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Zeros of higher order logarithmicderivativesJ.D. Hinchliffe aa School of Mathematical Sciences , University of Nottingham , NG72RD, UKb Communicated by B. GustafssonPublished online: 22 Aug 2006.

To cite this article: J.D. Hinchliffe (2004) Zeros of higher order logarithmic derivatives, ComplexVariables, Theory and Application: An International Journal, 49:2, 79-86

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Complex Variables, Vol. 49, No. 2, pp. 79–86, 10 February 2004

Zerosof Higher OrderLogarithmic Derivatives

J.D.HINCHLIFFE*

School of Mathematical Sciences,University of Nottingham, NG7 2RD,UK

Communicatedby B.Gustafsson

(Received 3 February 2003; In final form 22 June 2003)

We extend a result of Langley and Shea [J.K. Langley and D. Shea (1998). On multiple points ofmeromorphic functions. J. London Math. Soc., 57(2), 371–384.] concerning the distribution of zeros of thelogarithmic derivative f 0/f to higher order logarithmic derivatives of the form f (k)/f, k 2 N.

Keyword: Nevanlinna theory

MSC 2000 Subject Classification: 30D35

1 INTRODUCTION

For a meromorphic function f [2], the zeros of the logarithmic derivative f 0/f are thosezeros of f 0 at which f is non-zero. Clunie et al., proved in [1] that if f is transcendentaland meromorphic of at most order 1/2, minimal type, then f 0/f has infinitely manyzeros, and further that if f is transcendental and entire of at most order 1, minimaltype, then f 0/f has infinitely many zeros. These bounds are sharp; for example, ez isentire with order 1 but mean type, and its logarithmic derivative is identically l, andtan2

ffiffiffiz

pis meromorphic with order 1/2 and mean type, but its logarithmic derivative

omits the value 0.It is natural to attempt to generalise these results to f (k)/f, where k� 2. However, this

situation is more complicated, because while a zero of f 0/f must correspond to a zero off 0 at which f is non-zero, a zero of f (k)/f, for k� 2, need not exclude any value of f.Langley considers the problem of higher order logarithmic derivatives in [5], obtaining:

THEOREM A Suppose that k is a positive integer and f is meromorphic of finite order inthe plane such that f (k)/f is transcendental of order less than 1/2. Then f (k)/f has infinitely

* Corresponding author. E-mail: james.hinchliffe@maths.nott.ac.uk

ISSN 0278-1077 print: ISSN 1563-5066 online � 2004 Taylor & Francis Ltd

DOI: 1080/02781070310001634575

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many zeros. The same conclusion holds if f has only finitely many poles and

limr!1

infTðr, f Þ

r¼ 0: ð1Þ

Thus far, we have only considered results pertaining to the number of zeros, not theirfrequency. For the k¼ 1 case, Langley and Shea prove the following in [6]:

THEOREM B Suppose d1 is an odd positive integer and that f is a transcendental mero-morphic function with T(r, f )¼O(rd1 ) as r ! 1, and that f satisfies (1). Then either

lim infr!1

Tðr, f Þ �r1=2

2

Z 1

r

Nðt, f Þ

t3=2dt

� �< 1

or

lim infr!1

Tðr, f 0=f Þ �r1=2

2

Z 1

r

Nðt, f =f 0Þ

t3=2dt�

5þ d1

2log r

� �< 1:

We note that (1) ensures that f 0/ f is transcendental, and that the conclusionsshow that either f has a lot of poles, or f 0/ f has a lot of zeros. Combining thisresult with ideas from the proof of Theorem 1 of [4], we generalise this result forhigher derivatives of f.

THEOREM 1 With f and d1 as in Theorem B, in particular such that f satisfies (1), letk 2 N be such that f (k)/f is transcendental. Then either

lim infr!1

Tðr, f Þ �r1=2

2

Z 1

r

Nðt, f Þ

t3=2dt� ðk� 1Þ log r

� �< 1 ð2Þ

or

lim infr!1

Tðr, f ðkÞ=f Þ �r1=2

2

Z 1

r

Nðt, f =f ðkÞÞ

t3=2dt�

5þ d1 þ 4k

2log r

� �< 1: ð3Þ

Note that we must explicitly assume f (k)/f to be transcendental, whereas (1) guaran-tees f 0/f to be transcendental. It is perhaps worth noting that the proof of Theorem 1obviates the need to employ a rather complex lemma with multiple conclusionswhich was essential in the original proof of Theorem B. The following corollaryperhaps elucidates the implications of Theorem 1 more readily.

COROLLARY Suppose that f is transcendental meromorphic in the plane, of finite order,with lower order �< 1. If f (k)/f has lower order at least �, then N(r, f )þN(r, f/f (k))has order at least min{�, 1/2}.

This work was carried out as part of a Ph.D. thesis, and as such the author wouldlike to acknowledge the numerous useful conversations with J.K. Langley whichcontributed to this result.

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2 PRELIMINARIES

LEMMA 2.1 ([3]) Suppose that f is transcendental and meromorphic in the planesuch that

mðr, f Þ � r1=2Z 1

r

nðt, f Þ

t3=2dt ! 1 ð4Þ

as r ! 1, and suppose that | f (z)|�M<1 on a path � tending to infinity. Then thereexists a function v, non-constant, non-negative and sub-harmonic in the plane, such thatvðzÞ � logþ j f ðzÞj þOð1Þ for all z. Property (4) suffices to ensure that 1 is an asymptoticvalue of f.

LEMMA 2.2 ([6]) Let f be transcendental and meromorphic in the plane, and supposethat d� 1 is such that T(r, f )¼O(rd ) for all large r. Denote by L (r,R, f ) thetotal length of the level curves | f (z)|¼R lying in the region |z|< r. Then there existarbitrarily large positive R such that f (z) has no critical values on the circle S(0,R)¼ {z: |z|¼R} and

Lðr,R, f Þ ¼ Oðrd2 Þ

as r ! 1, where d2¼ (3þ d )/2.

LEMMA 2.3 Let v be a non-constant sub-harmonic function in the plane, and let � be apath in C tending to infinity such that v<T on � for some positive constant T. Then

Mðr, vÞ

log r! 1:

LEMMA 2.4 Let v1 and v2 be non-constant sub-harmonic functions in the plane, let s1 ands2 be real constants, and let U1 and U2 be disjoint, unbounded domains such that vj� sj on@Uj and vj(zj)> sj for at least one point zj in Uj. Then

logBðr, v1Þ þ logBðr, v2Þ � 2 log r�Oð1Þ

for all large r, in which B(r, v)¼ sup{v(z): |z|¼ r}.

Both Lemmas 2.3 and 2.4 can be proved by a standard application of Tsuji’s estimatefor harmonic measure [8] (see also [6]).

LEMMA 2.5 Let f be transcendental and meromorphic such that T(r, f )¼O(rd1 ) asr ! 1 for some odd positive integer dl. Let k 2 N, set

hðzÞ ¼ f ðzÞ=f ðkÞðzÞ, ð5Þ

and assume h is transcendental. Let

h1ðzÞ ¼ hðzÞ=zN1 and f1ðzÞ ¼f ðzÞ

zk�1, ð6Þ

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where N1¼ d2þ 2kþ 1, d2¼ (3þ d1)/2 is the constant associated with d1 via Lemma 2.2.Suppose further that the constant T> 0 is such that L(r,T, h1)¼Oðrd2Þ as r ! 1 andthere are no critical values of h1 on S(0,T ) (arbitrarily large such T exist by Lemma2.2), and that C is an unbounded component of the set {z: |h1(z)|>T} such thatC � fz: jzj � Rg, for some R> 1. Then f1(z)¼O(1) and f (z)¼O(|z|k� 1) as z ! 1 in C.

Proof The proof is based on part of the proof of Theorem 1 in [5]. We partition theboundary @C of C into its intersections with the annuli Am¼ {z: 2m�1

� |z|<2m}. On @C,we have |h1|¼T, so on Am \ @C, we have |h|>Tð2m�1Þ

N1 , and hence |h|�1� c2�N1ðm�1Þ,

using c here and henceforth to denote a positive constant, not necessarily the same ateach occurrence. Now,

ZAm\@C

jtj2kjhðtÞj�1jdtj � cLð2m,T , h1Þð2mÞ

2k2�N1ðm�1Þ

� cð2mÞd222km2�N1ðm�1Þ

by assumption, and so

Z@C

jtj2kjhðtÞj�1jdtj �X1m¼1

cð2mÞd222km2�N1ðm�1Þ ¼ c2N1

X1m¼1

2�m < 1: ð7Þ

Fix a point z0 in C. Any point � in the closure C of C can be joined to z0 by a path �&consisting of a part of the circle |z|¼ |z0| and part of the ray arg z¼ arg �. Replacing anypart of �& which leaves �CC with an arc of @C gives |h1|�T on ��, and we now have

Z��

jtj2kjhðtÞj�1jdtj � c < 1: ð8Þ

To see this, consider the contributions made to the integral by the part �1 of the ray,the part �2 of the circle and the arcs �3 of @C, separately. We have

Z�1

jtj2kjhðtÞj�1jdtj � c

Z 1

jz0j

s2ks�N1 ds � c

for �l,

Z�2

jtj2kjhðtÞj�1jdtj � 2�jz0j

2kþ1�N1

T� c

for �2, and finally

Z�3

jtj2kjhðtÞj�1jdtj � c

for �3, using Eq. (7). Note that the constants are independent of �. We now set

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vðzÞ ¼

Z z

z0

ðz� tÞk�1

ðk� 1Þ!f ðkÞðtÞ dt� f ðzÞ,

and note that, by expanding out the (z� t)k�1 term, it can be seen that v admits unres-tricted analytic continuation in C. We differentiate to obtain

v0ðzÞ ¼

Z z

z0

ðz� tÞk�2

ðk� 2Þ!f ðkÞðtÞ dt� f 0ðzÞ,

and so on, up to

vðk�1ÞðzÞ ¼

Z z

z0

f ðkÞðtÞ dt� f ðk�1ÞðzÞ and vðkÞðzÞ ¼ f ðkÞðzÞ � f ðkÞðzÞ � 0:

Hence v(z) is a polynomial P(z) of degree at most k� 1. Now, near z0,

f ðzÞ þ PðzÞ ¼

Z z

z0

ðz� tÞk�1

ðk� 1Þ!f ðkÞðtÞ dt ¼ uðzÞ, ð9Þ

i.e. fþP� u� 0 near z0.Now take any � in C, and suppose we have a path � joining z0 to � in C. Consider

the function f�P� u, which is identically zero on a neighbourhood of z0, and canbe analytically continued along � to �. Hence, since f and P are both single-valued,so too is u, and Eq. (9) holds throughout C.

We now estimate the function f on C. First note that since C � fz: jzj � R > 1g andthe degree of P is at most k� l, we have

PðzÞ

zk�1

�������� � c

on C. Now, for � 2 C, we parametrize �� with respect to arc length s. Since f1(z)¼f(z)/zk� 1, it follows from (9) that

j f1ð��ðsÞÞj � cþ

Z s

0

ð��ðsÞ � ��ðtÞÞk�1��ðtÞ

k�1f1ð��ðtÞÞ

ðk� 1Þ!��ðsÞk�1hð��ðtÞÞ

���������� dt, ð10Þ

and, using

j��ðsÞ � ��ðtÞÞk�1

j �Xk�1

p¼0

k� 1p

� �j��ðsÞ

p��ðtÞk�1�p

j � kðk� 1Þ! j��ðsÞk�1��ðtÞ

k�1j,

we arrive at

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j f1ð��ðsÞÞj � WðsÞ ¼ cþ

Z s

0

kj��ðtÞ2k�2f1ð��ðtÞÞhð��ðtÞÞ

�1j dt:

Hence,

dW

ds� kj��ðsÞj

2k�2j f1ð��ðsÞÞjjhð��ðsÞÞj�1 � kj��ðsÞj

2k�2jhð��ðsÞÞj�1WðsÞ

and so

WðsÞ � c exp

Z s

0

kj��ðtÞj2k�2jhð��ðtÞÞj

�1 dt

� �� c,

using (8), which gives f1(z)¼O(1),

f ðzÞ ¼ Oðjzjk�1Þ

for z 2 C:

3 PROOF OF THEOREM 1

We assume the existence of a transcendental meromorphic function f of finite orderas in the hypothesis, in particular satisfying (1) and such that, for some k 2 N, f ðkÞ=ftranscendental and meromorphic. We set h, h1 and f1 as in (5) and (6), and notethat Nl¼ (5þ d1þ 4k)/2.

We begin by assuming both (2) and (3) are false. Then

Tðr, f Þ �r1=2

2

Z 1

r

Nðt, f Þ

t3=2dt� ðk� 1Þ log r ! 1 ð11Þ

and

Tðr, f ðkÞ=f Þ �r1=2

2

Z 1

r

Nðt, f =f ðkÞÞ

t3=2dt�

5þ d1 þ 4k

2log r ! 1 ð12Þ

as r ! 1, and straightforward integration by parts gives

mðr, f1Þ � r1=2Z 1

r

nðt, f1Þ

t3=2dt ! 1 ð13Þ

and

mðr, h1Þ � r1=2Z 1

r

nðt, h1Þ

t3=2dt ! 1, ð14Þ

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respectively. Hence, by Lemma 2.1, we have paths �fl and �h1 tending to infinity onwhich f1 and h1 tend to infinity respectively.

Choose r0 2 ½1, 2� such that h1 is finite on |z|¼ r0, and choose R¼T1>M(r0, h1)large satisfying the conclusions of Lemma 2.2. By the existence of �h1 , there exists anunbounded component, V, say, of the set {z: |h1(z)>T1}, and V is contained in{z: |z|� r0} by the choice of T1. We consider two cases:

Case (i) @V has no unbounded component. Then, by the choice of T1, @V consistsof countably many Jordan curves, and V is the only unbounded component of{z: |h1(z)|>T1}. By Lemma 2.5, f1(z)¼O(1) as z ! 1 in V, and this contradicts theexistence of the path �f1

Case (ii) @V has an unbounded component. If this is so, we have a path tendingto infinity on which |hl (z)|¼T1, and this, together with (14), enables us to useLemma 2.1 to obtain a function vhl, sub-harmonic, non-constant and non-negative inthe plane, such that

vh1 ðzÞ � logþ jh1ðzÞj þOð1Þ ð15Þ

for all z. By Lemma 2.3 and a result of Lewis et al. [7], there exists a path �1 tending toinfinity such that

vh1 ðxÞ

log jzj! þ1 ð16Þ

as z tends to infinity on �1, and

Z�1

e��vh1 ðzÞjdzj < 1 ð17Þ

for all �>0. Equation (15) implies that

log jh1ðzÞj

log jzj! þ1

on �1 also. So there exists an unbounded component Ul of {z: |h1 (z)|>Tl } such that�1\Ul is bounded. On U1, we have by Lemma 2.5 that f1 is bounded by some finite posi-tive constant, M, say, and vh1 is unbounded since �1 \U1 is unbounded. However,vh1 � logT1 þOð1Þ � M1, say, on �U1. In particular, j f1ðzÞj � M as z ! 1, on �1,and so there exists, by Lemma 2.1, a non-constant, non-negative sub-harmonic functionvf1, such that

vf1 ðzÞ � logþ jf1ðzÞj þOð1Þ:

Lewis et al. results [7] show there is a path �2 tending to infinity on which vf1 tends toinfinity. On U1, clearly vf1 � M2, say, so we take an unbounded component U2 of theset {z: vf1 >M2þ 1} (such a component must exist by the existence of the path �2) and

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we have immediately that U1 and U2 are disjoint. We therefore have the requiredconditions for the employment of Lemma 2.4. We obtain, for large r,

logBðr, vh1 Þ þ logBðr, vf1 Þ � 2 log r�Oð1Þ, ð18Þ

where B(r,v)¼ sup{v(z): |z|¼ r}. Using a standard result for sub-harmonic functions, wehave

Bðr, vh1Þ �3

2�

Z 2�

0

vh1ð2rei�Þ d� � 3mð2r, hÞ þOð1Þ � OðTð2r, f ÞÞ,

with a similar result holding for B(r, vf1 ). We therefore obtain from (18) that

2 logTð2r, f Þ � 2 log r�Oð1Þ,

so

Tð2r, f Þ

r� c,

where the constant is positive. However, this contradicts (1) and so at least one of (2) or(3) holds.

References

[1] J. Clunie, A. Eremenko and J. Rossi (1993). On equilibrium points of logarithmic and Newtonian poten-tials. J. London Math. Soc., 47(2), 309–320.

[2] W.K. Hayman (1964). Meromorphic Functions. Clarendon Press, Oxford.[3] W.K. Hayman (1978). On Iversen’s theorem for meromorphic functions with few poles. Acta.Math., 141,

115–145.[4] J.K. Langley (2001). The second derivative of a meromorphic function. Proc. Edin. Math. Soc., 44,

455–478.[5] J.K. Langley (1998). On the Zeros of f (k)/f Complex Variables, 37, 385–394.[6] J.K. Langley and D. Shea (1998). On multiple points of meromorphic functions. J. London Math. Soc.,

57(2), 371–384.[7] J. Lewis, J. Rossi and A. Weitsman (1984). On the growth of subharmonic functions along paths. Ark.

Mat., 22(1), 109–119.[8] M. Tsuji (1959). Potential Theory in Modern Function Theory. Maruzen, Tokyo.

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