ON THE ZEROS OF LINEAR DIFFERENTIAL
POLYNOMIALS WITH SMALL RATIONAL COEFFICIENTS
J. K. LANGLEY
ABSTRACT
We prove the following: suppose that J{z) is transcendental and meromorphic of finite order in the
plane, and that the linear differential polynomial F(z) is defined by
and is non-constant, where ak_j(z), ...,ao(z) are rational functions vanishing at infinity. Then
N(r,l/(fFF')) =
implies that
N(r,f) = O(logr).
A corresponding result is proved for the case where F =f' + af, where a is a constant. The problem is
related to results of Frank and Hellerstein and others.
1. Introduction
Our starting point is the following question: when is the growth of a function f{z)
meromorphic in the plane determined by the number of zeros of its successive
derivatives? The following results, proved by Frank [3], and Frank, Hennekemper
and Polloczek [5], give an almost complete answer to this question, enlarging
considerably upon previous work of Polya and Saxer [12], Hayman [7], Clunie [2],
Mues [11], and others.
THEOREM A. Suppose thatf[z) is meromorphic and non-constant in the plane. Iff
andf" have no zeros, then fIf is constant or linear. Iffandf{k) have only finitely many
zeros, for some k~^2>, then f/f is rational.
Frank and Hellerstein [4] observed that the above question may be regarded
as a special case of the following: iff{z) is meromorphic in the plane, if k ^ 2 and
a
k-i(z)> ••••> ao(z) a r e polynomials, with
F{z)=r\z)+Zaj{z)f\z),
(1.1)
then what form may J[z) have if / and F have only finitely many zeros, and in
particular how fast may the counting function N(r,f) of the points at which / has
poles grow?
They proved the following results, noting first that eliminating ak_x(z) in (1.1)
amounts to multiplying / and F by a non-vanishing factor.
Received 5 February 1987.
1980 Mathematics Subject Classification 3OD35.
J. London. Math. Soc. (2) 36 (1987) 445-457
446
J. K. LANGLEY
THEOREM
that
B. Suppose thatj[z) is meromorphic and non-constant in the plane, and
k-2
is non-constant, where ak_2(z), ...,ao(z) are polynomials. Ifk^3,
T(r,f'/f) = O(N(r, \/f) + N(r, \/F) + r>)
then
(1.2)
at least outside a set of r of finite measure, where
X = 1 +max{deg(a,)/(A:-/) :j = 0,
...,k-2}.
IfF(z) = / w ( z ) , then r^ may be replaced by log r in (1.2). Also, iff[z) is entire then (1.2)
holds for k^2.
The paper [4] also includes further results in the case where / is entire. Here
N(r,f), T(r,f) etc. are the standard terms of Nevanlinna theory (see [8]), while deg (P)
denotes the degree of a polynomial P.
Thus Theorem B contains the second conclusion of Theorem A, and (1.2) gives a
bound on the counting function N(r,f) of the points at which/has poles if k ^ 3, and
/ and F have only finitely many zeros. Now it is easy to give examples where / and
F have finitely many zeros and
T(r,f) = (<x+o(\))N(r,f)
for some non-zero a. For example, if P is a non-constant polynomial, if H' = ep and
D denotes d/dz, then
/ = \/{H'kHL)
(1.3)
satisfies
(D + F)... (D + kF)f = c/HL+k,
where c is a constant. For constant coefficients, we have more general examples, for
instance,
satisfies
L+l)(D+l)f=c/(ez+\)L+2.
(D +
Notice that in these examples F' has no zeros, while the linear differential operator has
a factorisation similar to that in the following result.
THEOREM C [9]. Suppose that f{z) is transcendental and meromorphic in the plane,
and that F(z) is given by (I.I) and is non-constant. Then either
(r,f) < 3N(r, 1//H4#(r, \/{F-1))
at least outside a set of r of finite measure, or
F
wherefi= 2ak_Jk(k+ 1).
In the present paper we consider the case where a^z), ...,ao(z) in (1.1) are
rational functions vanishing at infinity, and where / has finite order. We prove the
following.
ZEROS OF LINEAR DIFFERENTIAL POLYNOMIALS
447
THEOREM 1. Suppose that f[z) is transcendental and meromorphic of finite order
in the plane, and that F(z) is given by (1.1) and is non-constant, where for k ^ 1 and
j = 0, ...,k— 1, the coefficient a}(z) is rational with fl/oo) = 0. Then
N(r,\/fFF')
= O(\ogr)
implies that
N(r,f) = O(\ogr).
Theorem 1 is sharp at least to the extent that the coefficients may not be replaced
by larger rational functions in view of the examples (1.3). Also the hypothesis
that F' has only finitely many zeros cannot be removed altogether. For example, if
g(z) = {zez + z)~l then g and g' + z~xg have no zeros. It seems possible that Theorem
1 remains true without the hypothesis t h a t / h a s finite order; it will be seen, however,
that the methods presented here do not apply in the infinite order case.
In the case where k = 1 and ao(z) is a non-zero constant we are able to determine
those functions/of finite order for w h i c h / F and F' have only finitely many zeros.
We shall prove the following.
THEOREM 2. Suppose that f{z) is meromorphic of finite order with infinitely many
poles
in the yplane, and that
^n
\
F
XT, t ir^r-,\
N(r,\/fFF')
= O(\ogr),
where F(z) =f'(z) + af{z) for some non-zero constant a. Then f satisfies one of the
following:
0)
M=
l
for some constants D, B and a positive integer L;
yv 7
(c+zvr
v
'
for some positive integer N and constants C, D, b.
The author would like to acknowledge valuable conversations with S. Bank,
G. Frank and S. Hellerstein.
2. Notation and preliminary results needed for the proofs of Theorems 1 and 2
For a non-constant polynomial
P(z) = a zn-\-... -\-a
we define, for each real 0,
We use the notation O(rM) to denote any term H(z) bounded by a power of \z\, not
necessarily the same power at each occurrence. If a specific power is intended it is
denoted by a different symbol.
For an increasing function S(r) on (0, -I- oo) we require the order of S(r) defined
and denote the order of a function/(z) meromorphic in the plane by a{f) (defined by
(2.1) with S(r) = T(r,f)).
448
J. K. LANGLEY
By a differential monomial we mean a constant multiple of a finite product
where h is meromorphic, and each qj is a non-negative integer.
We require the following 'small arcs' lemma of Fuchs [6].
LEMMA A. Suppose that h(z) is meromorphic in the plane of finite order p. Then,
given C > 0 and 8 with 0 < 8 < |, there exist a constant K(p, Q and a set of positive real
numbers H of lower logarithmic density at least 1 - ( such that ifO^02-O1^8
and
reH, then
dd < K(p, 0 (8 log(l/<5)) T(r, h).
h(reie)
We need also the following result of Lewis, Rossi and Weitst^an [10].
THEOREM D. Suppose that g(z) is a transcendental, entire function. Then there is a
path F extending to infinity such that
+ oo
,
•
•
l
,
and, for each positive X,
0
g
|
2
onY
1
I
We shall require the following lemma on asymptotic integration: the proof is by
repeated integration by parts and we omit it. It is in any case very similar to [1,
Lemma 2]; see also [8, p. 46].
LEMMA 1. Suppose that P is a non-constant polynomial with 8{P, 6) # 0 on [a, b],
and that H(z) is analytic on the sectorial set S given by
with
H'(z) = O(rM)ep
(2.2)
there. Then in a sectorial set
\z\^rv
a + e < a r g z < b — e,
where e may be chosen arbitrarily small and positive, we have
H(z) = O{rM)ep
if 3{P, d)>0on
[a, b], while if3(P, 6)<0on
(2.3)
[a, b] we have
H(z) = c + O(rM)ep
(2.4)
for some constant c. Also if H'{z) satisfies the stronger estimate
H'(z) = <xzp(l+o(\))ep
on S, where p is real and a is non-zero, then the term O(rM) in (2.3) and (2.4) may be
ZEROS OF LINEAR DIFFERENTIAL POLYNOMIALS
449
LEMMA 2. Ifn{t) is non-decreasing on [0, + oo), and m ^ 0, and
tmdn(t)
N(r)=
Jo
has order at most m, then n(t) has order zero.
Proof.
We have
tm dn(t) > (r/2)m (n(r) - n(r/2)).
N{r) >
Jr/2
Thus, given e > 0, there exists r0 such that if r ^ r 0 , then
n(r)-n(r/2) < r*.
k
Hence, with rk = 2 r0 and rk_1 ^ r ^ rk, we have
n(r) ^ n(rk) < £ (2V0)« + n(r0) = Oft) = 0{f).
LEMMA 3. Suppose that F(z) is transcendental and meromorphic of finite order p
in the plane, having no zeros or poles in the sectorial set
OL-E1 < a r g z < ^ + e 1 5
\z\ ^ r0,
where EX > 0.
(a) Then, if g = F'/F and e>0,we
g"\z)
have, for j = 0,1,2,...,
= OQzr-1-*)
(2.5)
in a sectorial set S given by
oc<argz</?,
\z\>rx.
(b) In addition, if h = 1 /g and zQeS with
\g(zo)\ > \zor-1
we have, for j = 0,1,2,...,
"-}+1)-
(2.6)
Proof. Equation (2.5) is a standard estimate (see for example [8, p. 21]). To prove
(2.6) we use induction o n / Since
0 = (hg)(j+1) = hu+1)g+
Y
m-0
we have, assuming that (2.6) holds for 0, ...J, and using (2.5),
LEMMA
^J^
'
4. (a) Suppose that k is a positive integer and that h = F/F'. Then, for
(Fhky»
Fhk-i(\+P(h'))
+ Fhk-i+lQ(hh',...),
(2.7)
where Pi and Qi are differential polynomials in h, with constant coefficients, which satisfy
the following.
(i) P} has no constant term and is independent of h.
(ii) Q} has no constant term and is a sum of monomials none of which is independent
of h' and its derivatives.
15
JLM 36
450
J. K. LANGLEY
(b) If p > s > 1, and F and z0 satisfy the hypotheses of Lemma 3(b) with e
sufficiently small, then for j = 1,..., k,
h(zoy = OQz0\1-8)
(2.8)
and
F\Fhk)"\z)
= hk-\z) + O(\z0\-g).
(2.9)
Proof. We prove part (a) by induction. Now
4- (Fhk) = F/ifc-1 + Fh"-lkh\
dz
so that we may set Px = kh' and QY = 0. Now suppose that (2.7) holds and that
1 ^j<k.
Then
(FhkY}+1) =
Thus we need only set
and
(2.10)
clearly no term in (2.10) has the form chq with q > 0 and ceC.
To prove part (b), (2.8) is obvious. Also, using (2.6),
and (2.9) follows immediately.
3. Proof of Theorem 1
Suppose that f{z) satisfies the hypotheses of Theorem
many
rem 1 and has infinitely mi
poles. We may write
^
^
^
1.1)
where R is rational and P is a polynomial. We consider two cases.
Case 1, in which P is non-constant. Now at a large pole o f / o f order q, we
have
Rep =
-1)!
'
and the right-hand side tends to 1 as q -> oo, so that poles o f / c a n only accumulate
near rays argz = 6 for which 5{P,0) = 0. Take a closed interval [a, b] on which
S(P, 0) 7* 0, and consider the sectorial set S given by
\z\ large,
argze[a,6].
If S(P, 6) < 0 on [a, 6] we have, using standard estimates for F/f
F'/F=O(rM)ep/k,
so that
F=J+o(l)
ZEROS OF LINEAR DIFFERENTIAL POLYNOMIALS
U)
451
M
for some non-zero J. Since f /f=
0{r ) on S for7 = 1,...,k, we obtain
+
log |l//| = O(logr)
(3.2)
on S. On the other hand, if S(P, 0) > 0 on [a, b] we have
F/f=O(rM)e-p
and hence
/<*> (z) +...+ K(z) + <K\z\-l))fLz) = 0
(3.3)
on S. We estimate Y = f'/f. We have, for some positive Mx,
Y(z) = O(\z\M>)
(3.4)
on S. Now it is easy to check by induction that, for n ^ 1,
(3.5)
where P n-1 is a differential polynomial in Y of degree at most n— 1 and which is a sum
of terms none of which is independent of the derivatives of Y. Taking a small positive
ex and a sectorial set 5X given by
|z| large,
a + e1 < arg z < b — e15
we obtain from (3.3), (3.4), Cauchy's estimate and (3.5),
Y{zf = Oflzl^*-"-1)
in Sj. Iterating this process we find that Y(z) = O(\z\~M) for some positive fi in a
sectorial set
|z| large,
a + e < arg z < b — e,
where e is arbitrarily small and positive. This and (3.2) imply that
off a set of arbitrarily small angular measure. Hence o{f) is less than 1. But this
contradicts the fact that P is non-constant in (3.1). Case 1 is therefore impossible.
Case 2, in which P = 0. We set
h = F/F',
(3.6)
so that h has only finitely many poles. We may assume that h is transcendental, for
otherwise there is nothing to prove. In fact, we have the following.
CLAIM.
a{h) ^ §.
Applying Theorem D to a suitable polynomial multiple of h, we obtain a path F
extending to infinity on which h is large and, recalling (3.6),
|F7F||<fe|<oo.
Hence 1/F approaches afinite,non-zero value, Jl say, as z tends to infinity on F. But
if o(h) < \, the classical cos np theorem implies that F meets arbitrarily large circles
\z\ = r on which \h{z)\ > r2 so that 1/F = Jx + o(l) on such a circle. Since 1/F has only
finitely many poles, this is a contradiction. This establishes the claim.
15-2
452
J. K. LANGLEY
k
We now substitute/= RFh into
to obtain an equation for h which we write in the form
£ (k )
» (Fhk)U) +... + Ra0Fhk = F.
(3.7)
i-o \J
We distinguish two subcases.
(a) Here R is constant. At a large pole of/of order q we have
The right-hand side of (3.8) may be written as a product of k functions of q, each
decreasing on (0, + oo). Hence all large poles of/have a fixed multiplicity which we
continue to call q. But then we may write
(<Fy+k/F"+k+1 = Rle\
(3.9)
where R1 is rational and Px is a polynomial. We may solve (3.9) in a simply-connected
region D consisting of an annulus \z\ > r2 slit along a ray argz = 0 for which
3(PV 0) = 0. This gives
F-ll(Q+k) = Cl+\ Tep*dt,
(3.10)
where T is algebraic, P2 is a polynomial which must be non-constant, c2 is a point of
Z) and cx is a constant. Determining h from (3.10) we see that since h is transcendental,
a{h) ^ 1. We now apply the Wiman-Valiron theory [13] to the equation (3.7) for h
and obtain a contradiction. There exist arbitrarily large points z = re** for which
\H(z)\ = M(r, H), where H = hk and simultaneously
(z)/H(z) ~ Mr)/ry,
/*<» (z)//2(z) ~ (v2(r)/r)>,
(3.11)
where v15 v2 are the central indices of H, h respectively. It is easy to see that, using
(16)
'
(Fhk)U) =
for 7 = \,...,k, where 7J is a differential polynomial in h, of total degree less than k.
Substituting in (3.7) gives
I)/fc
= 0(\)F
at such a point z, so that
Using (3.11) we now see that
,.,
„ , ..
vx(r)/r = 0(r A )
for some A > 0 contradicting the fact that a(//) ^ 1.
(b) A^ow suppose that R is non-constant. At a large pole of/ of order q, we have
((3.8) again)
ZEROS OF LINEAR DIFFERENTIAL POLYNOMIALS
453
if q is large, where y is positive. Since / has infinitely many poles we must therefore
have JKoo)-l, say
«(z) =
"-)
(3.12)
as z -• oo, and at a large pole z1 of/of multiplicity qv
(3.13)
for positive fi. Moreover all large poles lie near a finite set of rays on which <x/zm
is real, that is, there exist 0l5 ...,02m such that for a large pole zx, argzx = 0y + o(l)
for some/ Now suppose that the order of Fis pe(l, +00). Note that by the claim
above, Lemma 2 and (3.13) we cannot have p ^ 1, since if n{r) counts large zeros of
h, we have
n(rtl/F)>%\ tmdn{t).
(3.14)
Take s with 1 < s < p. Since 1/Fhas finite order and onlyfinitelymany poles, we can
find arbitrarily large points £ in sectors away from the rays argz = 0} such that
\KO\ =
and
for 1 ^ y ^ k. Here we invoke Lemma 3. Hence, by Lemma 4,
for j = \,...,k, and using Leibnitz's rule, if j < k, then
/<» (C) = {RFhky» (C) = F(Q Odd1-').
Also
since R'(oo) = 0. Since each coefficient a} vanishes at infinity we obtain, from the h
equation (3.7),
F(Q = R(QF(Q
+m
QQQ_S)
which gives, using (3.12),
Hence — m ^ —s. It follows that p ^ m. But then (3.13), (3.14), and Lemma 2 imply
that a{h) — 0, which contradicts the claim proved above. This completes the proof of
Theorem 1.
4. Proof of Theorem 2
Suppose that/satisfies the hypotheses of Theorem 2 and has infinitely many poles.
Then, as in the proof of Theorem 1, we may write
fF'/F2 = Rep,
(4.1)
where R is rational and P is a polynomial. Again we divide the proof into cases.
Case 1, in which P is non-constant. We may assume that
F = f'-Af
where A/Us positive, for otherwise we may make a change of independent variable,
andweset
O = e<O<<0
2n
454
J. K. LANGLEY
to be the set of 0 for which either 0e{O,7r,27r} or 3(P,6) = 0. We define, for small
positive e, and k = 2,...,L, intervals
Ak(e) = [0^ + 8,6,-8]
and sectorial sets
Sk(8) = {z:\z\ large, argze^(e)}.
We note that any Sk(e) is free of poles of/ since at a pole of/of order q we have
Rep = (q+\)/q.
Preliminary asymptotics for f and F
We take a small positive £ and assume that S(P,6) < 0 on ^(g/16). Then, using
standard estimates for F/f, we have
F'/F=O(rM)ep
on Sk(e/\6), so that by Lemma 1,
F = J1 + O(rM)ep
(4.2)
on Sk(e/S) for some non-zero constant Jv By Lemma 1 again we have, on Sk(e/4),
,
(4.3)
where Bx is a constant and Jz = —JJA.
Now suppose that S(P,6) > 0 on Ak(e/\6). Then on 5t(e/16)
so that, using Lemma 1,
F/f=O(rM)e-p
*
(4.4)
on 5fc(e/8), where B2 is a non-zero constant.
We deduce at once from (4.3) and (4.4) that
T{r,f)~m(r,\/f)
= O(r)
since e is arbitrary. From (4.1) it follows that deg(/>) = 1, say P(z) = bz. We now
establish the following.
CLAIM,
b is imaginary.
Assuming the claim to be false,/must have infinitely many poles in a sector S
given by
where 6k is not equal to 0, n or 2n. By Lemma A there exist arbitrarily large values
o rsuc t a t
iog|/{r^+e/2)]|-log|/[re(<<?*-£/2)]| = O(e*)T(r,f)
= 0(8>r).
Since e is arbitrary this can only imply that in the asymptotic formulae (4.3) and (4.4),
Bx is non-zero and eAz is large. But then there exist non-zero L15 L2 such that
as z tends to infinity with argz = 9k + (— l)/i(e/2). Again, since e is arbitrary and
eAi/J{z) has finite order and only finitely many poles, we conclude that by the
ZEROS OF LINEAR DIFFERENTIAL POLYNOMIALS
455
Az
Phragmen-Lindelof principle Lx = L2 and e /j{z) -+ Lx as z tends to infinity in the
sector S, so that/has onlyfinitelymany poles in S. This is a contradiction. It now
follows that the set {6X, ...,9L} is just {0,71,2^}.
Determination of possible asymptotics for f and F
We take a ray argze{0, n) and, assuming that/has infinitely many poles near this
ray, we consider various cases. Adjacent to this ray we have a sector S\e/4) on which
ebz is small and a sector SP(E/4) on which ebz is large (each ^(e/4) is Sk(e/4) for some
k), and we recall the asymptotic formulae (4.3) and (4.4) which hold in Sl(e/4) and
5^(e/4), respectively. We note that in the first sector either/is large o r / = 72 + o(l),
while in the second sector/is either small or large. Also/cannot be large in both
sectors.
We examine three subcases.
(a) We suppose that / = J2 + o(\) in S1(e/4) and is large in S2(e/4). But this is
impossible by the Phragmen-Lindelof principle, since e is arbitrary and l//has only
finitely many poles.
(b) Suppose that f is large in S1(e/4) and small in 52(e/4). Again this case is
impossible, for it implies that eAz/f{z) is rational, using the Phragmen-Lindelof
principle.
This leaves only one possibility.
(c) / = J2 + o(\) in S\e/4) and is small in S2(e/4).
We now have, from (4.1), (4.2) and (4.3),
F'/F=(J3 + o(\))Rebz
(4.5)
in S1(e/4), where in fact J3 = —A. By Lemma 1, in Sl(e/2),
and
)e 62 .
(4.6)
Since e~Az and ebz are both small locally, (4.6) and Lemma 1 yield
f=J2 + O(rM)eb*
(4.7)
in S^e). Now (4.5) and (4.7) imply that, in S\2e),
f'F/fF' = O(rM).
(4.8)
On the other hand, in S2(e/4) we have
/=(2
so that (4.1) yields
where 74 = l/5 2 . Since ebz and e~Az are both large locally, we obtain, using
L e m m a lf
F-1 = (1 +o(\))yz°e*-A)z
(4.9)
456
J. K. LANGLEY
in 52(e/2) for some real a and non-zero constant y. Now (4.4) and (4.9) imply that
f'F/fF'= (\+o{\))A/(A-b)
(4.10)
in S^e). But f'F/fF' has only finitely many poles, and must therefore be rational by
(4.8), (4.10) and the Phragmen-Lindeldf principle. Since A/(A — b) ^ 1 and / has
infinitely many poles, with f'F/fF' = qj(qx +1) at a pole of/of order qx, we must in
faCthaVe
rF/JF'm AHA-b).
Also all poles of/have a fixed multiplicity, K say, so that
F=/f~ K - 1
(4.11)
yw,/r*
(4.12)
and
for some meromorphic function H and non-zero constant Jy But then substituting
(4.11) and (4.12) into / ' - Af = F yields
for some constants D, B and d, and /satisfies (1.5).
Case 2, in which P = 0. In this case we set h = F/F' as in the proof of Theorem 1,
so that h has only finitely many poles but, by hypothesis, has infinitely many zeros.
Substituting/= RFh (from (4.1)) into / ' + af = F we obtain
^[hRez] = (\-R)eaz.
(4.13)
az
If R is constant it follows immediately that/satisfies (1.4). If R is non-constant then,
since at a pole of/of order q we have R = (q+ \)/q, we must have R{co) = 1. Also
the multiplicity px of a pole at zx must tend to infinity as zx becomes large. A partialfractions decomposition of 1 — R and integration of (4.13) by parts now yields, since
^ _ ( ^ + Ce-«,)jR(z)<
h is single-valued,
where T is rational and C is a constant. Setting g =feaz and using (4.1) we obtain
Now at a large pole w of/of large multiplicity « we have
say, where v(z) = Odzl"1) and where m must be positive. This gives
w = (\+o(l))cn1/m,
where cm = —ft,and writing w = cnllm+p, we have
(cnllm+p)m = -/?«(1 + v(w)) = -yfo
We therefore have p = 0(1), since p = o(nllm), so that
which is impossible. Therefore this case cannot occur and the proof of Theorem 2 is
complete.
ZEROS OF LINEAR DIFFERENTIAL POLYNOMIALS
457
REMARK. It is possible to prove some results in the case where F=f' + bf, where
b is a non-constant polynomial. The details are complicated, as there are more sectors
to consider than in Theorem 2. Moreover, importantly, it seems very difficult to rule
out the case where/has infinitely many poles of unbounded multiplicities, and I have
chosen not to pursue this.
References
1. S. B. BANK and J. K. LANGLEY, 'On the oscillation of solutions of certain linear differential equations
in the complex domain', Proc. Edinburgh Math. Soc. to appear.
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Department of Pure Mathematics
University of St Andrews
North Haugh
St Andrews KY16 9SS