Derivatives of meromorphic functions of finite order

A result is proved concerning meromorphic functions of finite order in the plane such that all but finitely many zeros of the second derivative are zeros of the first derivative.


Introduction
The starting point of this paper is the following theorem from [8]. (1) Then f has finitely many poles.
Condition (1) is evidently satisfied if f has finite order. Theorem 1.1 fails for k = 1, as shown by simple examples, and for k ≥ 2 and infinite lower order, in which case an example is constructed in [7] with infinitely many poles, all simple, such that f (k) has no zeros at all. The result was inspired by the conjecture made by A.A. Gol'dberg, to the effect that for k ≥ 2 and a meromorphic function f in the plane, regardless of growth, the frequency of distinct poles of f is controlled by the frequency of zeros of f (k) , up to an error term which is small compared to the Nevanlinna characteristic. Yamanoi has now proved this conjecture in a landmark paper [12]; however, because of the error terms involved, his result does not imply Theorem 1.1 directly.
This paper is concerned with a generalisation of Theorem 1.1 in a different direction.
The assumption there that f (k) has finitely many zeros is a strong one, so that it is natural to ask whether it may be replaced by something less restrictive. A reasonable candidate is the condition that all but finitely many zeros of f (k) have the same image under f (k−1) , which may then be assumed to be 0, but the following example shows that this does not by itself imply that f has finitely many poles. Set f (z) = z − tan z, f ′ (z) = 1 − sec 2 z = − tan 2 z, f ′′ (z) = −2 tan z sec 2 z.
Here all zeros of f ′′ are zeros of f ′ and fixpoints of f , all zeros and poles of f ′ have the same multiplicity, and 1 is an asymptotic value of f ′ . More generally it may be observed that, for any even positive integer n, the antiderivative of tan n z is meromorphic in C. The example (2) shows that the following theorem, which evidently implies Theorem 1.1, is essentially sharp.

Theorem 1.2 Let k ≥ 2 be an integer and let f be a meromorphic function of finite lower order
in the plane with the following properties: (i) the zeros of f (k−1) have bounded multiplicities; (ii) all but finitely many zeros of f (k) are zeros of f (k−1) ; (iii) there exists M ∈ (0, +∞) such that if ζ is a pole of f of multiplicity m ζ then (1) holds; (iv) for each ε > 0, all but finitely many zeros z of f (k) satisfy either |f (k−2) (z)| ≤ ε|z| or Then f (k) has a representation f (k) = Re P with R a rational function and P a polynomial. In particular, f has finite order and finitely many poles, and f (k) has finitely many zeros.
It suffices to prove Theorem 1.2 for k = 2 and, as already noted, condition (iii) holds when f has finite order. If f is a meromorphic function of finite lower order in the plane satisfying condition (ii) of Theorem 1.2, with k = 2, then f ′ has finitely many critical values and so finitely many asymptotic values, by a result of Bergweiler and Eremenko [2] and its extension by Hinchliffe [6] to functions of finite lower order (see Section 3). Therefore Theorem 1.2 follows from the next result, which fails for infinite lower order, because of the same example from [7]  2 Lemmas needed for Theorem 1.3 Throughout this paper B(z 0 , r) will denote the disc {z ∈ C : |z − z 0 | < r} and S(z 0 , r) will be the circle {z ∈ C : |z − z 0 | = r}. The following results are both well known.  For each Λ > 0 the estimate holds for all z outside a union of discs having sum of radii at most 6Λ.

Critical points and asymptotic values
Suppose that the function h is transcendental and meromorphic in the plane, and that h(z) tends to a ∈ C as z tends to infinity along a path γ. Then a is an asymptotic value of h, and the inverse function h −1 has a transcendental singularity over a [2,10]. For each t > 0, let C(t) be that component of C ′ (t) = {z ∈ C : |h(z) − a| < t} which contains an unbounded subpath of γ. The singularity of h −1 over a corresponding to γ is called direct [2] if C(t), for some t > 0, contains no zeros of h(z) − a. Singularities over ∞ are classified analogously.
Recall next some standard facts from [10, p.287]. Suppose that G is a transcendental meromorphic function with no asymptotic or critical values in 1 < |w| < ∞. Then every component C 0 of the set {z ∈ C : |G(z)| > 1} is simply connected, and there are two possibilities. Either (i) C 0 contains one pole z 0 of G of multiplicity k, in which case G −1/k maps C 0 univalently onto B(0, 1), or (ii) C 0 contains no pole of G, but instead a path tending to infinity on which G tends to infinity. In case (ii) the function w = log G(z) maps C 0 univalently onto the right half plane. Suppose that G is a transcendental meromorphic function in the plane and that G ′ has no asymptotic or critical values w with 0 < |w| < d 1 < ∞. Let D be a component of the set {z ∈ C : |G ′ (z)| < d 1 } on which G ′ has no zeros, but such that D contains a path tending to infinity on which G ′ (z) tends to 0. If z 1 is in D and log |d 1 /G ′ (z 1 )| ≥ 1 then in which the positive constant S depends on G and D but not on z 1 .
Suppose next that the function F is meromorphic of finite lower order in the plane, and that all but finitely many zeros of F ′ are zeros of F . Then F has finitely many critical values. By Hinchliffe's extension [6] to the finite lower order case of a theorem of Bergweiler and Eremenko [2], the function F has finitely many asymptotic values. Furthermore, all asymptotic values of The mapping h 1 may then be extended to be quasiconformal on the plane, fixing infinity, and there exist a meromorphic function G and a quasiconformal mapping ψ such that h 1 • F = G • ψ on C. It follows that for j = 1, 2 all components of F −1 (B j ) are simply connected and all but finitely many are unbounded, since all but finitely many zeros z of G ′ have G(z) = 0.

Proof of Theorem 1.3: first part
Let the function f be as in the hypotheses. If f ′′ /f ′ is a rational function then f ′ is a rational function multiplied by the exponential of a polynomial, and so is f ′′ . Assume henceforth that f ′′ /f ′ is transcendental: then obviously so is f . Apply the reasoning and notation of Section 3, with F = f ′ . The following is an immediate consequence of Lemma 3.1. properties. There exist finitely many unbounded simply connected domains U n , each of which to infinity on which f ′ (z) tends to the finite asymptotic value b n . Here f ′ (z) = b n on U n and |f (z) − b n z| < ε 2 |z| for all large z in U n . If Γ is a path tending to infinity on which f ′ tends to a finite asymptotic value α, then there exists n such that α = b n and Γ \ U n is bounded. The b n need not be distinct, and some of them may be 0. Proof. This follows from the Schwarz reflection principle and the fact that each non-zero b p lies on the polygonal Jordan curve J = ∂B 1 but is not a vertex of J.
Definitions 4.1 Fix positive real numbers ρ, σ and τ with τ < s 1 < ε 1 and σ/τ and ρ/σ small. having the following properties. Let z 0 be large with |f ′ (z 0 )| < τ and assume that z 0 lies in a component C of (f ′ ) −1 (B 1 ) satisfying one of the following two conditions: (B) the function f ′ is univalent on C, and C ∩ U p and C ∩ U q are both non-empty, where U p and U q are as in Lemma 4. Then Proof. Observe that conditions (A) and (B) are mutually exclusive. Denote positive constants by c j and small positive constants by δ j ; these will be independent of z 0 and C. In case (A) there is exactly one point in C at which f ′′ vanishes, and it must be a zero of f ′ . In both cases f ′ (C) = B 1 (see Section 3), and C contains precisely one zero z 1 of f ′ , of multiplicity m ≤ c 1 , by hypothesis (i) of the theorem, with m = 1 in Case (B). There exist only finitely many components which are bounded or have a zero of f ′′ on their boundary, and if one of these contains a zero of f ′ then the set {z ∈ C 1 : |f ′ (z)| ≤ τ } is compact. Therefore since z 0 is large the component C is unbounded and simply connected and its boundary ∂C contains no zeros of (5). Then v 0 maps C univalently onto B(0, 1), and Let Γ be a component of ∂C. Then Γ is a simple curve tending to infinity in both directions and, as z tends to infinity in either direction along Γ, the image f ′ (z) must tend to a finite non-zero asymptotic value of f ′ ; this is because v 0 is univalent on C. Hence there exists z 1 lying close to Γ, such that z 1 ∈ C ∩ U n , for some U n as in Lemma 4.1, with b n = 0 and |f ′ (z 1 ) − b n | < ε 1 . By construction, b n lies on the polygonal Jordan curve J but is not a vertex of J. Thus analytic continuation of (f ′ ) −1 along a path in the semi-disc B(b n , ε 1 ) ∩ B 1 then gives a point z 2 ∈ C ∩ U n with |f ′ (z 2 ) − b n | < ε 1 , as well as |h 1 (f ′ (z 2 ))| ≤ 1 − δ 2 , which implies in turn that |v 0 (z 2 )| ≤ 1 − δ 3 .
Let G 0 : B(0, 1) → C be the inverse function of v 0 , and suppose that G ′ 0 (u 0 ) = o(|z 0 |). Then Koebe's distortion theorem implies that G ′ 0 (u) = o(|z 0 |) for |u| ≤ 1 − δ 3 . In Case (A) this gives a path γ in C, of length o(|z 0 |), joining z 3 = G 0 (0) to z 2 via z 0 , and with |f ′ (z)| ≤ c 2 on γ. Since z 0 is large so are z 2 and z 3 . Thus Lemma 4.1 and integration of f ′ yield But by the assumption of Case (A), f ′′ has a zero in C, which must be at z 3 , so that, by the hypotheses of the theorem, |f (z 3 ) − b n z 3 | ≥ κ|z 3 |. This contradicts (7), since ε 2 is small. Next, in Case (B) the above analysis may be applied twice, to give a path γ in C of length o(|z 0 |), on which |f ′ (z)| ≤ c 2 , such that γ joins points w p ∈ C ∩ U p and w q ∈ C ∩ U q via z 0 , where b p and b q are distinct and non-zero, and |f ′ (w j ) − b j | < ε 1 for j = p, q. Therefore the w j satisfy Since ε 2 is small and integration of , this case also delivers a contradiction.
It will be seen that hypothesis (i) of Theorem 1.3 plays a key role in the above proof of Lemma 4.3, principally by preventing z 0 from lying too close to the boundary of C.

Lemma 4.4 With the notation of Lemma 4.1 and Definitions 4.1, let z 1 be large and satisfy
and let C be the component of (f ′ ) −1 (B 1 ) with z 1 ∈ ∂C. Assume that one of the following two mutually exclusive conditions holds: (a) the function f ′ is not univalent on C; (b) the function f ′ is univalent on C, and C ∩ U q is non-empty, for some q with 0 = b q = b p .

Then there exists an open set H 1 , with
such that f ′ maps H 1 onto an open disc K 1 ⊆ B 1 , of diameter less than ρ, which is tangent to H 1 contains an open disc L 1 of radius M 4 |z 1 | on which (6) holds; here both M 3 and M 4 are independent of z 1 and C.
Proof. The component C is unique because z 1 is large and f ′′ has finitely many zeros which are not zeros of f ′ . As in Lemma 4.3 denote small positive constants by δ j , and positive constants by c j ; these will again be independent of z 1 and C. Let γ 0 be the straight line segment where δ 1 is chosen sufficiently small that |h 1 (w)| ≤ δ 1 implies that |w| ≤ δ 2 < τ < ε 1 . Using (8) and the conformal extension of h 1 to B 1 ∪ B(b p , s 1 ) given by Lemma 4.2, define domains in which δ 3 is small compared to δ 1 , which ensures that 0 ∈ E 1 . Then F 1 contains no singular values of the inverse function (f ′ ) −1 , and z 1 lies in a component D of (

Repeated application of the Koebe distortion theorem yields
Suppose first that G ′ 1 (u 1 ) = o(|z 1 |). Then z 2 ∼ z 1 and G ′ 1 (u 2 ) = o(|z 1 |), from which it follows that z 2 f ′′ (z 2 ) is large. Hence Hence f ′ is not univalent on C but C contains no zero of f ′′ . Thus C must contain a path Γ tending to infinity on which f ′ (z) tends to 0, and C meets one of the components U n with b n = 0. Moreover, log(h 1 • f ′ ) maps C univalently onto the left half plane (see Section 3). Therefore, since |h 1 (f ′ (z 2 ))| ≤ δ 1 , there exists a path Γ ′ in C joining z 2 to some z 3 ∈ Γ on which |h 1 (f ′ (z))| ≤ δ 1 and |f ′ (z)| < ε 1 , and hence z 2 ∈ U n . Since z 1 is large, and z 2 ∼ z 1 , On the other hand |f ′ (z)| ≤ c 4 on σ 1 , and so integration yields f (z 1 ) = f (z 2 ) + o(|z 1 |) and a contradiction.
5 The frequency of poles of f and zeros of f ′′ Lemma 5.1 Let w 1 , . . . , w Q be pairwise distinct poles of f with |w j | large. For 1 ≤ j ≤ Q let D j be the component of (f ′ ) −1 (B 2 ) in which w j lies. Then for each j there exists p j ∈ Z such that ∂D j contains a Jordan arc λ j which is mapped univalently by f ′ onto a line segment µ j of length at least σ, and these may be chosen so that where U p j and b p j are as in Lemma 4.1, while σ and τ are as in Definitions 4.1.
Moreover, if points z j are chosen such that z j ∈ λ j for 1 ≤ j ≤ Q, then each |z j | is large and for each j there exists an open disc L j ⊆ B z j , 1 2 |z j | of radius M 4 |z j |, on which (6) holds, where M 4 is as in Lemma 4.4. The L j are pairwise disjoint.
Proof. By the discussion in Section 3, each D j is unbounded and simply connected and the boundary ∂D j contains no zeros of f ′′ . Each component of ∂D j is a simple path tending to infinity in both directions, and there exists a component Γ j of ∂D j which separates w j from the point W 0 chosen in Definitions 4.1. Since D j contains a pole of f it follows that f ′ is finite-valent on D j . Thus as z tends to infinity in either direction along Γ j the image f ′ (z) must tend to a non-zero finite asymptotic value of f ′ . In particular, Γ j meets some U p as in Lemma 4.1 with b p = 0, and following Γ j while staying in U p gives λ j and µ j as in (10). Furthermore, each w j is large and, for any M 5 > 0, the disc B(0, M 5 ) meets only finitely many components of (f ′ ) −1 (B 2 ), each of which contains at most one pole of f . Hence if z j ∈ λ j then z j is large.
To prove the existence of the L j , choose for each j a component E j of (f ′ ) −1 (B 1 ) with Γ j ⊆ ∂E j . Since Γ j separates the pole w j of f from W 0 it follows that Γ j is not the whole boundary ∂E j . In particular, if f ′ is univalent on E j then Γ j must meet components U p and U q with b p and b q distinct and non-zero. Thus each of these components E j of (f ′ ) −1 (B 1 ) satisfies one of the conditions (a), (b) of Lemma 4.4, which may now be applied with z 1 replaced by each z j . This gives open sets H j ⊆ B z j , 1 2 |z j | ∩ E j , each containing an open disc L j of radius M 4 |z j | on which (6) holds. Moreover, f ′ maps H j onto a disc K j ⊆ B 1 which is tangent to J at f ′ (z j ) and has diameter less than ρ.
To show that the L j are disjoint, suppose that 1 ≤ j < j ′ ≤ Q and that H j ∩ H j ′ = ∅, from which it follows of course that K j ∩ K j ′ = ∅. Since ρ is small compared to σ and z j ∈ λ j , the open disc U = B(f ′ (z j ), 3ρ) contains no singular value of (f ′ ) −1 , by (10). But K j and K j ′ have diameter less than ρ, and so their closures lie in U. Thus H j and H j ′ both lie in the same component of (f ′ ) −1 (U), as do z j and z j ′ , which forces Γ j = Γ j ′ and gives a contradiction.
Proof. Assume that r is large and that A(1) contains Q = 2N distinct poles w 1 , . . . , w 2N of f , with φ(r) = o(N). For j = 1, . . . , Q let D j be the component of (f ′ ) −1 (B 2 ) in which w j lies, let q j be the multiplicity of the pole of f ′ at w j . Each D j is unbounded and simply connected and may be assumed not to contain the origin. Let v j = (h 2 • f ′ ) 1/q j , so that v j maps D j conformally onto B(0, 1), with v j (w j ) = 0.
For 0 < t < ∞ let θ j (t) be the angular measure of D j ∩ S(0, t). Let c denote positive constants, not necessarily the same at each occurrence, but not depending on r, L(r) or N. For m ∈ N the Cauchy-Schwarz inequality gives m 2 ≤ 2π m j=1 1/θ j (t) so that, as in [7], at least N of the D j have > cNL(r), It may be assumed after re-labelling if necessary that (12) holds for D 1 , . . . , D N . Since w j lies in A(1), it follows from Lemma 2.1 that .
Combining this with (11), (12) and condition (iii) of the theorem shows that ω(w j , σ j , D j ) = o(1/q j ) for j = 1, . . . , N, where σ j = ∂D j \ A(2). But Lemma 5.1 gives an arc λ j ⊆ ∂D j , mapped by f ′ onto a line segment µ j ⊆ J as in (10), of length at least σ. Since b p j in (10) is not a vertex of J, while τ is small, an application of the Schwarz reflection principle to h 2 shows that h 2 • f ′ maps λ j to an arc of S(0, 1) of length at least c, and v j (λ j ) has angular measure at least c/q j . The conformal invariance of harmonic measure under v j implies that λ j cannot be contained in σ j , and so there exists z j ∈ λ j ∩ A(2). The corresponding N pairwise disjoint discs 5.1 give a positive real number d 1 and w ∈ C with |w| = r arbitrarily large, such that (6) holds on the disc B(w, d 1 r). Let ε and K be positive, with ε small, and let Here K is chosen so large that the harmonic measure with respect to U K satisfies ω(z, S(0, 1/K) ∪ S(0, K), U K ) < ε for z ∈ U K , Denote by d j positive constants which are independent of r, ε, K and S. Standard estimates from [3] give a real number S = S r such that K < S < 2K and |h(z)| ≤ |z| d 2 for |z| = r S and |z| = rS, in which h(z) = zf ′′′ (z)/f ′′ (z) as in Lemma 5.5 and d 2 = ρ(f ) + 1. Let w 1 , . . . , w Q be the poles of h in r/S ≤ |z| ≤ rS. Applying Lemma 5.2 with L(r) = (log r) 1/2 shows that Q ≤ d 3 (log r) 1/2 .
On the annulus A given by r/S ≤ |z| ≤ rS set where M 3 is as in (6) and may be assumed to be at least 1, and the sum is empty if there are no poles w j . Then u is subharmonic on A, with u(z) ≤ 0 on the closure of B(w, d 1 r) by (6), and u(z) ≤ log |h(z)| ≤ d 2 log |z| ≤ d 2 log(2Kr) for z ∈ S(0, r/S) ∪ S(0, rS), by (15). Hence (14) and the monotonicity of harmonic measure yield u(z) ≤ εd 2 log(2Kr) for r 2 < |z| < 2r.
Since ε may be chosen arbitrarily small, while s is large, this gives a contradiction and the proof of Theorem 1.3 is complete. .
Remark. Hypothesis (iii) on the multiplicities of poles may not be really essential for Theorem 1.3 but it does play a key role in the above proof. If it is assumed merely that f has finite lower order, then techniques such as Pólya peaks should give annuli on which the analysis of Lemma 5.2 can be applied, but it seems difficult to ensure that these contain enough distinct poles of f that the discs on which (6) holds are not so remote that the method of Section 6 fails.