be the sequence of real poles of L enumerated in increasing order.Then for |k|large,a k and a k+1are real and of the same sign,and are both simple poles of L with positive residue,so that there is at least one zero of L in the interval(a k,a k+1).We choose one such zero in each such interval and denote it by b k.Then we set,for some large k0,
ψ(z)= |k|≥k01?z/b k
a k ?
1
a k?z
<πand so Imψ(z)>0.Finally,we de?neφby(2),and properties(a)–(d)follow(for the details see[11,18,24]).
Theorem1.3Let L be a function meromorphic in the plane,real on the real axis,such that all but?nitely many poles of L are real and simple and have positive residues.Letψ,φbe as in(2)and(a),(b),(c),(d).Ifφis transcendental then L+L′/L has in?nitely many non-real zeros.
Theorem1.3is proved in§3and§4,while Theorem1.2is deduced from Theorem1.3in§5.
We thank Iosif Ostrovskii,Terry Sheil-Small and Misha Sodin for their valuable comments.
5
2Preliminaries
We will require the following well known consequence of Carleman’s estimate for harmonic measure.
Lemma2.1Let u be a non-constant continuous subharmonic function in the plane.For r>0let B(r,u)=max{u(z):|z|=r},and letθ(r)be the angular measure of that subset of the circle C(0,r)={z∈C:|z|=r} on which u(z)>0.De?neθ?(r)byθ?(r)=θ(r),except thatθ?(r)=∞if u(z)>0on the whole circle C(0,r).Then if r>2r0and B(r0,u)>1we have
log u+(4re iθ) ≥log B(2r,u)?c1≥ r2r0πdt
2π π?πmax{u(re iθ),0}dθ.
The?rst inequality follows from Poisson’s formula,and for the second we refer to[26,Thm III.68].Note that in the case that u=log|f|where f is an entire function, u+(re iθ) coincides with the Nevanlinna characteristic T(r,f).
Next,we need the characteristic function in a half-plane as developed by Tsuji[25]and Levin and Ostrovskii[18](see also[7]for a comprehensive treatment).Let g be a meromorphic function in a domain containing the closed upper half-plane
t2
dt,r≥1.
The Tsuji characteristic is de?ned as
T(r,g)=m(r,g)+N(r,g),
where
m(r,g)=1
r sin2θ
dθ.
6
The upper half-plane is thus exhausted by circles of diameter r≥1tangent to the real axis at0.For non-constant g and any a∈C the?rst fundamental theorem then reads[7,25]
T(r,g)=T(r,1/(g?a))+O(1),r→∞,(3) and the lemma on the logarithmic derivative[18,p.332],[7,Theorem5.4] gives
m(r,g′/g)=O(log r+log+T(r,g))(4) as r→∞outside a set of?nite measure.Further,T(r,g)di?ers from a non-decreasing function by a bounded additive term[25].Standard inequalities give
T(r,g1+g2)≤T(r,g1)+T(r,g2)+log2,T(r,g1g2)≤T(r,g1)+T(r,g2),(5) whenever g1,g2are meromorphic in
H,and for r≥1set
1
m0π(r,Q)=
dr≤ ∞R m(r,Q)
r3
H,with g(k)≡1.Then T(r,g)≤ 2+1g + 2+2g(k)?1 +
O(log r+log+T(r,g))
as r→∞outside a set of?nite measure.
7
Lemma2.3is established by following Hayman’s proof exactly,but using the Tsuji characteristic and the lemma on the logarithmic derivative(4).
We also need the following result of Yong Xing Gu(Ku Yung-hsing,[14]).
Lemma2.4For every k∈N,the meromorphic functions g in an arbitrary domain with the properties that g(z)=0and g(k)(z)=1form a normal family.
A simpli?ed proof of this result is now available[27].It is based on a rescaling lemma of Zalcman–Pang[20]which permits an easy derivation of Lemma2.4 from the following result of Hayman:Let k∈N and let g be a meromorphic function in the plane such that g(z)=0and g(k)(z)=1for z∈C.Then g=const,see[8]or[9,Corollary of Thm3.5].
3Proof of Theorem1.3
Let L,ψ,φbe as in the hypotheses,and assume thatφis transcendental but L+L′/L has only?nitely many non-real zeros.Condition(b)implies the Carath′e odory inequality:
1
r <|ψ(re iθ)|<5|ψ(i)|
r
Sinceφhas?nitely many poles and is real on the real axis there exist a real entire functionφ1and a rational function R1with
φ=φ1+R1,R1(∞)=0.(9) Lemma3.2The entire functionφ1has order at most1.
Proof.Again,this proof is almost identical to the corresponding argument in[18].Lemmas2.2and3.1give
∞R m0π(r,L)
r2
dr=O(R?1log R),R→∞.
Since m0π(r,1/ψ)=O(log r)by(8),we obtain using(2)
∞
R
m0π(r,φ)
2<|z|<2,
δ2
2
}.
For r≥r0,with r0large,let g r(z)=1/(rL(rz)).Then g r(z)=0on?0, provided r0is large enough,since all but?nitely many poles of L are real. Further,
g′r(z)=?L′(rz)/L(rz)2.
9
Since L has?nitely many poles in H and L+L′/L has?nitely many zeros in H it follows that provided r0is large enough the equation g′r(z)=1has no solutions in?0.Thus the functions g r(z)form a normal family on?0,by Lemma2.4with k=1.
Suppose that|w0|=r≥r0,andδ1≤arg w0≤π?δ1,and that
|w0L(w0)|≤K.(11) Then
|g r(z0)|≥1/K,z0=
w0
|ψ(i)|sinδ2.
(13)
Thus(11)implies(13).For t≥r0let
E2(t)={w∈C:|w|=t,|φ1(w)|>K2}.
Further,letθ(t)be the angular measure of E2(t),and as in Lemma2.1let θ?(t)=θ(t),except thatθ?(t)=∞if E2(t)=C(0,t).Let
E3={t∈[r0,∞):θ(t)≤4δ2}.
Since(11)implies(13),we have(10)for t∈[r0,∞)\E3.Applying Lemma2.1 we obtain,sinceφ1has order at most1by Lemma3.2,
(1+o(1))log r≥ r r0πdt4δ2t,
from which it follows that E3has upper logarithmic density at most4δ2/π. Sinceδ2may be chosen arbitrarily small,the lemma is proved.
10
The estimates(8)and(10)and the fact thatφis real now give |φ(z)|>
K sinδ1
L(z),F′(z)=1+
L′(z)
Lemma3.6The function F has?nitely many critical points over C\R,i.e. zeros z of F′with F(z)non-real.
Lemma3.7There existsα∈H with the property that F(z)→αas z→∞along a pathγαin H.
Lemma3.7is a re?nement of Theorem4of[24],and will be proved in§4.
Now set
g(z)=z2L(z)?z=zF(z)
F(z)?α
,(15)
in whichαis as in Lemma3.7.Then g has?nitely many poles in H and(5), (14)and Lemma3.1give
T(r,g)+T(r,h)=O(log r),r→∞.
Hence Lemma2.2leads to
∞1m0π(r,g)
r3
dr<∞,(16)
in which m0π(r,g)and m0π(r,h)are as de?ned in(6).
Lemma3.8The function F has at most four?nite non-real asymptotic values.
Proof.Assume the contrary.Since F(z)is real on the real axis we may take distinct?nite non-realα0,...,αn,n≥2,such that F(z)→αj as z→∞along a simple pathγj:[0,∞)→H∪{0}.Here we assume thatγj(0)=0, thatγj(t)∈H for t>0,and thatγj(t)→∞as t→∞.We may further assume thatγj(t)=γj′(t′)for t>0,t′>0,j=j′.
Re-labelling if necessary,we obtain n pairwise disjoint simply connected domains D1,...,D n in H,with D j bounded byγj?1andγj,and for t>0 we letθj(t)be the angular measure of the intersection of D j with the circle C(0,t).Since g has?nitely many poles in H there exists a rational function R2,with R2(∞)=0,such that g2=g?R2is analytic in H∪{0}.By(15), the function g2(z)tends toαj as z→∞onγj.Thus g2(z)is unbounded on each D j but bounded on the?nite boundary?D j of each D j.
Let c be large and positive,and for each j de?ne
u j(z)=log+|g2(z)/c|,z∈D j.(17)
12
Set u j(z)=0for z∈D j.Then u j is continuous,and subharmonic in the plane since g2is analytic in H∪{0}.
Lemma2.1gives,for some R>0and for each j,
r
R
πdt
tθj(t)
≤log m0π(4r,g2)+O(1)≤log m0π(4r,g)+O(1),r→∞,(18) for all j∈{1,...,n}.However,the Cauchy-Schwarz inequality gives
n2≤
n
j=1θj(t)n j=11θj(t)
which on combination with(18)leads to,for some positive constant c3, n log r≤log m0π(4r,g)+O(1),m0π(r,g)≥c3r n,r→∞.
Since n≥2this contradicts(16),and Lemma3.8is proved.
From Lemmas3.6and3.8we deduce that the inverse function F?1has ?nitely many non-real singular https://www.sodocs.net/doc/133888556.html,ing Lemma3.7,takeα∈H such that F(z)→αalong a pathγαtending to in?nity in H,and takeε0with 0<ε0u(z)=log ε0
Lemma3.9For large z with|zL(z)|>3we have|F(z)?α|>|z|/2and u(z)=0.
Lemma3.10We have
log u(re iθ)
lim
r→∞
≥ [R,r]\E1πdt4δ1(1?o(1))log r
tσ(t)
as r→∞.Sinceδ1may be chosen arbitrarily small the lemma follows.
Now(19)gives
u(re iθ) ≤m0π(r,h)+O(1),
from which we deduce using(20)that
log m0π(r,h)
lim
r→∞
Lemma4.1All but?nitely many components C of Y are unbounded and satisfy
Im L(z)>0.(21)
lim sup
z→∞,z∈C
Proof.Suppose?rst that C is a component of Y such that?C contains no pole of L.Then Im L(z)is harmonic and positive in C,and vanishes on?C. Thus C satis?es both conclusions of the lemma by the maximum principle.
Since each pole of L belongs to the closure of at most?nitely many components C of Y,it su?ces therefore to show that L has at most?nitely many poles in the closure of Y.To see this,let x0be a pole of L,with |x0|large.Then x0is real,and is a simple pole of L with positive residue. Hence lim y→0+Im L(x0+iy)=?∞and since L is univalent on an open disc N0=B(x0,R0)it follows that Im L(z)<0on N0∩H.Thus N0∩Y=?. Lemma4.2To each component A of W corresponds a?nite number v(A) such that F takes every value at most v(A)times in A and has at most v(A)distinct poles on?A.Moreover,v(A)=1for all but?nitely many components A of W.
Proof.By Lemma3.6,F has?nitely many critical points in W,so only ?nitely many components A of the set W can contain critical points of F. Further,the assumption made in the beginning of this section implies that there is noα∈H such that F(z)tends toαalong a path tending to in?nity in W.
Suppose?rst that A is a component of W which contains no critical points of F.Then every branch of F?1with values in A can be analytically continued along every path in H.This implies that F maps A univalently onto H,and we set v(A)=1in this case.
Now consider a component A of W on which F is not univalent.Then A contains?nitely many critical points of F,which we denote by z1,...,z p. We connect the points0,F(z1),...,F(z p)by a simple polygonal curveΓ?H∪{0},so that the region D=H\Γis simply connected.Let X={z∈A: F(z)∈D}.Then every branch of F?1with values in X can be analytically continued along every curve in D,so every component B of X is conformally equivalent to D via F.
If?B∩A contains no critical points of F then the inverse branch F?1
B which maps D onto B can be analytically continued into H,so in this case
15
F:A→H is a conformal equivalence which contradicts our assumption that F is not univalent in A.
As every critical point of F can belong to the boundaries of only?nitely many conponents B,we conclude that the set X has?nitely many compo-nents.Denoting the number of these components by v(A)we conclude from the open mapping theorem that F takes every value at most v(A)times in A.
To show that F has at most v(A)poles on?A,it is enough to note that if z0∈?A is a pole of F,then for every neighbourhood N of z0,F assumes in N∩A all su?ciently large values in H.
Remark.Once it is established that F takes every value?nitely many times in A,the Riemann–Hurwitz formula shows that one can take v(A)= p+1,where p is the number of critical points of F in A,counting multiplicity, but we don’t use this observation.
Lemma4.3There are in?nitely many components A of W which satisfy all of the following conditions:(i)A contains a component C of Y;(ii)?A∩?C contains a zero of L;(iii)F is univalent on A;(iv)C is unbounded and satis?es(21).
Proof.We recall from Lemma3.5that L has in?nitely many zerosη,satis-fying one of the conditions(I),(II)or(III)of Lemma3.5.Fix such a zero η.We are going to show that there exists a component C of Y such that η∈?C.We write
L(z)=(z?η)m(ae iθ+O(|z?η|)),z→η,a>0,θ∈[?π,π).(22) Let t0be small and positive,and set
ζ(t)=η+t exp i2?θ ,t∈(0,t0].
Then arg L(ζ(t))→π/2as t→0,and so L(ζ(t))∈H for t∈(0,t0]if t0is small enough.We claim thatζ(t)∈H for t∈(0,t0],provided t0is small enough.In Case(I)this evidently holds if t016
our claim,and thusζ((0,t0])?Y.Let C be that component of Y which contains the curveζ((0,t0]).Thenη∈?C sinceζ(t)→ηas t→0.
Thus there are in?nitely many zerosηof L that belong to the boundaries of components C of the set Y.As F(η)=∞,by(14),and Y?W,we have η∈?C∩?A,where A is a component of the set W containing C.Lemma4.2 implies that in?nitely many zerosηof L cannot belong to the boundary of the same component A,and thus there are in?nitely many such components A.Finally,(iii)follows from Lemma4.2and(iv)from Lemma4.1.
We now complete the proof of Lemma3.7.Applying Lemma4.3we obtain at least one zeroηof L,withη∈?A∩?C,in which A,C are compo-nents of W,Y respectively,satisfying C?A and conditions(iii)and(iv)of Lemma4.3.Since F(η)=∞by(14),it follows that for an arbitrarily small neighbourhood N ofη,all values w of positive imaginary part and su?ciently large modulus are taken by F in A∩https://www.sodocs.net/doc/133888556.html,ing(iii)we deduce that F(z)is bounded as z→∞in A.Now(14)gives L(z)→0as z→∞in A,and hence as z→∞in C.This contradicts(21).
5Proof of Theorem1.2
Suppose that f is a real entire function,and that f has?nitely many non-real zeros.Then L=f′/f has?nitely many non-real poles,and all poles of L are simple and have positive residues.Thus L has a representation(2).
Lemma5.1Suppose thatφis a rational function.Then f has?nite order.
Proof.Lemma5.1may be proved by modifying arguments of Levin-Ostrovskii [18,pp.336-337]or of Hellerstein and Williamson[12,pp.500-501]based on the residues ofψ,or by the following argument using the Wiman-Valiron theory[10].Denote by N(r)the central index of f.By[10,Theorems10 and12],provided r lies outside a set E4of?nite logarithmic measure and |z0|=r,|f(z0)|=M(r,f)=max{|f(z)|:|z|=r},we have
f′(z)
(1+o(1)),z=z0e it,t∈[?N(r)?2/3,N(r)?2/3].
z
This leads to
2π
|f′(re it)/f(re it)|5/6dt≥N(r)1/6r?5/6,r→∞,r∈E4.
17
Sinceφis by assumption a rational function,(2)and(8)give
2π
|f′(re it)/f(re it)|5/6dt=O(r M),r→∞,
for some positive M.We deduce that N(r)=O(r6M+5)as r→∞,and thus f has?nite order[10,(1.8)and Theorem6].This proves Lemma5.1.
Since L+L′/L=f′′/f′,Theorem1.2now follows from Theorem1.3and Lemma5.1.
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W.B.:Mathematisches Seminar,
Christian–Albrechts–Universit¨a t zu Kiel,
Ludewig–Meyn–Str.4,
D–24098Kiel,
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bergweiler@math.uni-kiel.de
A.E.:Purdue University,
West Lafayette IN47907
USA
eremenko@https://www.sodocs.net/doc/133888556.html,
J.K.L.:School of Mathematical Sciences,
University of Nottingham,
Nottingham NG72RD
UK
jkl@https://www.sodocs.net/doc/133888556.html,
20
