Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 24, 1999, 409–436

ON THE ZEROS OF PAIRS OF LINEAR DIFFERENTIAL POLYNOMIALS

G. Frank and J.K. Langley

Technische Universität Berlin, Fachbereich Mathematik Strasse des 17. Juni 136, D-10623 Berlin, Germany University of Nottingham, Department of Mathematics Nottingham NG7 2RD, England; jkl@maths.nott.ac.uk

Abstract. Suppose that f is meromorphic in the plane and that F and G are given by

F = f^{(k)} + ∑_{j=0}^{k−1} a_{j}f^{(j)}, G = f^{(k)} + ∑_{j=0}^{k−1} b_{j}f^{(j)},

with k ≥ 1 and the a_{j}, b_{j} rational functions, such that a_{j} ≢ b_{j} for at least one j. We classify those f for which F and G have only finitely many zeros.

1. Introduction

The study of zeros of linear differential polynomials has a long history, going back to the fundamental work of Pólya [28] on entire and meromorphic functions and their derivatives. The following theorem was proved by the first author and Hennekemper and Polloczek [5], [7] for k ≥ 3 and by the second author [20] for k = 2, and confirmed a conjecture of Hayman [9], [10], [11] from 1959.

Theorem A. Suppose that f is meromorphic in the plane and that f and f^{(k)} have only finitely many zeros, for some k ≥ 2. Then we have f(z) = R(z)e^{P (z)} , with R a rational function and P a polynomial. In particular, f has finite order and finitely many poles.

Refinements of this theorem may be found in [6], [20], [21], [23], while simple examples show that no comparable result holds for k = 1 (see however [4]). A natural generalization of Theorem A involves replacing the k ’th derivative f^{(k)} by a linear differential polynomial

(1)  F = f^{(k)} + ∑_{j=0}^{k−1} a_{j}f^{(j)},

1991 Mathematics Subject Classification: Primary 30D35.

with coefficients a_{j} which are rational functions. Thus the first author and Hellerstein proved in [6] that if f is meromorphic in the plane and

N(r, 1/f) + N(r, 1/F) = o(T (r, f′/f)), r → ∞,

in which k ≥ 3 and F is given by (1) with polynomial coefficients a_{j}, and in which the notation is that of [10], then f′/f has finite order. Subsequent papers [3], [29] determined all functions f meromorphic in the plane for which f and F, subject to the above assumptions, have no zeros, while the papers [20], [22] give a rather more complicated classification of all functions f meromorphic in the plane such that f and f′′+a_{1}f′+a_{0}f have only finitely many zeros, for any rational functions a_{1} , a_{0} . Related results appear in [14], [19], [27] and elsewhere.

With regard to these results, it seems reasonable to ask how essential the hypothesis on the zeros of f really is. Of course, it is easy to give examples of entire f for which F, as given by (1), has no zeros: just set F = e^{P} , with P a polynomial, and solve the resulting differential equation for f. However, some conclusion regarding poles might be expected, and the following theorem [24], [25],

[26] summarizes some results in this direction.

Theorem B. Suppose that f is meromorphic of finite order in the plane, and that f′′ has only finitely many zeros. Then

\overline{N}(r, f) = O(log r)^{3}, r → ∞.

If, in addition, T (r, f) = O(r) or N(r, 1/f′) = o(r^{1/2}) as r → ∞, then f has only finitely many poles.

On the other hand, examples of meromorphic f having infinite order, such that f′ and f′′ have no zeros, while f has an arbitrary set of poles, were given in [24], and we show in the next section how to construct examples of functions f and linear differential polynomials F in f, such that F and F′ have no zeros, while f has an arbitrary set of poles. Thus the zeros of a single linear differential polynomial in f do not suffice to determine f.

In the present paper, we consider two linear differential polynomials

F = L_{k}(f) = f^{(k)} + ∑_{j=0}^{k−1} a_{j}f^{(j)}, G = M_{k}(f) = f^{(k)} + ∑_{j=0}^{k−1} b_{j}f^{(j)},

in a meromorphic function f, with k a positive integer and the a_{j} and b_{j} rational functions, and with a_{j} ≢ b_{j} for at least one j. There is a well-known reduction procedure [17], described in Lemma 1 below, to obtain linear differential operators P, Q, H with coefficients which are rational functions, such that L_{k} = P (H) and M_{k} = Q(H) and the common (local) solutions of the homogeneous equations

(2)  L_{k}(w) = 0,  M_{k}(w) = 0

are precisely the (local) solutions of H(w) = 0. This allows us to concentrate on the case where the equations (2) have no non-trivial common (local) solution, that is, no common (local) solution other than the trivial solution w ≡ 0, for in the contrary case we may regard F and G as linear differential polynomials in H(f). Our main result is then the following.

Theorem 1. Let k be a positive integer and let a_{0}, …, a_{k−1} and b_{0}, …, b_{k−1} be rational functions with a_{j} ≢ b_{j} for at least one j. Assume that the equations

w^{(k)} + ∑_{j=0}^{k−1} a_{j}w^{(j)} = 0, w^{(k)} + ∑_{j=0}^{k−1} b_{j}w^{(j)} = 0,

have no non-trivial common (local) solution. Let f be meromorphic in the plane such that

F = f^{(k)} + ∑_{j=0}^{k−1} a_{j}f^{(j)}, G = f^{(k)} + ∑_{j=0}^{k−1} b_{j}f^{(j)}

both have only finitely many zeros. Then f has finite order and finitely many zeros and f′/f has a representation

(3)  \frac{f′(z)}{f(z)} = Y (z) + \frac{P_{0}(Q(z) + log S(z))(Q′(z) + S′(z)/S(z))}{S(z)e^{Q(z)} − 1},

in which S and Y are rational functions and Q and P_{0} are polynomials, and at least one of P_{0} and S is constant.

In the next section we will give examples showing that (3) can indeed occur. Our approach to proving Theorem 1 exploits the fact that F and G have, with finitely many exceptions, the same poles, and proceeds via the rather surprising conclusion that f itself has finitely many zeros. This allows us to use the machinery developed in [5], [6], [7].

This research was carried out during the second author’s visit to the Technische Universität, Berlin, funded by a grant from the Deutscher Akademischer Austauschdienst (DAAD).

2. Examples

2.1. Example. Let a_{0}, …, a_{k−2} and P be polynomials, and let f be a solution of the equation

f^{(k−1)} + ∑_{j=0}^{k−2} a_{j}f^{(j)} = K = e^{P} .

Let c, d be distinct constants. Then F = K′ + cK and G = K′ + dK are both linear differential polynomials of order k in f, having finitely many zeros. However, here F and G should, according to the reduction procedure referred to in the discussion of the system (2), more correctly be regarded as linear differential polynomials in K.

2.2. Example. Setting f(z) = tan z we find that F = f′′ − 2if′ and G = f′′ + 2if′ are both zero-free. This example does not, however, contradict Theorem 1 since the equations w′′ − 2iw′ = 0, w′′ + 2iw′ = 0 have the non-trivial common solution w = 1, and F and G are more properly regarded as linear differential polynomials in f′.

2.3. Example. Let P and P_{1} be polynomials, with P non-constant and P_{1} not identically zero, chosen so that P_{1}(P) is a non-positive integer at every zero of e^{P} − 1. For example, we may take P_{1}(P) = P^{2}4^{−1}π^{−2} . Then

f′/f = T (e^{P} − 1)^{−1} = −P_{1}(P)P′ + P_{1}(P)P′e^{P} (e^{P} − 1)^{−1}, T = P_{1}(P)P′,

defines a meromorphic function having no zeros, and poles at all but finitely many zeros of e^{P} − 1, while the equations

f′(z) = 0, f′(z) + T (z)f(z) = 0

each have only finitely many solutions z. Further, with a and b rational functions we define L by

L/f = f′′/f + af′/f + b = S(e^{P} − 1)^{−2},

where

S = be^{2P} + e^{P} (T′ − T P′ + aT − 2b) + (T^{2} − T′ − aT + b),

and L cannot vanish identically, since f has infinitely many poles. There are thus three ways to ensure that L/f has only finitely many zeros, the same then being true of L. We can either solve simultaneously both equations

(4)  aT − 2b = T P′ − T′,  −aT + b = T′ − T^{2},

for a and b, using the fact that the determinant of the coefficients is −T, which is not identically zero, or we can set b = 0, and solve either of the equations (4) for a. To see that a non-zero rational function Y can arise in (3), we need only write f = Ue^{V} g, with U a rational function and V a polynomial, so that there are linear differential polynomials G_{1} , G_{2} in g, with coefficients which are rational functions and with G_{1}/G_{2} non-constant, each having finitely many zeros.

2.4. Example. Let c be a constant, let k ≥ 1 and let A_{0}, …, A_{k} be polynomials with A_{k} = 1, and define the operator L by

L = ∑_{j=0}^{k} A_{j}D^{j}, D = d/dz.

Let a_{n} and M_{n} be sequences, such that each M_{n} is a positive integer, while the complex sequence (a_{n}) tends to infinity, without repetition, as n → ∞. Define rational functions R_{n}(z) by

R_{n}(z) = L((z − a_{n})^{−M_{n}}).

Then R_{n} has a pole of order M_{n} + k at a_{n}. Let g be an entire function having a simple zero at each a_{n}, and no other zeros. Using Mittag–Leffler interpolation, choose an entire function h such that we have, for each n,

c + g(z)^{−1}e^{h(z)} = R′_{n}(z)/R_{n}(z) + O(|z − a_{n}|^{M_{n}+k−1})

as z tends to a_{n}. Define H by

H′/H = c + g^{−1}e^{h}.

Then there are non-zero constants b_{n} such that we have, for each n,

H(z) = b_{n}R_{n}(z)(1 + O(|z − a_{n}|^{M_{n}+k})) = b_{n}R_{n}(z) + O(1)

as z tends to a_{n}. Hence there is a function h_{n} analytic at a_{n} such that H(z) − b_{n}R_{n}(z) = h_{n}(z) on a punctured neighbourhood U_{n} of a_{n}. It follows that if w is a solution of the equation L(w) = H(z) − b_{n}R_{n}(z) on a simply connected subdomain V_{n} of U_{n} then w has an analytic extension to a neighbourhood of a_{n}. If f_{1} is a solution of the equation L(f_{1}) = H on V_{n} then f_{1} may be written in the form

f_{1}(z) = b_{n}(z − a_{n})^{−M_{n}} + w(z) + v(z),

in which L(v) = 0 so that v is the restriction to V_{n} of an entire function. It follows that f_{1} has an analytic extension to U_{n} with a pole at a_{n}. Therefore every local solution f of L(f) = H extends to a function meromorphic in the plane and, since every zero of g is a pole of H, both H = L(f) and H′ −cH have no zeros.

3. Preliminaries

The following lemma is well known [17, p. 126].

Lemma 1. Let k, n be non-negative integers with k ≥ n and let D denote d/dz, and let linear differential operators L_{1} , L_{2} of orders k, n be defined by

L_{1} = ∑_{j=0}^{k} a_{j}D^{j}, L_{2} = ∑_{j=0}^{n} b_{j}D^{j},

in which a_{0}, …, a_{k}, b_{0}, …, b_{n} are rational functions with a_{k}b_{n} ≢ 0. Then there exist an integer q with 0 ≤ q ≤ n and an operator H = ∑_{j=0}^{q} c_{j}D^{j}, with the coefficients c_{j} rational functions and c_{q} ≢ 0, and linear differential operators Q_{1} , Q_{2} , P_{1} , P_{2} with rational functions as coefficients, such that

L_{1} = Q_{1}(H),  L_{2} = Q_{2}(H),  P_{1}(L_{1}) + P_{2}(L_{2}) = H,

in which the parentheses denote composition. Further, if w is meromorphic on some domain U, we have H(w) ≡ 0 on U if and only if L_{1}(w) ≡ L_{2}(w) ≡ 0 on U. Moreover, the operators Q_{1} , Q_{2} have orders k − q, n − q respectively, while the operators P_{1} , P_{2} both have order at most k.

Proof. This is just the Euclidean algorithm for linear differential operators but, since we need the estimate for the orders of P_{1} and P_{2} , we present a proof. We proceed by induction on n, there being nothing at all to prove when n = 0, as in this case H is the identity operator. Assuming the result true when one of the operators has order less than n, we apply the division algorithm [17, p. 126] for linear differential operators in order to write

L_{1} = L(L_{2}) + M_{1}

with L and M_{1} each a linear differential operator with rational functions as coefficients, and in which M_{1} either is the zero operator or has order less than n. Plainly, the order of L is k − n. If M_{1} is the zero operator we write H = L_{2} and Q_{1} = L, and P_{1} is the zero operator, with P_{2} and Q_{2} the identity.

Now assume that M_{1} is not the zero operator. The induction hypothesis gives us operators H, p_{1} , p_{2} , q_{1} , q_{2} such that the orders of p_{1} and p_{2} are at most n, and such that

L_{2} = q_{2}(H),  M_{1} = q_{1}(H),  p_{1}(M_{1}) + p_{2}(L_{2}) = H.

Now we set Q_{1} = L(q_{2}) + q_{1} , Q_{2} = q_{2} and we have

H = p_{1}(L_{1}) + (p_{2} − p_{1}(L))(L_{2}).

Thus P_{1} = p_{1} and P_{2} = p_{2} −p_{1}(L) have order at most k. The remaining assertion is obvious.

The next lemma is also fairly standard.

Lemma 2. There exists a positive constant c with the following properties. Suppose that f is transcendental and meromorphic in the plane, and that r is large and N > 1. Then we have

|log |f(z)|| ≤ cN^{2}T (r, f)

for all z with \frac{1}{2}r ≤ |z| ≤ re^{−2/N} and lying outside a union of discs having sum of radii at most 4erN^{−2} .

Proof. We denote by d positive constants not depending on f, r, N. Let r_{j} = re^{−j/N}, j = 1, 2. Then provided r is large enough we have

(5) n(r_{1}, f) + n(r_{1}, 1/f) ≤ (log(r/r_{1}))^{−1}(2T (r, f) + log |1/f(0)|) ≤ dNT (r, f),

with minor modifications if f(0) = 0, ∞. Let the zeros and poles of f in \frac{1}{4}r ≤ |z| ≤ r_{1} be a_{1}, …, a_{n} and b_{1}, …, b_{m}, respectively, repeated according to multiplicity, and write

f(z) = g(z)F (z)G(z)^{−1},  F (z) = ∏_{j=1}^{m}(1 − z/b_{j}),  G(z) = ∏_{k=1}^{n}(1 − z/a_{k}),

so that g is analytic and non-zero in \frac{1}{4}r ≤ |z| ≤ r_{1} . For |z| ≤ r we have, using (5),

(6)  log^{+} |F (z)| + log^{+} |G(z)| ≤ d(m + n) ≤ dNT (r, f).

We also have [13, p. 366]

log|∏_{k=1}^{n}(z − a_{k})| ≥ n log(rN^{−2})

outside a union E_{1} of discs having sum of radii at most 2erN^{−2} , so that for z satisfying \frac{1}{2}r ≤ |z| ≤ r_{1} but lying outside E_{1} we have

(7)

log |G(z)| ≥ n log(rN^{−2}) − ∑_{k=1}^{n} log |a_{k}|
≥ n log(rN^{−2}) − n log r ≥ −dN^{2}T (r, f),

using (5). Using the fact that F (0) = G(0) = 1, we clearly have

(8)  T (r, g) ≤ dNT (r, f),

by (6). Finally, a standard application of the Poisson–Jensen formula to g(ζ) in |ζ| ≤ r_{1} gives, for \frac{1}{2}r ≤ |z| ≤ r_{2} ,

\begin{aligned}|log |g(z)|| & ≤ \frac{r_{1} + r_{2}}{r_{1} − r_{2}} (m(r_{1}, g) + m(r_{1}, 1/g)) + dn(r_{1}, g) + dn(r_{1}, 1/g) + O(log r) \\ & ≤ dN^{2}T (r, f), \end{aligned}

using (5) and (8) and, combining the last estimate with (6) and (7), the result follows.

We use the following notation in the next lemma and henceforth. If g is meromorphic in 0 ≤ r_{1} ≤ |z| < ∞ then by a result of Valiron [30, p. 15] we may write g(z) = z^{N}h(z)g_{1}(z), in which g_{1} is meromorphic in the plane, N is an integer, and h is analytic in |z| ≥ r_{1} with h(∞) = 1. With n(r, g) the number of poles of g in r_{1} ≤ |z| ≤ r, the Nevanlinna characteristic is defined for r ≥ r_{1} by [2, p. 89]

(9)  \begin{aligned}T (r, g) & = m(r, g) + N(r, g) = \frac{1}{2π} ∫_{0}^{2π} log^{+} |g(re^{iθ})| dθ + N(r, g), \\ N(r, g) & = ∫_{R}^{r} n(t, g) \frac{dt}{t} \end{aligned}

and we have T (r, g) = T (r, 1/g) + O(log r). Further, S(r, g) will denote any quantity such that

(10)  S(r, g) = O(log^{+} T (r, g) + log r)  (n.e.),

in which (n.e.) (‘nearly everywhere’) means as r tends to infinity outside a set of finite measure. In particular, m(r, g′/g) = S(r, g).

We denote sectorial regions using

S^{∗}(r, α, β) = {z : |z| > r, α < arg z < β},

this a region on the Riemann surface of log z if β − α > 2π.

Lemma 3. Suppose that f_{1}, …, f_{N} are functions analytic in the region S = S^{∗}(r_{0}, −π, π) and each admitting unrestricted analytic continuation in |z| > r_{0} , the continuations satisfying

(11)  log^{+} log^{+} |f_{j}(z)| = O(log |z|)

on S^{∗}(r_{0}, −2π, 2π). Suppose that g is meromorphic in |z| > r_{0} . Suppose further that for some positive integer Q, each of the functions g_{1}, …, g_{k} on S is a polynomial in the f_{j}^{(m)} , g^{(m)} , 1 ≤ j ≤ k, 0 ≤ m ≤ Q. Suppose finally that g_{1}, …, g_{k} are linearly independent solutions in S of an equation

(12)  w^{(k)} + ∑_{j=0}^{k−1} A_{j}w^{(j)} = 0,

in which the A_{j} are meromorphic in |z| > r_{0} . Then we have, for j = 0, …, k − 1,

(13)  m(r, A_{j}) = S(r, g).

Proof. In this proof we use characteristic functions defined as in (9). When the f_{j} are meromorphic in |z| > r_{0} , the estimate (13) is well known [6]. We first note that each g_{μ}, being a polynomial in the f_{j}^{(m)} and g^{(m)} , may be written as a quotient of function elements admitting unrestricted analytic continuation in |z| > r_{0} , and that continuing g_{j} in this way around any curve homotopic to zero in |z| > r_{0} leads back to g_{j}. Since each g_{j} solves (12), there exist constants c_{j,m} such that continuing g_{j} once counter-clockwise around any circle |z| = r > r_{0} leads to

∑_{m=1}^{k} c_{j,m}g_{m}.

The usual eigenvalue argument [17, p. 358] then gives a solution g^{∗} of (12), which we can assume without loss of generality is g_{1} , such that under this continuation g_{1} leads to a constant multiple of itself. Thus, for some constant c, the function

h_{1} = z^{c}g_{1}

is meromorphic in |z| > r_{0} . We then have, for j = 1, …, k,

(14) m(r, g_{1}^{(j)}/g_{1}) ≤ ∑_{m=1}^{j} m(r, h_{1}^{(m)}/h_{1}) + O(log r)

≤ O(log^{+} T (r, h_{1}) + log r) (n.e.).

Further, (11) gives, for 1 ≤ j ≤ k,

log^{+} log^{+} |f_{j}^{(m)}(z)| = O(log |z|)

on S^{∗}(2r_{0}, −π, π) and recalling the representation of the g_{μ} as polynomials in the f_{j}^{(m)} and g^{(m)} we get, for some positive constant M,

(15)  T (r, h_{1}) ≤ m(r, h_{1}) + O(N(r, g)) ≤ O(T (r, g) + r^{M})  (n.e.).

We now proceed by induction on k, and for k = 1 the result already follows, since A_{0} = −g′_{1}/g_{1} . Assuming now that the result is true for k − 1, we apply the familiar reduction of order method [17] to write v_{j} = g_{j}/g_{1} for j = 2, …, k, so that each v_{j} solves an equation

(16)  v^{(k)} + ∑_{m=1}^{k−1} B_{m}v^{(m)} = 0,

and the B_{m} are meromorphic in |z| > r_{0} and can be calculated from the coefficients A_{1}, …, A_{k−1} and the g_{1}^{(j)}/g_{1} as follows. We have, with A_{k} = 1,

(17)  B_{m} = A_{m} + ∑_{j=m+1}^{k} A_{j}P_{j,m}(g′_{1}/g_{1}),  m = 1, …, k − 1.

Here each P_{j,m}(g′_{1}/g_{1}) is a differential polynomial in g′_{1}/g_{1} with constant coefficients. We regard (16) as an equation of order k − 1 in the w_{j} = v′_{j} and we then write y_{j} = w_{j}g_{1}^{2} = g′_{j}g_{1} − g_{j}g′_{1} . The y_{j} solve

y^{(k−1)} + ∑_{m=0}^{k−2} C_{m}y^{(m)} = 0,

with coefficients C_{m} meromorphic in |z| > r_{0} , and the y_{j} are themselves polynomials in the f_{j}^{(m)} and g^{(m)} . Further, with B_{k} = 1 we have

(18)  C_{m} = B_{m+1} + ∑_{j=m+2}^{k} B_{j}Q_{j,m}(g′_{1}/g_{1}),  m = 0, …, k − 2,

in which each Q_{j,m}(g′_{1}/g_{1}) is a differential polynomial in g′_{1}/g_{1} , with constant coefficients.

From the induction hypothesis we deduce that m(r, C_{m}) = S(r, g) for each m, so that the same is true of the B_{m}, using (14), (15) and (18). We now have m(r, A_{m}) = S(r, g) for m = 1, …, k − 1, using (17), and (12) and (14) and (15) give m(r, A_{0}) = S(r, g). The induction is complete and the lemma is proved.

Lemma 4. Let k ≥ 1 be an integer and let f_{1}, …, f_{k}, G, H and a_{0}, …, a_{k−1} , A_{0}, …, A_{k−1} all be meromorphic in a domain U. Suppose that f_{1}, …, f_{k} are linearly independent solutions in U of

L_{k}(w) = w^{(k)} + ∑_{j=0}^{k−1} a_{j}w^{(j)} = 0.

Suppose further that the functions f_{1}H + f′_{1}G, …, f_{k}H + f′_{k}G are linearly independent solutions in U of

M_{k}(w) = w^{(k)} + ∑_{j=0}^{k−1} A_{j}w^{(j)} = 0.

Then we have, in U, setting A_{k} = 1 and A_{−1} = a_{−1} = 0,

kH′ + (A_{k−1} − a_{k−1})H = −(\frac{1}{2}k(k − 1)G′′ + (k − 1)A_{k−1}G′ + A_{k−2}G)

+ a_{k−1}(A_{k−1}G + kG′) + G(a′_{k−1} + a_{k−2} − a_{k−1}^{2}). (19)

Proof. When a_{k−1} = 0 this is a special case of Lemma 6 of [6]. Since M_{k}(f_{j}H + f′_{j}G) = 0 we have

(20)  M_{k}(f_{j}H) = −M_{k}(f′_{j}G).

For integers n and m, we use the notation

nC_{m} = \frac{n!}{m!(n − m)!}

when 0 ≤ m ≤ n, and

nC_{m} = 0

otherwise. We also write, for 0 ≤ μ ≤ k,

M_{k,μ}(w) = ∑_{m=μ}^{k}(^{m}C_{μ})A_{m}w^{(m−μ)}, M_{k,−1}(w) = 0.

Thus, for j = 1, …, k,

(21)  \begin{aligned}M_{k}(f_{j}H) & = ∑_{m=0}^{k} A_{m} ∑_{μ=0}^{k}(^{m}C_{μ})f_{j}^{(μ)}H^{(m−μ)} = ∑_{μ=0}^{k} f_{j}^{(μ)}M_{k,μ}(H) \\ & = ∑_{μ=0}^{k−1} f_{j}^{(μ)}(M_{k,μ}(H) − a_{μ}H). \end{aligned}

We also have

M_{k}(f′_{j}G) = ∑_{μ=0}^{k} f_{j}^{(μ+1)}M_{k,μ}(G)

= ∑_{μ=0}^{k−2} f_{j}^{(μ+1)}M_{k,μ}(G) + f_{j}^{(k)}M_{k,k−1}(G) + f_{j}^{(k+1)}G (22)

= ∑_{μ=0}^{k−1} f_{j}^{(μ)}(M_{k,μ−1}(G) − a_{μ}M_{k,k−1}(G) + (a_{μ}a_{k−1} − a′_{μ} − a_{μ−1})G).

Since the Wronskian determinant of the f_{j} is not identically zero, the coefficient of f_{j}^{(μ)} on the right-hand-side of (21) and that on the right-hand-side of (22) must have sum 0, by (20). Now μ = k − 1 gives (19).

The following lemma is from [22].

Lemma A. Let c, M, N be positive constants and let Q(z) be analytic and satisfy |Q(z)| ≤ M + |z|^{M} in a half-plane Re(z) ≥ c. Suppose that Q(n) is an integer for all integers n ≥ N. Then Q is a polynomial.

Lemma 5. Suppose that R, S are rational functions, with R not identically zero, that P, P_{1} are polynomials, with P_{1} non-constant. Suppose that we have P_{1}(P (z) + log R(z)) ≡ S(z) in some domain U. Then R is constant.

Proof. By the hypotheses there is an equation

(23)  ∑_{j=0}^{q} a_{j}(z)w^{j} = 0,

with polynomial coefficients a_{j}, not all 0, having a local solution w = log R(z). The analytic continuations of log R(z) all satisfy the same equation. But log R(z) adds an integer multiple of 2πi as we continue once around a zero or pole of R and, since the solution of (23) has at most q branches, we conclude that R has no zeros or poles and is constant.

4. An estimate for logarithmic derivatives

Lemma 6. Suppose that k ≥ 1 and that f is meromorphic in the plane and that

(24)  F = f^{(k)} + ∑_{j=0}^{k−1} a_{j}f^{(j)},  G = f^{(k)} + ∑_{j=0}^{k−1} b_{j}f^{(j)},

with the a_{j} and b_{j} rational functions. Then either F/G is constant or

(25)  m(r, f′/f) ≤ \overline{N}(r, 1/F) + \overline{N}(r, 1/G) + S(r, f′/f).

Proof. Let f_{1}, …, f_{k} be linearly independent solutions of the equation

(26)  w^{(k)} + ∑_{j=0}^{k−1} a_{j}w^{(j)} = 0

in the domain S = S^{∗}(r_{0}, −π, π). Then the f_{j} all admit unrestricted analytic continuation in |z| > r_{0} , provided r_{0} is large enough. Let

(27)  W = W (f_{1}, …, f_{k}),  W′/W = −a_{k−1}

in S. Then we have, in S,

W (f_{1}, …, f_{k}, f) = W F

and so

W ((f_{1}/f)′, …, (f_{k}/f)′) = (−1)^{k}W F f^{−k−1}

and

W (w_{1}, … w_{k}) = W F/f, w_{j} = −f′_{j} + f_{j}f′/f.

Thus the w_{j} are linearly independent solutions in S of an equation

(28)  w^{(k)} + ∑_{j=0}^{k−1} A_{j}w^{(j)} = 0,  A_{k−1} = −W′/W + f′/f − F′/F.

We assert that the A_{j} are meromorphic in |z| > r_{0} , establishing this in the standard way by noting that if A_{j}^{∗} and w_{m}^{∗} are respectively the function elements obtained by analytically continuing A_{j} and w_{m} once counter-clockwise around |z| = r > r_{0} , then w_{m}^{∗} is a linear combination of the w_{j} in S, and w_{1}^{∗}, …, w_{k}^{∗} are linearly independent by the law of permanence of functional relations. Since we have

∑_{j=0}^{k−1}(A_{j}^{∗} − A_{j})w_{m}^{(j)} = 0

in S for 1 ≤ m ≤ k, we deduce that A_{j}^{∗} = A_{j} on S. Our assertion established, we have, by Lemma 3,

(29)  m(r, A_{j}) = S(r, f′/f),  j = 0, …, k − 1.

We may apply Lemma 4, with H = f′/f and G = −1, to obtain, using (27), (28) and (29),

\begin{aligned}k(f′/f)′ + (A_{k−1} − a_{k−1})f′/f & = k(f′/f)′ + (f′/f − F′/F)f′/f = C, \\ m(r, C) & = S(r, f′/f). \end{aligned}

The same argument with the a_{j} replaced by b_{j} gives

\begin{aligned}k(f′/f)′ + (B_{k−1} − b_{k−1})f′/f & = k(f′/f)′ + (f′/f − G′/G)f′/f = D, \\ m(r, B_{k−1}) + m(r, D) & = S(r, f′/f). \end{aligned}

We therefore have

\begin{aligned}A^{∗}f′/f & = E^{∗}, \\ A^{∗} & = (A_{k−1} − a_{k−1}) − (B_{k−1} − b_{k−1}) = G′/G − F′/F, \\ m(r, A^{∗}) + m(r, E^{∗}) & = S(r, f′/f), \end{aligned}

and either F/G is constant or A^{∗} ≢ 0, in which case we obtain (25), on writing f′/f = E^{∗}/A^{∗} and

m(r, f′/f) ≤ m(r, E^{∗}) + m(r, A^{∗}) + N(r, A^{∗}) + O(1).

5. Estimates for counting functions

We use the following notation throughout this section. Let k be a positive integer and let f be a meromorphic function in the plane. Let F and G be given by (24), with the a_{j}, b_{j} rational functions and a_{j} − b_{j} ≢ 0 for at least one j. Assume that neither F nor G vanishes identically. Define V , E by

(30)  F = V G,  E = G − F = (1 − V)G = ∑_{j=0}^{k−1}(b_{j} − a_{j})f^{(j)}.

We begin with some basic estimates. Dividing the first relation of (30) through by f and writing each f^{(j)}/f as a differential polynomial in f′/f, we see at once that

(31)  T (r, F/f) + T (r, G/f) ≤ O(T (r, f′/f)) + S(r, f′/f).

Since all but finitely many zeros and poles of V arise from zeros of F and G, we also have

(32)  \begin{aligned}m(r, V′/V) & = S(r, f′/f), \\ T (r, V′/V) & ≤ \overline{N}(r, 1/F) + \overline{N}(r, 1/G) + S(r, f′/f). \end{aligned}

We now estimate m(r, F′/F) and m(r, G′/G). From (31) and the relations

F′/F = (F/f)′/(F/f) + f′/f, V′/V = F′/F − G′/G,

we see that any term which is S(r, F′/F), or S(r, G′/G) or S(r, V′/V), is an S(r, f′/f), while if V is non-constant then, using Lemma 6,

(33) m(r, F′/F) ≤ m(r, f′/f) + S(r, f′/f) ≤ \overline{N}(r, 1/F) + \overline{N}(r, 1/G) + S(r, f′/f).

The same estimate plainly holds with F′/F replaced by G′/G.

Lemma 7. Let V be as in (30). Then there exists an integer q with 1 ≤ q ≤ k such that G satisfies

(34)  0 = (1 − V)G^{(q)} + ∑_{j=0}^{q−1} T_{j}(V)G^{(j)},

in which each T_{j}(V) has a representation

T_{j}(V) = α_{j} + ∑_{m=0}^{q} β_{m,j}V^{(m)},

with the coefficients α_{j} and β_{m,j} rational functions.

Proof. We begin by recalling (30). Thus, for s = 0, …, k, we have

(35)  E^{(s)} = (1 − V)G^{(s)} − ∑_{j=1}^{s} \frac{s!}{j!(s − j)!}V^{(j)}G^{(s−j)}.

Using the division algorithm for linear differential operators, each E^{(s)} may be written in the form

E^{(s)} = ∑_{m=0}^{s−1} c_{m,s}G^{(m)} + ∑_{m=0}^{k−1} d_{m,s}f^{(m)}, s = 0, …, k,

the first sum not appearing when s = 0, and with the c_{m,s} and d_{m,s} rational functions. Now the k by k + 1 matrix (d_{m,s}) has rank at most k and so by elementary linear algebra its columns are linearly dependent over the field of rational functions. Hence there are rational functions δ_{s}, not all identically zero, such that

∑_{s=0}^{k} δ_{s}d_{m,s} ≡ 0

for 0 ≤ m ≤ k − 1. Thus, if q is the largest s such that δ_{s} ≢ 0, then q > 0 because, by hypothesis, at least one d_{m,0} is non-zero, and we have

∑_{s=0}^{q} δ_{s}E^{(s)} = ∑_{t=0}^{q−1} d_{t}G^{(t)},

with coefficients d_{t} which are rational functions. Replacing the E^{(s)} using (35), we have an equation as asserted.

We now make some estimates for the number of zeros and poles of f, under certain assumptions on the coefficients.

Lemma 8. If a_{k−1} ≡ b_{k−1} then

\overline{N}(r, f) ≤ \overline{N}(r, 1/F) + \overline{N}(r, 1/G) + S(r, f′/f).

To prove Lemma 8 we write, using (30),

(36)  (1 − V)f^{(k)} + ∑_{j=0}^{k−1}(a_{j} − V b_{j})f^{(j)} = 0.

Suppose that z is large and that f(z) = ∞. Then V (z) = 1 and, by the hypothesis that a_{k−1} ≡ b_{k−1} , we have a_{k−1}(z) − V (z)b_{k−1}(z) = 0. Dividing (36) through by f^{(k−1)} , we see that (1 − V)f^{(k)}/f^{(k−1)} vanishes at z. Thus V′(z) = V (z) − 1 = 0 and the result follows from (32).

Lemma 9. Suppose that the equations

L_{1}(w) = w^{(k)} + ∑_{j=0}^{k−1} a_{j}w^{(j)} = 0, L_{2}(w) = w^{(k)} + ∑_{j=0}^{k−1} b_{j}w^{(j)} = 0

have no non-trivial common (local) solution, and that V is non-constant. Then we have

(37) N(r, 1/f) ≤ 2N(r, 1/F)+4k\overline{N}(r, 1/F)+5k\overline{N}(r, 1/G)+S(r, f′/f)  (n.e.).

Proof. Since the equations L_{1}(w) = 0, L_{2}(w) = 0 have no non-trivial common (local) solution, it follows using Lemma 1 that there exist a rational function b, not identically zero, and linear differential operators L, M with coefficients which are rational functions and with order at most k, such that we have b = L(L_{1}) + M(L_{2}) and hence

bf = L(F) + M(G),

and using (30) we write the last relation in the form

bf/F = L(F)/F + M(G)/F = L(F)/F + M(G)/GV.

But L(F)/F may be written as a polynomial of degree at most k in F′/F and its derivatives, with coefficients which are rational functions. Using (33) applied to F and G and the remark preceding (33), this gives

(38)  m(r, f/F) ≤ m(r, 1/V) + 2k(\overline{N}(r, 1/F) + \overline{N}(r, 1/G)) + S(r, f′/f).

We now write the equation (34) in the form

\begin{aligned}1/V & = V_{1}/V_{2}, \\ V_{1} & = (G^{(q)}/G + ∑_{j=0}^{q−1} M_{j}G^{(j)}/G), \\ V_{2} & = (G^{(q)}/G + ∑_{j=0}^{q−1} N_{j}G^{(j)}/G). \end{aligned}

Here each M_{j} is a differential polynomial in V′/V , with coefficients which are rational functions, and each N_{j} is a rational function. We now have, by (32) and (33), applied to G,

m(r, V_{j}) ≤ q(\overline{N}(r, 1/F) + \overline{N}(r, 1/G)) + S(r, f′/f)

for j = 1, 2, as well as

N(r, V_{2}) ≤ q(\overline{N}(r, f) + \overline{N}(r, 1/G)) + S(r, f′/f).

Combining these estimates with (38), we have

(39) m(r, f/F) ≤ q\overline{N}(r, f)+(2k+2q)\overline{N}(r, 1/F)+(2k+3q)\overline{N}(r, 1/G)+S(r, f′/f).

Since 1/f = (F/f)(1/F) and since each pole z of f with z large is a pole of F/f of order k, we can now write

N(r, 1/f) + k\overline{N}(r, f) ≤ N(r, F/f) + N(r, 1/F) + O(log r)
≤ T (r, f/F) + N(r, 1/F) + O(log r)
≤ m(r, f/F) + 2N(r, 1/F) + O(log r)

to obtain (37), using (39) and the fact that q ≤ k.

Lemma 10. With the hypotheses of Lemma 9, suppose that

N(r, 1/F) + N(r, 1/G) = S(r, f′/f).

Then

N(r, 1/f) = S(r, f′/f),

and

T (r, f′/f) ≤ \overline{N}(r, f) + S(r, f′/f) ≤ T (r, V) + S(r, f′/f).

The proof is obvious, using Lemmas 6 and 9 and recalling that V = 1 at all but finitely many poles of f.

6. A growth lemma

Lemma 11. Suppose that f is meromorphic in the plane and that F and G are given by (24), with k ≥ 1 and with the a_{j} and b_{j} rational functions, such that a_{j} ≢ b_{j} for at least one j. Suppose that F and G have only finitely many zeros. Then f has finite order.

We remark that when k ≥ 3 and at least one of F and G has polynomial coefficients, it already follows from the hypotheses and Theorem 2 of [6] that f′/f has finite order.

Proof of Lemma 11. Suppose that k and the functions f(z), F (z), G(z) are as in the hypotheses, and suppose that f is meromorphic of infinite order in the plane. Then F/G is transcendental, because otherwise f would be a solution of a homogeneous linear differential equation with rational functions as coefficients, and f would have finite order. With the notation of Lemma 1, and with a_{k} = b_{k} = 1, we may write F = L_{1}(f), G = L_{2}(f). Let H be the operator of Lemma 1.

Suppose first that H has positive order. In this case we set g_{1} = H(f) = ∑_{j=0}^{q} c_{j}f^{(j)} , in which 0 ≤ q ≤ k − 1, and the c_{j} are rational functions, with c_{q} ≢ 0. By Lemma 1 there are differential operators Q_{j} and P_{j}, with coefficients which are rational functions, such that

g_{1} = P_{1}(F) + P_{2}(G),  F = Q_{1}(g_{1}),  G = Q_{2}(g_{1}),

and, dividing through by the leading coefficients, there are linear differential operators Q_{1}^{∗}, Q_{2}^{∗}, each of form

Q_{j}^{∗} = D^{k−q} + ∑_{m=0}^{k−q−1} a_{j,m}D^{m}

and having coefficients which are rational functions, such that Q_{1}^{∗}(g_{1}) and Q_{2}^{∗}(g_{1}) both have finitely many zeros.

Lemma 12. If g_{1} has finite order then so has f.

Proof. Assume that g_{1} has finite order. Then since all but finitely many poles of f are poles of g_{1}/f of multiplicity q, it follows that N(r, f) has finite order and we can write f = f_{1}/f_{2} = f_{1}u_{2} , with the f_{j} entire, f_{2} of finite order. For |z| outside a set F_{0} of finite measure, standard estimates from [10, p. 22] or [8] give f_{2}^{(m)}(z)/f_{2}(z) = O(|z|^{d_{1}}), for 1 ≤ m ≤ q, in which d_{1} is a positive constant. Substituting f = f_{1}u_{1} into the equation H(f) = g_{1} and dividing through by f_{1}u_{1} , we obtain an equation

f_{1}^{(q)}/f_{1} + ∑_{j=0}^{q−1} A_{j}f_{1}^{(j)}/f_{1} = g_{1}/f = g_{1}f_{2}/f_{1},

in which the coefficients A_{j} satisfy A_{j}(z) = O(|z|^{d_{2}}), for 0 ≤ j ≤ q − 1, and for |z| outside F_{0} , where d_{2} is a positive constant. A standard application of the Wiman–Valiron theory [12, Theorem 12] (see also [30]) now shows that f_{1} has finite order and so has f. This proves Lemma 12.

Returning to the proof of Lemma 11, we may assume henceforth that H has order 0, that is, that the equations

L_{1}(w) = 0, L_{2}(w) = 0

have no non-trivial common (local) solution. Then by Lemma 10 we have

(40)  T (r, f′/f) ≤ cT (r, F/G)  (n.e.),

using c throughout this proof to denote a positive constant, not necessarily the same at each occurrence.

Since F and G are given by (24), we may write

(41)  \begin{aligned}F (z) & = R(z)e^{P (z)}G(z), \\ E(z) & = G(z) − F (z) = (1 − R(z)e^{P (z)})G(z) = ∑_{j=0}^{k−1} B_{j}(z)f^{(j)}(z), \end{aligned}

with P entire, and with R and the B_{j} rational functions. If |z| is large and f has a pole of multiplicity n at z then, dividing the equation F = Re^{P} G through by f^{(k−1)} , we obtain

(42) R(z)e^{P (z)} = 1,  (n + k − 1)(R′(z)/R(z) + P′(z)) = b_{k−1}(z) − a_{k−1}(z)

and so

(43)  log^{+} |P′(z)| = O(log^{+} |z|).

Also, if R(z)e^{P (z)} = 1 and |z| is large, then either f has a pole at z, or E(z) = 0.

Suppose that P is a polynomial. Since R(z)e^{P (z)} = 1 at all but finitely many poles of f we have \overline{N}(r, f) ≤ T (r, Re^{P}) + O(log r). But by (42) the multiplicity n of a pole of f at z is bounded by a power of |z|. Therefore

log^{+} N(r, f) = O(log r)

and we have f = f_{1}/f_{2} in which the f_{j} are entire and f_{2} is not identically zero but has finite order. There then exists a subset E^{∗} of (1, ∞) of infinite logarithmic measure such that for |z| = r in E^{∗} we have

|R(z)e^{P (z)} − 1| ≥ r^{−c},  |f_{2}^{(j)}(z)/f_{2}(z)| ≤ r^{c},  1 ≤ j ≤ k,

the easiest way to establish this being to write

1/(Re^{P} − 1) = R^{−1}e^{−P} (1 − R^{−1}e^{−P})^{−1}

and then use standard estimates [10, p. 22] for the logarithmic derivative of the function 1 − R^{−1}e^{−P} . As in the proof of Lemma 12 a standard application of the Wiman–Valiron theory [12, Theorem 12] to the relation G = (1 − Re^{P})^{−1}E shows that f_{1} has finite order and so does f. Therefore we may assume for the rest of the proof that P is transcendental.

Take a large positive r_{0} , normal for P with respect to the Wiman–Valiron theory [12], [30], and such that, using (40) and (41),

(44)  T (r_{0}, f^{(j)}/f) < cT (r_{0}, F/G) = cT (r_{0}, Re^{P})

for j = 1, …, k. For non-zero complex v, and positive K, we define the logarithmic rectangle

D(v, K) = {u = ve^{τ} : | Re(τ)| ≤ KN^{−2/3}, | Im(τ)| ≤ KN^{−2/3}},

in which N = ν(r_{0}, P) is the central index of P, and is large if r_{0} is large. By Lemma 2, (41) and (44) we have, for j = 1, …, k,

(45) log |f^{(j)}(z)/f(z)| ≤ cN^{2}T (r_{0}, Re^{P}), log |E(z)/f(z)| ≥ −cN^{2}T (r_{0}, Re^{P}),

for all z with \frac{1}{2}r_{0} ≤ |z| ≤ r_{0}e^{−2/N} and lying outside a union D_{0} of open discs having sum of radii at most cr_{0}N^{−2} , so that there is a subset D_{1} of [0, 2π], having measure at most cN^{−2} , such that some determination of arg ζ is in D_{1} for every ζ in D_{0} .

Choose z_{0} with |z_{0}| = r_{0} and |P (z_{0})| = M(r_{0}, P). On D(z_{0}, 128) we have [12, Theorem 12]

(46) P (z) + log R(z) = P (z)(1 + o(1)) = P (z_{0})(z/z_{0})^{N}(1 + o(1)) = αζ^{N},

(47) α = P (z_{0})z_{0}^{−N},  ζ = z(1 + o(1/N)),  P′(z)/P (z) = (1 + o(1))N/z.

In particular, P′(z) is large on D(z_{0}, 128) so that, using (43), there are no poles of f in D(z_{0}, 128), and by (41) every zero of R(z)e^{P (z)} −1 in D(z_{0}, 128) is simple and is a zero of E.

On D(z_{0}, 128) we write

z = z_{0}e^{τ} , ζ = z_{0}e^{σ}, σ = τ + o(1/N),

so that

\frac{dσ}{dτ} = 1 + o(N^{−1/3})

and, by convexity, σ is a univalent function of τ for | Re(τ)| ≤ 64N^{−2/3} , | Im(τ)| ≤ 64N^{−2/3} . Further,

(48)  \frac{dζ}{dz} = \frac{ζ}{z} \frac{dσ}{dτ} = 1 + o(N^{−1/3})

on D(z_{0}, 64). In addition, the image of D(z_{0}, 64) under ζ = ζ(z) contains D(z_{0}, 32), and αζ^{N} is large for ζ in D(z_{0}, 32).

If c_{0} is a positive constant there exists a positive constant c_{1} such that on each circle |w| = (2n + 1)π, with n a positive integer, and on the ray arg w = 0, |w| ≥ c_{0} , we have |e^{w} − 1| ≥ c_{1} . We choose σ_{0} such that

σ_{0} ∈ [−16N^{−1}, −8N^{−1}], |αz_{0}^{N}e^{Nσ_{0}}| = (2n + 1)π

for some integer n, and we choose m_{1} , n_{1} , m_{2} , n_{2} such that

m_{1}, n_{1}, m_{2}, n_{2} ∈ [4N^{−2/3}, 8N^{−2/3}]

and

arg(αζ^{N}) = 0 for ζ = z_{0}e^{σ}, Im(σ) ∈ {−n_{1}, n_{2}},

and such that |(α/π)z_{0}^{N}e^{Nσ}| is an odd integer for Re(σ) in {σ_{0} − m_{1}, σ_{0} + m_{2}}. Thus on the boundary of the logarithmic rectangle

B = {ζ = z_{0}e^{σ} : σ_{0} − m_{1} ≤ Re(σ) ≤ σ_{0} + m_{2}, −n_{1} ≤ Im(σ) ≤ n_{2}}, and on the arc L_{0} given by

(49) ζ = z_{0}e^{σ_{0}+iλ}, −n_{1} ≤ λ ≤ n_{2},

we have

(50)  |e^{αζ^{N}} − 1| ≥ c_{1} > 0.

Now L_{0} lies in |ζ| ≤ r_{0}e^{−8/N} and so using (47) the image z(L_{0}) of L_{0} under the mapping z = z(ζ) lies in

|z| ≤ r_{0}e^{−8/N}(1 + o(1)/N) ≤ r_{0}e^{−2/N}.

Further, if L_{1} is the sub-arc of L_{0} given by −1/N ≤ λ ≤ 1/N in (49), then the variation of arg z on z(L_{1}) is, by (47), at least (c−o(1))/N, and using the remark following (45) we may therefore choose ζ_{1} lying on L_{1} , such that the inequalities of (45) all hold at z_{1} = z(ζ_{1}). Note that D(ζ_{1}, 2) is contained in B, provided r_{0} is large enough, while B in turn lies in D(z_{0}, 16), and z(B) lies in D(z_{0}, 32).

Lemma 13. The number of zeros of e^{αζ^{N}} − 1 in D(ζ_{1}, 1) is at least ce^{N^{1/3}}M(r_{0}, P).

Proof. We have |ζ_{1}| = r_{0}e^{σ_{0}} and σ_{0} is in [−16/N, −8/N], so that |αζ_{1}^{N}| = |P (z_{0})|e^{γ}, for some γ in [−16, −8], and the image of D(ζ_{1}, 1) under w = αζ^{N} covers the annulus

|P (z_{0})|e^{−N^{1/3}+γ} ≤ |w| ≤ |P (z_{0})|e^{N^{1/3}+γ},

so that the number of zeros of e^{αζ^{N}} − 1 in D(ζ_{1}, 1) is at least ce^{N^{1/3}}|P (z_{0})| = ce^{N^{1/3}}M(r_{0}, P). This proves Lemma 13.

We may now complete the proof of Lemma 11. Let g(z) = f(z)/f(z_{1}), and let C be the union of L_{0} and the boundary of B. Using (46) and (50) and the relation G = (1 − Re^{P})^{−1}E, we have, on z(C),

(51)  g^{(k)}(z) = ∑_{j=0}^{k−1} s_{j}(z)g^{(j)}(z),  s_{j}(z) = O(|z|^{c}).

Since |dz| ≤ 2|dζ|, by (48), the arc length of z(C) is o(r_{0}). We write the equation (51) in vector form as

I′(z) = A(z)I(z), I(z) = (g^{(k−1)}(z), …, g(z))^{T} ,

in which the k by k matrix A has entries which are O(r_{0})^{c} on z(C). Writing

I(z) = I(z_{1}) + ∫_{z_{1}}^{z} A(u)I(u) du, S(z) = max{|g^{(j)}(z)| : j = 0, …, k − 1},

we have

S(z) ≤ V (z) = S(z_{1}) + ∫_{z_{1}}^{z} r_{0}^{c}S(u) |du|,

and the standard Gronwall method [1, p. 35] (see also [15], [16], [20]) gives, with t denoting arc length on z(C),

\frac{d}{dt}(V (z(t))) ≤ r_{0}^{c}S(z(t)) ≤ r_{0}^{c}V (z(t)),

and so

S(z(t)) ≤ V (z(t)) ≤ V (z_{1}) exp(r_{0}^{c}t) ≤ S(z_{1}) exp(r_{0}^{c}).

We thus have, for j = 0, …, k − 1,

|g^{(j)}(z)| ≤ S(z) ≤ S(z_{1}) exp(r_{0}^{c}) ≤ exp(N^{c}T (r_{0}, Re^{P}) + r_{0}^{c}),

using (45), and so

|E(z)/f(z_{1})| ≤ exp(N^{c}T (r_{0}, Re^{P}) + r_{0}^{c})

on z(C). Hence the function H_{1} defined by H_{1}(ζ) = E(z)/f(z_{1}) satisfies

log |H_{1}(ζ)| ≤ N^{c}T (r_{0}, Re^{P})

for all ζ on C, and so for all ζ in B, and hence for all ζ in D(ζ_{1}, 2), by the maximum principle. But we also have, by (45),

log |H_{1}(ζ_{1})| = log |E(z_{1})/f(z_{1})| ≥ −N^{c}T (r_{0}, Re^{P}).

Mapping D(ζ_{1}, 2) to the unit disc, using w = ϕ(ζ), with ζ_{1} mapped to 0, and writing J(w) = H_{1}(ζ), we have, for 0 < r < 1,

T (r, 1/J) ≤ T (r, J) + log |1/J(0)| ≤ log M(r, J) + log |1/H_{1}(ζ_{1})| ≤ N^{c}T (r_{0}, Re^{P}).

Thus the number of zeros of H_{1}(ζ) in D(ζ_{1}, 1) is at most N^{c}T (r_{0}, Re^{P}). Hence, using (41) and (46) and the remark following (47), the number of zeros of e^{αζ^{N}} −1 in D(ζ_{1}, 1) is at most

N^{c}T (r_{0}, Re^{P}) ≤ N^{c} log M(r_{0}, Re^{P}) ≤ N^{c}(M(r_{0}, P) + O(log r)) ≤ N^{c}M(r_{0}, P).

This contradicts Lemma 13 and Lemma 11 is proved.

7. Proof of Theorem 1

Suppose that f and F and G are as in the hypotheses. Then we know by Lemma 11 that f has finite order. If F/G is constant then f has finitely many poles and since F has finitely many zeros we have F = R_{1}e^{V} with R_{1} a rational function and V a polynomial. Since G is a constant multiple of F and since Lemma 1 gives f = V_{1}(F) + V_{2}(G), in which V_{1} and V_{2} are linear differential operators, the coefficients of which are rational functions, we deduce that f = R_{2}e^{V} with R_{2} a rational function, and f′/f is a rational function.

Assume henceforth that F/G is non-constant. It follows from Lemma 10 and (10) that f has only finitely many zeros. If f has only finitely many poles then again f′/f is a rational function. We assume henceforth that f has infinitely many poles.

We have (41), with R a rational function and P a non-constant polynomial. Since

m(r, f′/f) + N(r, 1/f) = O(log r),

the order ρ of T (r, f′/f) is the same as that of \overline{N}(r, f). Writing F/f and G/f as differential polynomials in f′/f with coefficients which are rational functions, it is now clear that

(52)  deg (P) ≤ ρ = lim sup_{r→∞} \frac{log \overline{N}(r, f)}{log r}.

Let r_{0} be large and positive. We define, in the domain U = S^{∗}(r_{0}, −π, π), linearly independent solutions f_{1}, …, f_{k} of the equation (26), and the Wronskian W = W (f_{1}, …, f_{k}) satisfies (27) in U. We further define g, h in U by

(53)  g^{−k} = F/f,  h = (−f′/f)g.

Then g and h are analytic in U and g, h, W and the f_{j} all admit unrestricted analytic continuation in |z| > r_{0} , the continuations of these functions H_{m} all satisfying

(54)  log^{+} log^{+} |H_{m}(z)| = O(log |z|)

on S^{∗}(r_{0}, −2π, 2π).

We have

W (f_{1}, …, f_{k}, f) = W F = W fg^{−k}

and hence

W ((f_{1}/f)′, …, (f_{k}/f)′) = (−1)^{k}W f^{−k}g^{−k}

and

(55)  W (f_{1}h + f′_{1}g, …, f_{k}h + f′_{k}g) = (−1)^{k}W

in U. Thus the functions f_{j}h + f′_{j}g, for j = 1, …, k, are linearly independent solutions in U of an equation

(56)  w^{(k)} + ∑_{j=0}^{k−1} A_{j}w^{(j)} = 0,  A_{k−1} = −W′/W.

We assert that the A_{j} are rational functions. First, if E_{1} is the set of all singular points of the equation (26) as well as of all zeros of f and F then E_{1} is finite and the f_{j} and g and h all admit unrestricted analytic continuation in the complement Ω of E_{1} in the plane. Further, since g^{k} and h^{k} are meromorphic, and since the f_{j} form a fundamental solution set of (26), analytic continuation of any of the functions f_{1}h + f′_{1}g, …, f_{k}h + f′_{k}g once around any point of E_{1} leads back to a linear combination of the same functions. By (55) and the standard representation for the A_{j} as quotients of determinants, we deduce that the A_{j} are analytic in Ω. By (54) and Lemma 3, they satisfy

m(r, A_{j}) = O(log r), r → ∞.

Thus the A_{j} each have at most a pole at infinity, and a similar analysis in a punctured neighbourhood of each point of E_{1} shows that the A_{j} are rational functions.

We denote henceforth by d_{j} rational functions. Since each f_{j}h + f′_{j}g satisfies (56) we obtain, using (26), (27), (56) again and Lemma 4,

(57)  h′ = −\frac{1}{2}(k − 1)g′′ + d_{1}g′ + d_{2}g.

However, we may define Y and g_{1} , h_{1} on U by

(58)  Y^{k} = Re^{P} ,  g_{1} = Y g,  h_{1} = Y h,

and using (41) and (53) we have

G = Y^{−k}F = g_{1}^{−k}f.

The same method as above gives us an equation

h′_{1} = −\frac{1}{2}(k − 1)g′′_{1} + d_{3}g′_{1} + d_{4}g_{1}

in U, which leads at once to

(59)  h′ + (Y′/Y)h = −\frac{1}{2}(k − 1)g′′ + d_{5}g′ + d_{6}g,

using (58). Thus (57) and (59) give

(60)  h = d_{7}g′ + d_{8}g,  −f′/f = d_{7}g′/g + d_{8}.

The equations (60) continue to hold under analytic continuation of g and h. Further, d_{7}(z) is a positive integer at a pole z of f with |z| large. Hence d_{7} ≢ −\frac{1}{2}(k − 1). Therefore (57) and (60) together give

g′′ + D_{1}g′ + D_{0}g = 0,

with coefficients D_{j} which are rational functions.

Writing

(61)  g = uv,  2v′/v = −D_{1},

the function u admits unrestricted analytic continuation in |z| > r_{0} and solves an equation

(62)  u′′(z) + a(z)u(z) = 0,

in which a is a rational function. We assume that either a(z) ≡ 0 or

(63)  a(z) = α_{m}z^{m}(1 + o(1)),  z → ∞,

in which m is an integer and α_{m} ≠ 0. If a(z) ≡ 0 or m ≤ −2 we can take any sectorial region U_{1} given by |z| > r_{1} , | arg z − θ_{1}| ≤ \frac{1}{2}π. We can estimate the number n(r, U_{1}, 1/u) of zeros of u, and hence zeros of g, in the set {z ∈ U_{1} : |z| ≤ r} as follows. Under the assumption m ≤ −2 the equation (62) has a regular singular point at infinity [17], and there exist a constant d and a solution u_{1} of (62), such that in the sectorial region U_{1} we have

u_{1}(z) = z^{d}ϕ(z) = z^{d}(1 + o(1)),

in which ϕ(z) is analytic in |z| > r_{0} with ϕ(∞) = 1. A second solution of (62) may be obtained by writing

(u_{2}/u_{1})′ = u_{1}^{−2},

so that, subtracting a constant if necessary,

u_{2}(z)/u_{1}(z) = (1 + o(1))(1 − 2d)^{−1}z^{1−2d}

in U_{1} , provided d ≠ \frac{1}{2} , while if d = \frac{1}{2} we get

u_{2}(z)/u_{1}(z) = (1 + o(1)) log z.

Writing u as a linear combination of u_{1} and u_{2} in U_{1} we deduce that

n(r, U_{1}, 1/u) = O(log r), r → ∞,

which contradicts (52). We may assume henceforth that a(z) ≢ 0 and m ≥ −1 in (63).

Now asymptotic representations for the solutions of (62) are obtained by the method of Hille [15], [16], as follows. The critical rays for (62) are those rays arg z = θ_{0} for which

arg α_{m} + (m + 2)θ = 0 mod 2π.

If arg z = θ_{0} is a critical ray and ε is a positive constant then in the sectorial region

S_{0} = S^{∗}(r_{0}, θ_{0} + ε − 2π/(m + 2), θ_{0} − ε + 2π/(m + 2))

we write z^{∗} = 2r_{0}e^{iθ_{0}} and

(64)  Z = ∫_{z^{∗}}^{z} a(t)^{1/2} dt = 2α_{m}^{1/2}(m + 2)^{−1}z^{(m+2)/2}(1 + o(1)),  z → ∞,

and we have principal solutions u_{1} , u_{2} of (62) satisfying

(65)  u_{j}(z) = a(z)^{−1/4} exp(iZ(−1)^{j} + o(1))

in S_{0} . In one of the sectorial regions

S_{1} = S^{∗}(r_{0}, θ_{0}+ε, θ_{0}−ε+2π/(m+2)), S_{2} = S^{∗}(r_{0}, θ_{0}+ε−2π/(m+2), θ_{0}−ε),

we have u_{1}(z)/u_{2}(z) → 0 as |z| → ∞, and we refer to u_{2} as dominant and u_{1} as sub-dominant in that sectorial region, while in the other we have u_{2}(z)/u_{1}(z) → 0 and u_{1} is dominant. If u^{∗} is any solution of (62), then u^{∗} has at most finitely many zeros in S_{1}∪S_{2} . Both principal solutions u_{1} , u_{2} admit unrestricted analytic continuation in |z| > r_{0} , although not generally without zeros.

It follows from these asymptotics that we have

n(r, U_{1}, 1/u) = O(r^{(m+2)/2}), r → ∞,

for any sectorial region U_{1} as above. Hence the degree n of P satisfies, by (52),

(66)  n ≤ \frac{1}{2}(m + 2).

We take a critical ray arg z = θ_{0} of (62) such that f has infinitely many poles in |z| > r_{0} , | arg z − θ_{0}| ≤ π/(m + 2), and we write

u = C_{1}u_{1} − C_{2}u_{2}

there, with C_{1} , C_{2} constants, both necessarily non-zero. The function ζ = ±(1/2πi) log(C_{2}u_{2}/C_{1}u_{1}) maps the sectorial region S_{0} conformally onto a region containing a half-plane Re(ζ) ≥ c. At each point in S_{0} where ζ is an integer, we have u = 0 and hence f = ∞ and hence Re^{P} = 1, so that (P + log R)/2πi is an integer. Writing P + log R as a function of ζ and applying Lemma A, we obtain a polynomial P_{1} such that we have

P + log R = P_{1}(ζ).

But (66) and the asymptotics (64), (65) for u_{1} , u_{2} and ζ force P_{1} to be linear. Consequently there exist constants c, c^{∗} such that C_{2}u_{2}/C_{1}u_{1} = c^{∗}(Re^{P})^{c}. Hence u′_{2}/u_{2} − u′_{1}/u_{1} is a rational function, and so are u_{2}u_{1} and u′_{1}/u_{1} and u′_{2}/u_{2} . So using (61) there exist rational functions T_{j} such that we have

g′/g = T_{1} +u′/u = T_{2} +T_{3}(u_{2}/u_{1})′(1−C_{2}u_{2}/C_{1}u_{1})^{−1} = T_{4} +T_{5}(c^{∗}(Re^{P})^{c} −1)^{−1},

On the zeros of pairs of linear differential polynomials 435

and, using the second equation of (60),

f′/f = T_{6} + T_{7}(c^{∗}(Re^{P})^{c} − 1)^{−1}.

By analytic continuation, R^{c} must be a rational function, and we can write

f′/f = T_{6} + T_{7}(Se^{Q} − 1)^{−1},

with S a rational function and Q a non-constant polynomial. Examining the residue of f′/f at a zero of Se^{Q} − 1, a further application of Lemma A shows that T_{7} has a representation

T_{7} = P_{2}(Q + log S)(Q′ + S′/S),

with P_{2} a polynomial, and by Lemma 5 either S or P_{2} is constant.

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Received 26 September 1997