Meromorphic solutions of some linear functional equations

Aequationes Math. 60 (2000) 148–1660001-9054/00/010148-19 $ 1.50+0.20/0

c© Birkhauser Verlag, Basel, 2000

Aequationes Mathematicae

Meromorphic solutions of some linear functional equations

Janne Heittokangas, Ilpo Laine, Jarkko Rieppo and Degui Yang

Summary. We consider meromorphic solutions in the complex plane to functional equations ofthe form

∑n

j=0aj(z)f(cjz)=Q(z), where 0 < |c| < 1, n ∈ N and a0, . . . , an, Q are meromorphic

functions. If the coefficients a0, . . . , an are constants and∑n

j=0ajc

jk 6=0 for all k ∈ Z, thenexactly one meromorphic solution exists. In the general case, we give growth estimates for thesolutions f as well as for the exponent of convergence λ(1/f) of poles and λ(f) of zeros of f .Similar results hold in the case 1 < |c| < +∞. Concluding remarks show that the case |c| = 1 isdifferent.

Mathematics Subject Classification (2000). Primary 39B32; Secondary 30D35.

Keywords. Complex functional equations, meromorphic solutions, Nevanlinna theory.

1. Introduction

This article is devoted to considering solutions meromorphic in the complex planeC to functional equations of type

n∑j=0

aj(z)f(cjz) = Q(z), (1.1)

where c ∈ C \ {0}, n ∈ N, and the coefficients a0, . . . , an, Q are complex functionsto be specified later on. Moreover, we may assume that a0an 6≡ 0. Throughout, weuse the notation s = 1/c. Unless otherwise specified, we assume that 0 < |c| < 1.

We also need to consider the corresponding homogeneous equation

n∑j=0

aj(z)f(cjz) = 0. (1.2)

Reminiscent to the standard theory of linear differential equations, the meromor-phic solutions to (1.2) form a linear space with the scalar field C. Moreover, if f0is a particular meromorphic solution to (1.1), then any meromorphic solution f

Vol. 60 (2000) Meromorphic solutions of some linear functional equations 149

to (1.1) may be represented in the form f = fh + f0, where fh is a meromorphicsolution to (1.2).

For f meromorphic in C, let ρ(f) be its order of growth, λ(f) the exponent ofconvergence of its zeros, and λ(1/f) the exponent of convergence of its poles. Inaddition we assume that the reader is familiar with the fundamental results andthe standard notations of the Nevanlinna theory, see e.g. [3] or [5].

The following two results were proved in [2].

Theorem 1.1. Let the coefficients a0, . . . , an, Q of (1.1) be rational functions.Then all meromorphic solutions of (1.1) satisfy

T (r, f) = O((log r)2).

Theorem 1.2. Let the coefficients a0, . . . , an, Q of (1.1) be rational functions.Then all transcendental meromorphic solutions of (1.1) satisfy

(log r)2 = O(T (r, f)).

It was also remarked in [2] that all meromorphic solutions of (1.1) are rationalfunctions, if the coefficients a0, . . . , an, Q in (1.1) are complex constants.

In Section 2, we assume that the coefficients a0, . . . , an in (1.1) are complexconstants and consider the existence of meromorphic solutions for certain simpletypes of Q. In Section 3, the main result is that all meromorphic solutions of (1.1)are of finite order≤ ρ, provided that the coefficients a0, . . . , an, Q are meromorphicand of finite order ≤ ρ. In Section 4, we make use of the Weierstrass factorizationto obtain results on the order of the solutions as well as on the number of zerosand poles. Observe that replacing z by z

cn in (1.1) results in

n∑j=0

aj

( zcn

)f( z

cn−j

)= Q

( zcn

).

Therefore, most of the results in Sections 2–4 can be rewritten for the case 1 <|c| <∞ too. In Section 5, some concluding remarks are given for the case |c| = 1.

Finally, we remark that the existence of meromorphic solutions to (1.1) in thecomplex plane is not always true. As a simple example, consider

f(cz)− f(z) =1

1− z , (1.3)

which has no meromorphic solutions in the plane. In fact, let f be a meromorphicsolution to (1.3). If f has a pole at z = 0, then (1.3) implies that |c| = 1, acontradiction. Hence f(0) is finite, and now the left-hand side of (1.3) wouldvanish at z = 0, while

(1

1−z

)z=0

= 1, again a contradiction.

150 J. Heittokangas et al. AEM

2. The case of constant coefficients aj

Theorem 2.1. Let the coefficients a0, . . . , an of (1.1) be complex constants andlet Q be of the reduced form Q(z) = P1(z)

zl, where P1 is a polynomial of degree d

and l ∈ N ∪ {0}. Then all meromorphic solutions f of (1.1) are of the reducedform f(z) = P2(z)

zp , where P2 is a polynomial and p ≥ l.

Proof. Suppose that (1.1) possesses a meromorphic solution f . If z0 6= 0 is a poleof f , then by looking at

f(z0) =1a0Q(z0)−

n∑j=1

aja0f(cjz0), (2.1)

we see that at least one of the values f(cjz0) must be infinite. Let j0 ∈ {1, . . . , n}be the maximum j such that f(cj0z0) =∞. Then by looking at

f(cj0z0) =1a0Q(cj0z0)−

n∑j=1

aja0f(cj0+jz0),

we see that at least one of the values f(cj0+jz0) must be infinite. Let j1 ∈{1, . . . , n} be the maximum j such that f(cj0+j1z0) = ∞. By repeating thisprocess we see that there is a sequence of poles of f accumulating to the origin,which is a contradiction. Hence f may only have a pole at the origin.

Suppose now that f has a pole of order p at the origin. It follows immediatelyfrom (1.1) that p ≥ l. Then g(z) := f(z)zp is entire and satisfies

n∑j=0

ajcjp

g(cjz) = Q(z)zp = P1(z)zp−l. (2.2)

We now differentiate (2.2) d+ p− l + 1 times with respect to z to obtainn∑j=0

aj

cj(l−d−1) g(d+p−l+1)(cjz) = 0. (2.3)

Now g(d+p−l+1) is also an entire function. Equation (2.3) yields

M(r, g(d+p−l+1)) ≤n∑j=1

∣∣∣∣ aj

a0cj(l−d−1)

∣∣∣∣M(r, g(d+p−l+1)(cjz))

=n∑j=1

∣∣∣∣ aj

a0cj(l−d−1)

∣∣∣∣M(|c|jr, g(d+p−l+1))

≤ n max1≤j≤n

∣∣∣∣ aj

a0cj(l−d−1)

∣∣∣∣M(|c|r, g(d+p−l+1)). (2.4)


Hence there exist B > 0 and R > 0 such that

logM(r, g(d+p−l+1)) ≤ logM(|c|r, g(d+p−l+1)) +B, r ≥ R. (2.5)

For any r ≥ R there exists a positive integer m such that

R

|c|m−1 ≤ r <R

|c|m . (2.6)

Now (2.5) and (2.6) imply

logM(r, g(d+p−l+1)) ≤ logM(|c|mr, g(d+p−l+1)) +Bm

≤ logM(R, g(d+p−l+1)) +B

(log r

log 1|c|− logR

log 1|c|

+ 1

).

SincelogM(r, g(d+p−l+1)) = O(log r),

g(d+p−l+1) is a polynomial. Hence g is a polynomial and f(z) = g(z)zp , the solution

of (1.1), is a rational function having no non-zero poles. �

Remark. It is possible to prove Theorem 2.1 without making use of the maximummodulus. Namely, one supposes that f is a transcendental meromorphic solutionof (1.1) (having no non-zero poles). Then, after using the series expansions for fand Q, one obtains a system of n+ 1 linear equations that finally gives a0 = · · · =an = 0, a contradiction.

Example. The rational function f(z) = 1zp , p ∈ N, is a solution of

f(cz)− 1cpf(z) = 0.

Before proceeding, we apply Theorem 2.1 to the homogeneous case (1.2).

Theorem 2.2. Let the coefficients a0, . . . , an of (1.2) be complex constants. Thenthe meromorphic solutions of (1.2) form a complex linear space of dimension d ≤ n.Moreover, some power functions zk1 , . . . , zkd with k1, . . . , kd ∈ Z form a base ofthis linear space.

Proof. By Theorem 2.1, all meromorphic solutions of (1.2) must be of the form

f(z) =m(f)∑j=0

γjzkj , γj ∈ C, kj ∈ Z.


Since the power functions zk1 , . . . , zkm(f) are linearly independent over the com-plex field, we see that all meromorphic solutions of (1.2) are linear combinationsof some power functions also solving (1.2). Substituting such a function, sayf(z) = zt, into (1.2), we obtain

a0 + a1ct + · · ·+ an(ct)n = 0. (2.7)

Obviously, at most n integers t1, . . . , tn may be found to satisfy (2.7). �

Remarks. (1) The upper bound n for the dimension of the solution base maybe achieved. In fact, given c, fix distinct integers t1, . . . , tn and an = 1. Thenit is elementary to find a0, . . . , an−1 such that ct1 , . . . , ctn are the roots of (2.7).Hence, the functions fj(z) := ztj , j = 1, . . . , n, are linearly independent solutionsto (1.2). To give a concrete example, f1(z) = 1

z and f2(z) = 1z2 are solutions to

f(z

4

)− 6f

(z2

)+ 8f(z) = 0.

(2) Of course, the upper bound for the dimension of the solution base decreasesprovided (2.7) has multiple roots.

Theorem 2.3. Let the coefficients a0, . . . , an of (1.1) be complex constants andlet Q be meromorphic. If Q has exactly one nonzero pole, then all meromorphicsolutions f of (1.1) have infinitely many poles. If Q has more than one nonzeropole, then f may have finitely many poles.

Proof. Let Q be a meromorphic function in C having exactly one nonzero pole atz0. Suppose that (1.1) possesses a meromorphic solution f . By looking at

n∑j=0

ajf(cjz0) = Q(z0), (2.8)

it is clear that at least one of the values f(cjz0) must be infinite. Let j0 ∈{0, . . . , n} be the maximum j such that f(cj0z0) = ∞. Suppose that j0 ≥ 1. Weshall show that this leads to a contradiction. Substitute cj0z0 for z in (1.1) toobtain

∞ = f(cj0z0) =1a0Q(cj0z0)−

n∑j=1

aja0f(cj+j0z0).

As in the proof of Theorem 2.1, we obtain a sequence of poles of f accumulatingto the origin, which is a contradiction. Hence j0 = 0 and z0 is a pole of f .

Denote m1 = 0. Now substitute z0cn for z in (1.1) to obtain

n∑j=0

ajf(cj−nz0) = Q(z0cn

), (2.9)


and we see that there exists an integer m2 > m1 such that z0cm2 is a pole of f .

By continuing in this fashion, we get an increasing sequence {mj} of integers suchthat the sequence

{ z0cmj

}is a pole sequence of f outside the origin, i.e., f has

infinitely many poles.For the second assertion it suffices to observe that f(z) = 1

1−z is a solution to

n∑j=0

f( z

2j)

=n∑j=0

11− z

2j. �

Theorem 2.4. Let the coefficients a0, . . . , an of (1.1) be complex constants sat-isfying

n∑j=0

ajcjk 6= 0 for all k ∈ Z, (2.10)

and let Q be of the reduced form Q(z) = g1(z)zl

, where g1 is entire and l ∈ N∪ {0}.Then (1.1) possesses exactly one meromorphic solution f . The solution f is of thereduced form f(z) = g2(z)

zl, where g2 is entire.

Proof. By (2.10) and Theorem 2.2, the homogeneous equation (1.2) has no non-trivial solutions. Hence (1.1) possesses at most one meromorphic solution. Wefirst obtain a formal solution of (1.1) by a series expansion at the origin. To thisend, set

f(z) =∞∑

m=−qαmz

m, q ∈ N ∪ {0}, αm ∈ C, (2.11)

and

Q(z) =∞∑

m=−lβmz

m, βm ∈ C. (2.12)

Substitute f and Q into (1.1) to obtain

n∑j=0

aj

( ∞∑m=−q

αm(cjz)m)

=∞∑

m=−lβmz

m.

We immediately see that q ≥ l. Comparing the coefficients of zm and making useof the assumption (2.10), we get q = l and

αm =βm∑n

j=0 ajcjm

=βmγm

for m = −q,−q + 1, . . . , (2.13)

where γm =∑nj=0 ajc

jm. Since γm 6= 0 for all m and γm −→ a0 as m −→∞, theanalytic function zlf(z) has the same radius of convergence as g1. Hence, zlf(z)is entire. �


Corollary 2.5. If Q is entire in Theorem 2.4 (i.e. if l = 0), then (1.1) possessesexactly one meromorphic solution f , which is entire.

Remark. Observe that the condition (2.10) fails with k = 0 for the equation(1.3).

Our next result extends Theorem 2.4.

Theorem 2.6. Let the coefficients a0, . . . , an of (1.1) be complex constants sat-isfying (2.10) and let Q be a meromorphic function. Then (1.1) possesses exactlyone meromorphic solution f .

Proof. As in the proof of Theorem 2.3, at most one meromorphic solution exists.Denote the order of the pole of Q at the origin by l ∈ N∪{0}. We again substitutethe series (2.11) and (2.12) into (1.1) to obtain (2.13). Thus zlf(z) has a positiveradius of convergence R around the origin which is equal to that of zlQ(z). Thenthe global solution can be obtained by meromorphic continuation using (1.1). Infact, we define Dm = {|z| < R/|c|m} and note that ∪∞m=0Dm = C. We first showthat the local solution existing in D0 can be extended into D1. To this end, wedefine f(z) in D1 \D0 by

f(z) =1a0Q(z)−

n∑j=1

aja0f(cjz), z ∈ D1 \D0.

Since cjz ∈ D0 for all j = 1, . . . , n, the functions f(cjz) are meromorphic in D1.Since Q is also meromorphic in D1, it follows that f must be meromorphic in D1.By repeating this process we construct the global solution of (1.1). �

Example. Consider the functional equation

f(cz)− af(z) =1

1− z , (2.14)

where c ∈ (−1, 1) \ {0}, a ∈ C, and Im a 6= 0. For any integer k, ck − a 6= 0, andso the condition (2.10) holds. The series expansion for f around the origin is

f(z) =∞∑j=0

1cj − az

j .

It follows that f is analytic in |z| < 1. The global meromorphic solution f of(1.1) follows by meromorphic continuation. To prove that f has infinitely manypoles, z = 1 is a pole of f by (2.14). Inductively, all points in the sequence {sn},n ∈ N ∪ {0}, are poles of f .


Theorem 2.7. Let the coefficients a0, . . . , an of (1.1) be complex constants sat-isfying the condition (2.10) and let Q be of the form Q(z) = eP (z), where P is apolynomial of degree d ∈ N ∪ {0}. Then (1.1) possesses exactly one meromorphicsolution f , which is an entire function of order of growth ρ(f) = d.

Proof. By Corollary 2.5, (1.1) possesses exactly one meromorphic solution f , whichis entire. We proceed to show that ρ(f) = d.

Similarly as to (2.4), we get

M(r, f) ≤ n max1≤j≤n

∣∣∣∣aja0

∣∣∣∣M(|c|r, f) + eCrd

,

where C > 0 is a finite constant. Hence there exist B > 0 and R > 0 such that

logM(r, f) ≤ logM(|c|r, f) + Crd +B, r ≥ R.For any r ≥ R, take an integer m such that (2.6) holds. Then

logM(r, f) ≤ logM(|c|mr, f) + (Crd +B)m

≤ logM(R, f) + (Crd +B)

(log r

log 1|c|− logR

log 1|c|

+ 1

).

We conclude that ρ(f) ≤ d. On the other hand, the converse inequality ρ(f) ≥ dis immediate by (1.1). �

Example. An entire solution for

f(z

2) + f(z) = ez (2.15)

may be constructed by the power series method. In fact, substituting f(z) =∑∞j=0 αjz

j into (2.15), we get∞∑j=0

(αj2j

+ αj)zj = ez =

∞∑j=0

1j!zj ,

and so

f(z) =∞∑j=0

2j

2j + 11j!zj.

It is immediate that this is an entire solution. By the standard formula to de-termine the order of growth of an entire function from its Taylor coefficients, andmaking use of the Stirling formula

j! ∼(j

e

)j√2πj,

it is easy to conclude that ρ(f) = 1.


3. The case of meromorphic coefficients aj

By making some additional assumptions, we obtain the following extension ofTheorem 2.3:

Theorem 3.1. Let the coefficients a0, . . . , an, Q of (1.1) be meromorphic. If Qhas a non-zero pole at z0 satisfying

a) clz0 is not a pole of Q for any l ∈ Z \ {0},b) clz0 is neither a pole nor a zero of aj for any l ∈ Z and j = 0, . . . , n,

then all meromorphic solutions of (1.1) have infinitely many poles correspondingto every such z0.

Proof. As in the proof of Theorem 2.3, we see that z0 must be a pole of f . Then wecan get an increasing integer sequence {mj}, mj = mj(z0), such that the pointsin the sequence

{z0cmj

}are poles of f . �

We next proceed to show that Theorem 2.7 is typical in the sense that thegrowth of the coefficients in (1.1) determines the growth of all meromorphic so-lutions of (1.1). We invite the reader to compare this situation with the case ofcomplex differential equations, see [1, p. 292].

Lemma 3.2. Let the coefficients a0, . . . , an, Q of (1.1) be meromorphic and sup-pose that (1.1) possesses a meromorphic solution f . Let z0 be a pole of f such thatf(cjz0) is finite for all j ∈ N. If f has a pole at skz0 of order µk, then

µk ≤k∑l=0

ql +n+1∑j=0

αj,l

≤ n(|skz0|, Q) +n∑j=0

n(|skz0|, aj) + n

(|skz0|,

1a0

),

(3.1)where αj,l is the order of the pole of aj, j = 0, . . . , n, ql is the order of the pole ofQ, and αn+1,l is the order of the zero of a0 at slz0.

Proof. Denote γk := max{α0,k, . . . , αn,k, qk}. For k = 0, we consider

(z − z0)γ0a0(z)f(z) = (z − z0)γ0Q(z)−n∑j=1

aj(z)(z − z0)γ0f(cjz).

The right-hand side here is finite at z = z0, hence γ0 + αn+1,0 − µ0 ≥ 0 and so

µ0 ≤ γ0 + αn+1,0 ≤ q0 +n+1∑j=0

αj,0.

Assume now that the first inequality in (3.1) has been proved for all l ≤ k, and


consider

(z − sk+1z0)γk+1a0(z)f(z) = (z − sk+1z0)γk+1Q(z)

−n∑j=1

aj(z)(z − sk+1z0)γk+1f(cjz). (3.2)

Obviously, the first term on the right-hand side is finite at sk+1z0. If the left handside is finite at sk+1z0, then

µk+1 ≤ γk+1 + αn+1,k+1 ≤ qk+1 +n+1∑j=0

αj,k+1 ≤k+1∑l=0

ql +n+1∑j=0

αj,l

.

If the left-hand side of (3.2) is not finite at sk+1z0, at least one term in the sumon the right-hand side of (3.2) has a pole of higher order than the left-hand sideof (3.2). Therefore, for some j, 1 ≤ j ≤ n,

µk+1 − γk+1 − αn+1,k+1 ≤ µk+1−j + αj,k+1 − γk+1,

and so

µk+1 ≤ αj,k+1 + αn+1,k+1 + µk+1−j

≤ αj,k+1 + αn+1,k+1 +k∑l=0

ql +n+1∑j=0

αj,l

≤ k+1∑l=0

ql +n+1∑j=0

αj,l

proving the first inequality in (3.1).

The second inequality in (3.1) is obvious. �

Lemma 3.3. Let the coefficients a0, . . . , an, Q of (1.1) be meromorphic. If f is ameromorphic solution of (1.1), then there exists a finite constant A > 0 such that

n(r, f) ≤ A

n∑j=0

n(r, aj) + n(r,Q) + n(r,

1a0

) log r

for sufficiently large values of r.

Proof. We may assume that f has no non-zero poles in a disc |z| < r0, wherer0 > 0 is a finite constant. For any |z| ≥ r0 satisfying |skz| ≤ r, we have

k ≤ log r − log |z|log |s| ≤ B log r, (3.3)


where B > 0 is a constant, provided that r is sufficiently large.Let I(r) := {z0 ∈ C \ {0} | z0 is a pole of f and for any j ∈ N, cjz0 is not a

pole of f , and |z0| ≤ r}. For every z0 ∈ I(r), we see that |z0| ≥ r0 and that thereexists an integer k = k(z0) such that

|sk(z0)z0| ≤ r < |sk(z0)+1z0|.

By Lemma 3.2 and the inequality (3.3),

n(r, f) = n(0, f) +∑

z0∈I(r)

(µ0(z0) + · · ·+ µk(z0)(z0)

)

≤ n(0, f) +∑

z0∈I(r)

k(z0)∑p=0

p∑l=0

ql +n+1∑j=0

αj,l

≤ n(0, f) +

∑z0∈I(r)

k(z0)∑p=0

k(z0)∑l=0

ql +n+1∑j=0

αj,l

= n(0, f) +

∑z0∈I(r)

(k(z0) + 1)k(z0)∑l=0

ql +n+1∑j=0

αj,l

≤ n(0, f) + (B log r + 1)

∑z0∈I(r)

k(z0)∑l=0

ql +n+1∑j=0

αj,l

≤ A

n∑j=0

n(r, aj) + n(r,Q) + n(r,1a0

)

log r,

where A > 0 is a finite constant and r is sufficiently large. �

Proposition 3.4. Let the coefficients a0, . . . , an, Q of (1.1) be meromorphic. As-sume that

λ := max{λ

(1a0

), . . . , λ

(1an

), λ(a0), λ

(1Q

)}.

Then for any non-constant meromorphic solution f of (1.1), λ( 1f ) ≤ λ.

Proof. The assertion follows from

λ

(1f

)= lim sup

r→∞

logn(r, f)log r

and Lemma 3.3. �


Remark. If the coefficients in (1.1) are of finite order ≤ ρ, then λ( 1f ) ≤ ρ, and

so N(r, f) = O(rρ+ε

).

Theorem 3.5. If the coefficients a0, . . . , an, Q in (1.1) are meromorphic and offinite order ≤ ρ, then all meromorphic solutions of (1.1) are of finite order ≤ ρ.

Proof. Assume first that the coefficients a0, . . . , an, Q of (1.1) are entire. Withoutloss of generality we may assume that a0 equals to the canonical product formedwith its zeros. We next apply the Cartan lemma (see e.g. [9, p. 185]) to fix certaindiscs of radius |zj|−(ρ+ε) around the zeros zj of a0 such that outside of these discs,|a0(z)| > exp(−rρ+ε) for sufficiently large values of r. By [4, p. 75], there existst ∈ (1, |s|) such that (i) f has no poles on the circles |z| = |s|kt, k ≥ 0, and (ii)these circles are outside of the Cartan discs. For k ∈ N, define

Mk := 1 + max0≤j≤k

M(|s|jt, f)

andM ′(r) := max

1≤j≤n{M(r, aj),M(r,Q)}.

Given ε > 0,M ′(r) ≤ exp(rρ+ε)

for r large enough. Therefore

M(|s|kt, f) ≤M(|s|kt, 1

a0

)M(|s|kt,Q) +n∑j=1

M(|s|kt, aj)M(|s|k−jt, f)

≤ exp(|s|kt)ρ+εM ′(|s|kt)

1 +k−1∑

j=k−nM(|s|jt, f)

≤(exp(|s|kt)ρ+ε

)21 +k−1∑

j=k−nMj

≤(exp(|s|kt)ρ+ε

)2(n+ 1)Mk−1

for k ≥ n large enough, say k ≥ k0 ≥ n. Hence,

Mk ≤ A(exp(|s|kt)ρ+ε

)2(n+ 1)Mk−1,

where A > 0 is a finite constant. By Lk := logMk, this takes the form

Lk ≤ 2(|s|kt|)ρ+ε + Lk−1 + logA(n+ 1).


By induction, we get

Lk ≤ 2k(|s|kt)ρ+ε +B + Lk0

for some B > 0, and so

Lk ≤ Ck(|s|kt)ρ+ε

for some C > 0 and for k ≥ k0. This implies now

m(|s|kt, f) ≤ logM(|s|kt, f) ≤ Ck(|s|kt)ρ+ε

= Clog(|s|kt)− log t

log |s| (|s|kt)ρ+ε.

By the remark to Proposition 3.4,

N(|s|kt, f) ≤ D(|s|kt)ρ+ε

holds for some D > 0 and for k large enough. Combining these inequalities, weget

T (|s|kt, f) ≤ E(log(|s|kt))(|s|kt)ρ+ε

for some E > 0 and for k large enough. Since T (r, f) and rρ+ε log r are increasingand continuous functions, it is easy to prove (by using (2.6)) that

T (r, f) ≤ Frρ+ε log r, F = 2E|s|ρ+ε,

holds for all r sufficiently large. The assertion now follows immediately.Finally, let the coefficients in (1.1) be meromorphic functions. Given ε > 0,

they may be represented by quotients of entire functions of order ≤ ρ+ ε by [8, §14]. Multiplying away denominators, we may apply the first part of the proof toget the assertion for meromorphic coefficients. �

If ρ(aj) < ρ(Q) in (1.1) for j = 0, . . . , n, it is easily seen that ρ(f) ≥ ρ(Q).Hence we obtain the following extension of Theorem 2.7:

Corollary 3.6. Let the coefficients a0, . . . , an, Q of (1.1) be meromorphic func-tions. If ρ(Q) > ρ(aj) for all j = 0, . . . , n, then all meromorphic solutions f of(1.1) satisfy ρ(f) = ρ(Q).


4. Meromorphic coefficients and the Weierstrass factorization

We begin by a lemma due to Hiromi and Ozawa, see [6, Lemma 1].

Lemma 4.1. Let a0, . . . , an be non-trivial meromorphic functions and g1, . . . , gnbe non-trivial entire functions such that

T (r, aj) = o

(n∑i=1

m(r, egi)

), j = 0, . . . , n,

outside of a set of finite linear measure. If an identityn∑i=1

aiegi = a0

holds, then there exists constants α1, . . . , αn such that they are not all equal tozero and

n∑i=1

αiaiegi = 0.

As an application of the Hiromi–Ozawa lemma above, we have the followinglemma. The case where the functions gi are polynomials has been treated in [7,Lemma 4.2].

Lemma 4.2. Let g1, . . . , gn be transcendental entire functions such that gi − gjis transcendental for every i, j ∈ {1, . . . , n}, i 6= j. If there exists meromorphicfunctions a1, . . . , an of finite order of growth such that

n∑i=1

aiegi = 0,

then a1 = a2 = · · · = an = 0.

Proof. We will use an inductive argument to prove the lemma. The case n = 1is clear. Suppose next that the assertion is true for 1 ≤ n ≤ k, k ∈ N. Assumecontrary to the assertion that

k+1∑i=1

aiegi = 0,

for some non-trivial meromorphic functions a1, . . . , ak+1 of finite order of growth.It follows that

k∑i=1

aieGi = −ak+1,


where Gi = gi − gk+1, i = 1, . . . , k. Since the coefficients ai are of finite orderand the functions Gi are not polynomials, we have T (r, aj) = o(m(r, exp(Gi))),j = 1, . . . , k + 1, i = 1, . . . , k. Thus we may apply the Hiromi–Ozawa lemma toobtain constants αi, i = 1, . . . , k, not all equal to zero, such that

k∑i=1

αiaieGi = 0.

We easily see that Gi−Gj is transcendental for all i, j ∈ {1, . . . k}, i 6= j, and thuswe have a contradiction to the induction hypothesis. �

To make the notations shorter, we replace Q in the equation (1.1) by an+1.Hence, we consider the equation

n∑j=0

aj(z)f(cjz) = an+1(z), n ∈ N, (4.1)

where a0, a1, . . . , an+1 are meromorphic functions such that a0an 6≡ 0.Let us define

M := max0≤j≤n+1

{ρ(aj)}, and

I :={j ∈ {0, . . . , n+ 1} | ρ(aj) = M}.With these notations, we state

Proposition 4.3. Suppose that c ∈ C \ {0}, M = ∞, and that aj = Ajegj for

all j ∈ I, where gj is an entire transcendental function and Aj is a meromorphicfunction of finite order. Furthermore, suppose that gi−gj is transcendental for alli, j ∈ I, i 6= j. Then any nontrivial meromorphic solution f to (4.1) is of infiniteorder of growth.

Proof. Assume contrary to the assertion that f is of finite order. We may writethe equation (4.1) in the form∑

j∈IBj(z)egj(z) = S(z),

where the Bj , j ∈ I, and S are meromorphic functions of finite order. If S 6≡ 0,we may eliminate it from the equation above by using the Hiromi–Ozawa lemma.Hence we obtain an equation which is impossible by Lemma 4.2. �

By a similar reasoning as to above and making use of [7, Lemma 4.2] insteadof Lemma 4.2, we easily obtain

Proposition 4.4. Suppose that c ∈ C \ {0}, M ∈ N, and that for all j ∈ I,aj(z) = Aj(z)epjz

M

, where pj ∈ C \ {0} and Aj is a meromorphic function such


that ρ(Aj) < M . If the constants pj are distinct, then ρ(f) ≥ M for a nontrivialmeromorphic solution f of the equation (4.1).

By combining Proposition 4.4 and Theorem 3.5 we make the following obser-vation.

Corollary 4.5. In addition to the assumptions in Proposition 4.4, assume that0 < |c| < 1. Then ρ(f) = M for a nontrivial meromorphic solution f of theequation (4.1).

Lemma 4.6. Assume that M ∈ N ∪ {∞} and ρ(aj) < M for j = 0, . . . , nand max(λ(an+1), λ(1/an+1)) < ρ(an+1) = M . If (4.1) admits a meromorphicsolution f , then max(λ(f), λ(1/f)) = ρ(f).

Proof. Assume contrary to the assertion that max(λ(f), λ(1/f)) < ρ(f).(1) Assume first M <∞. By the Weierstrass factorization, we get

an+1(z) = q(z)ebzM

, b ∈ C \ {0},

where q is a meromorphic function such that ρ(q) < M . By Corollary 3.6,

f(z) = H(z)epzM

, p ∈ C \ {0},

where H is meromorphic such that ρ(H) < M . Substituting the above presenta-tions into (4.1), we obtain

n∑j=0

aj(z)H(cjz)epcjMzM = q(z)ebz

M

. (4.2)

We easily see that polynomials pcjMzM , j = 0, . . . , n, are nontrivial anddivisible by zM . Also, the polynomial bzM is nontrivial and divisible by zM .The functions ebz

M

and epcjMzM , j = 0, . . . , n, satisfy the linear equation (4.2)

over the field Lexp(zM ), which is the field of meromorphic functions g such thatT (r, g) = S(r, exp(zM)).

Applying [7, Lemma 4.2] to the equation (4.2), we now obtain three possibilities.(i) There exists j1, j2 ∈ {0, . . . , n}, j1 6= j2 and pcj1M = pcj2M . Consequently,

c(j1−j2) = 1 and thus |c| = 1, a contradiction.(ii) We have a0 = · · · = an = q = 0, contradicting the assumption.(iii) There exists an index j0 ∈ {0, . . . , n} such that pcj0M = b. Hence, the

equation reduces to the form

n∑j=0j 6=j0

aj(z)H(cjz)epcjMzM =

(q(z)− aj0H(cj0z)

)epc

j0MzM .


Since at least two of the functions aj were nontrivial, we see that thereexists j1 ∈ {0, . . . , n}, j1 6= j0, and j2 ∈ {0, . . . , n}, j2 6= j1 such thatpcj1M = pcj2M . This gives us a contradiction as in the case (i).

(2) Assume next M =∞. By the Weierstrass factorization and Corollary 3.6,we may write

an+1 = q(z)eh(z),

where h is transcendental entire and q a meromorphic function such that ρ(q) <∞,and

f(z) = H(z)eg(z),

where H is meromorphic such that ρ(H) < ∞ and g is a transcendental entirefunction. Substituting the above presentations into the equation (4.1), we obtain

n∑j=0

aj(z)H(cjz)eg(cjz) = q(z)eh(z). (4.3)

We may write the equation (4.3) in the form

n∑j=0

aj(z)H(cjz)eg(cjz)−h(z) = q(z).

If g(cjz)−h(z) and g(ciz)−h(z) are polynomials for distinct i and j, then g(ciz)−g(cjz) is a polynomial. However, this is impossible by Theorem 2.1. Thus we mayapply the Hiromi–Ozawa lemma to eliminate q(z) from the right-hand side of theabove equation. This results to an equation impossible by Lemma 4.2. �

Theorem 4.7. Under the assumptions of Lemma 4.6, the meromorphic solutionsf of (1.1) satisfy λ(f) = ρ(f).

Proof. By Lemma 4.6, max{λ(f), λ( 1f )} = ρ(f). To show that λ( 1

f ) < ρ(f), itsuffices to make use of Proposition 3.4. �

5. Remarks on the case |c| = 1

In the previous sections we assumed c to be a non-zero complex constant with|c| 6= 1. In some cases, we had a pole sequence of f accumulating to the origin,hence leading to a contradiction. In some cases, the pole sequence of f accumulatedto infinity causing no problems.

The following result is a partial description as to what may happen if |c| = 1.

Theorem 5.1. Let the coefficients a0, . . . , an, Q of (1.1) be meromorphic and letc = eiθ, where θ ∈ R. Then the following two conclusions hold.


1) Suppose that θ is of the form θ = 2πpq , where p ∈ Z and q ∈ N, q ≥ 2. Also

suppose that Q has a non-zero pole at z0 satisfyinga) clz0 is not a pole of Q for any l ∈ {1, . . . , q − 1},b) clz0 is neither a pole nor a zero of aj for any l ∈ {0, . . . , q − 1} and j =

0, . . . , n.Then every meromorphic solution f of (1.1) has at most q poles on the circle|z| = |z0| corresponding to every such z0.

2) Suppose that θ is of the form θ = 2πr, where r ∈ R \Q. Suppose the polesof Q have been arranged by increasing moduli and let z0 be the first non-zero pole.Assume now that

c) Q has no other poles than z0 on the circle |z| = |z0|,d) the coefficients aj have neither poles nor zeros on the circle |z| = |z0|.

Then every locally meromorphic solution f of (1.1) around the origin has infinitelymany poles on the circle |z| = |z0|.

Proof. We prove the part 1) only since the part 2) then easily follows. So, supposethat the assumptions of part 1) hold and that (1.1) possesses a meromorphicsolution f . Then all the poles of f on the circle |z| = |z0| are contained in

A = {clz0 | l ∈ N ∪ {0}} = {ei2πpq lz0 | l ∈ N ∪ {0}}.

Sinceei2π

pq l = ei2π

pq (q+l) for all l ∈ N ∪ {0},

we see that A has exactly q elements. Hence f has at most q poles on the circle|z| = |z0|. �

Acknowledgement. The authors would like to thank Gary Gundersen for anumber of useful comments.

References

[1] S. Bank, A note on algebraic differential equations whose coefficients are entire functionsof finite order, Ann. Scuola Norm. Sup. Pisa 26 (3) (1972), 291–297.

[2] W. Bergweiler, K. Ishizaki and N. Yanagihara, Meromorphic solutions of some func-tional equations, Methods Appl. Anal. 5 (3) (1998), 248–258.

[3] W. K. Hayman, Meromorphic Functions, Clarendon Press, Oxford, 1964.[4] W. K. Hayman, Slowly growing integral and subharmonic functions, Comment. Math.

Helv. 34 (1960), 75–84.[5] I. Laine, Nevanlinna Theory and Complex Differential Equations, Walter de Gruyter,

Berlin, 1993.[6] G. Hiromi and M. Ozawa, On the existence of analytic mappings between two ultrahyper-

elliptic surfaces, Kodai Math. Sem. Report 17 (1965), 281–306.[7] J. Rieppo, Differential fields and complex differential equations, Ann. Acad. Sci. Fenn.

Math. Diss. 118 (1998).


[8] L. Rubel, Entire and Meromorphic Functions, Springer, New York–Berlin–Heidelberg,1995.

[9] W. A. Veech, A Second Course in Complex Analysis, Benjamin, New York–Amsterdam,1967.

J. Heittokangas, I. Laine, J. Rieppo and D. YangUniversity of JoensuuDepartment of MathematicsP. O. Box 111FIN–80101 JoensuuFinland

Manuscript received: July 6, 1999 and, in final form, October 1, 1999.

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Meromorphic solutions of some linear functional equations