ) tending to infinity such that (4.5) Let m be large and let w 1 , . . . , w q m be the zeros and poles of A k in r m /4 ≤ |z| ≤ 4r m , repeated according to multiplicity: then (4.5) and standard estimates yield Let U m be the union of the discs D(w j , r −μ m ). Since the sum of the radii of the discs of U m is o(r m ) by (4.6), there exists a set E m ⊆ [r m /2, 2r m ], of linear measure at least r m , and so logarithmic measure l m ≥ 1/2, such that for r ∈ E m the circle |z| = r does not meet U m . A standard application of the Poisson-Jensen formula [8] on the disc |ζ | ≤ 4r m then yields Since m is large and l m ≥ 1/2, there exists s m ∈ E m \F 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 m ≥ r m /2, yields points z j with |z j | = s m and, for at least one j, On combination with (4.7), this forces 2μ ≥ n.

Indirect Singularities
Proposition 5.1 Let f be a transcendental meromorphic function in the plane such that f (k) / f has finite lower order μ for some k ∈ N. Assume that f −1 has an indirect transcendental singularity over α ∈ C\{0}. Then for each δ > 0, the neighbourhood C(δ) of the singularity contains infinitely many zeros of f f (k) .
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 ε > 0, contains finitely many zeros of f f (k) . It may be assumed that α = 1, and that C(ε) contains no zeros of f f (k) . Choose positive integers N 1 , N 2 , . . . , N 9 with 5μ + 12 < N 1 and N j+1 /N j large for each j.

Lemma 5.1
For each j ∈ {1, . . . , N 9 } there exist z j ∈ C(ε) and a j ∈ C with 0 < r j = |1 − a j | < ε/2, as well as a simply connected domain D j ⊆ C(ε), with the following properties. The a j are pairwise distinct and the D j pairwise disjoint. Furthermore, the function f maps D j univalently onto D(1, r j ), with z j ∈ D j and f (z j ) = 1. Moreover, 0 / ∈ D j but D j contains a path σ j tending to infinity, which is mapped by  f onto the half-open line segment [1, a j ), with f (z) → a j as z → ∞ on σ j . This is proved exactly as in [2]. If 0 < T j < ε/2 and z j ∈ C(T j ) is such that f (z j ) = 1, let r j be the supremum of t > 0 such that the branch of f −1 mapping 1 to z j admits unrestricted analytic continuation in D (1, t). Then r j < T j because f is not univalent on C(T j ), and there is a singularity a j of f −1 with |1 − a j | = r j ; moreover, a j must be an asymptotic value of f . The z j and T j are then chosen inductively: for the details see [2] (or [13,Lemma 10.3]).

Lemma 5.2 Let the z j , a j , σ j and D j be as in Lemma 5.1. For t > 0, let tθ j (t) be the length of the longest open arc of S(0, t) which lies in D j . Then f satisfies, as z tends to infinity on σ j ,
. D(1, r j ) onto D j . For z ∈ σ j , the distance from z to ∂ D j is at most |z|θ j (|z|). Thus Koebe's quarter theorem [10,Ch. 1] implies that Hence, for large z ∈ σ j and w = f (z), Since N 1 > 5μ, there exists a positive sequence (s n ) tending to infinity such that 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 as n → ∞, (5.4) using (5.2) and the fact that N 1 > 12. Assume henceforth that n is large. and there exist t n , T n satisfying such that Proof (5.6) follows from (5.2). Let U n be the union of the discs D(w q , s −N 1 −1 n ): these discs have sum of radii at most s −1 n and so since n is large there exist t n , T n satisfying (5.7) such that the circles S(0, t n ), S(0, T n ) do not meet U n . Hence the Poisson-Jensen formula gives (5.8).
Then the number of components E q of E ∩ K n which meet L n is at most s N 1 n .
Proof If the closure F q of E q lies in K n , then E q must contain a zero of G, whereas if F q K n then ∂ E q ∩ K n has arc length at least s n /8. Thus the lemma follows from (5.4) and (5.6).

Lemma 5.6
Let u lie on σ j with s n /4 ≤ |u| ≤ 4s n . Then, with d k as in Lemma 3.2, there exists v on σ j such that: Proof Starting at u, follow σ j in the direction in which | f (z) − a j | decreases. Then σ j describes an arc γ joining the circles S(0, |u|) and S(0, |u| + s −N 3 n ), such that the first two inequalities of (5.9) hold for all v ∈ γ . Since f maps D j univalently onto D(1, r j ), the inverse function H of f maps a proper sub-segment I of the half-open For each j ∈ {1, . . . , N 8 } choose λ = s N 2 n points u j,1 , . . . , u j,λ on σ j , each with s n /2 ≤ |u j,κ | ≤ s n and such that |u j,κ+1 | ≥ |u j,κ |+2s −N 3 n . Applying Lemma 5.6 with u = u j,κ gives points v j,κ ∈ σ j with s n /2 ≤ |u j,κ | ≤ |v j,κ | ≤ |u j,κ | + s −N 3 n ≤ 2s n and, using (5.3), (5.5) and (5.9), These points v j,κ satisfy |v j,κ+1 | ≥ |v j,κ | + s −N 3 n , and each lies in a component of E ∩ K n which meets L n . Since there are s N 2 n of these v j,κ for each j, but at most s N 1 n available components E p by Lemma 5.5, it must be the case that for each j there are at least k points v j,κ lying in the same component E p . Lemma 5.7 then implies that E p ⊆ C(ε) and f (z) = a j + o(1) on E p . Thus for j = 1, . . . , N 8 the following exist: a component C j = E p j ⊆ C(ε) of E ∩ K n which meets L n and on which f (z) = a j + o(1); a point v j ∈ C j such that, by (5.12), (5.13) Since C j ⊆ C(ε), the function log |1/G(z)| is subharmonic on C j . Moreover, because j = j gives f (z) → a j = a j as z → ∞ on σ j , the C j are pairwise disjoint and none of them contains a circle S(0, t) with t ∈ [t n , T n ]. For t > 0 let φ j (t) be the angular measure of C j ∩ S(0, t). Then (5.7) and [20, p. 116] give a harmonic measure estimate for j = 1, . . . , N 8 , in which c 1 is a positive absolute constant. By Lemma 3.3 and (5.7), there exists at least one j for which ω(v j , C j , S(0, T n ) ∪ S(0, t n )) ≤ 2c 1 s −N 7 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 j ∩ K n , a contradiction since n is large.

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 as in the hypotheses, but there exists ε > 0 such that in the neighbourhood C(ε) of the singularity the function f f (k) 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 singularity is indirect, f must have infinitely many α-points in C(ε). Since f (k) / f has finite lower order, f −1 cannot have infinitely many direct transcendental singularities over finite non-zero values, by Theorem 1.
In either case, it may be assumed that ε is so small that A(ε) ⊆ C\{0} and there is no w in A(ε) such that f −1 has a direct transcendental singularity over w. Take z 0 ∈ C(ε), with f (z 0 ) = w 0 = α, and let g be that branch of f −1 mapping w 0 to z 0 . If g admits unrestricted analytic continuation in A(ε) then, exactly as in [2], the classification theorem from [18, p. 287] shows that z 0 lies in a component C 0 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 0 . But any z 1 ∈ C(ε) can be joined to z 0 by a path λ on which f (z) ∈ A(ε) ∪ {α}, which gives λ ⊆ C 0 and hence C(ε) ⊆ C 0 , a contradiction.
Hence there exists a path γ : [0, 1] → A(ε), starting at w 0 , 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 2 ∈ C(ε) of f with f (z 2 ) = 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 1 ∈ 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 {z ∈ C : | f (z) − w 1 | < δ}, with δ so small that f f (k) has no zeros on C . Further, the singularity over w 1 must be indirect, since direct singularities over values in A(ε) have been excluded, and this contradicts Proposition 5.1.

Proof of Theorem 1.6
Assume that f and S f are as in the hypotheses, but that the inverse function of S f has a direct transcendental singularity over 0. Then evidently so has that of A = S f /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).