Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 31, 2006, 265–286

ON ENTIRE FUNCTIONS THAT SHARE A VALUE WITH THEIR DERIVATIVES

Jianming Chang and Mingliang Fang

Nanjing Normal University, Department of Mathematics, Nanjing 210097 and Changshu Institute of Technology, Department of Mathematics Changshu, Jiangsu 215500, P. R. China; jmwchang@pub.sz.jsinfo.net South China Agricultural University, Department of Applied Mathematics Guangzhou 510642, P. R. China; hnmlfang@hotmail.com

Abstract. Let f be a nonconstant entire function, a a finite complex number, k and m two distinct positive integers, and (k, m) the greatest common divisor of k and m. If f, f^{(k)} and f^{(m)} share a CM, then

f(z) = \frac{c − 1}{c} a + ∑_{j=1}^{q} C_{j}e^{λ_{j}z},

where q is a positive integer with q ≤ (k, m), c and C_{j} for 1 ≤ j ≤ q are nonzero constants, and λ_{j} for 1 ≤ j ≤ q, are distinct nonzero constants satisfying (λ_{j})^{k} = (λ_{j})^{m} = c, for a ≠ 0, and (λ_{j})^{k} = c, (λ_{j})^{m} = d, for a = 0, where d is a nonzero constant. This answers a question of Yang and Yi [14] for entire functions, and extends a result of Csillag [2].

1. Introduction and main results

Let f and g be two nonconstant meromorphic functions in the complex plane, and let a be a finite complex number. If f(z) − a and g(z) − a have the same zeros with the same multiplicities, then we say that f and g share a CM.

In 1986, Jank–Mues–Volkmann [8] proved

Theorem A. Let f be a nonconstant meromorphic function and a a nonzero finite complex number. If f, f′ and f′′ share a CM, then f ≡ f′.

By Theorem A, the following question was posed.

Question 1 (see [6], [7], [13], [14]). Let f be a nonconstant meromorphic function, a a nonzero finite complex number, and k, m two distinct positive integers. Suppose that f, f^{(k)} and f^{(m)} share a CM. Can we get f ≡ f^{(k)} ?

The following example [12] shows that the answer to Question 1 is, in general, negative.

2000 Mathematics Subject Classification: Primary 30D35.

Supported by the NNSF of China (Grant No. 10471065), the NSF of Education Department of Jiangsu Province (Grant No. 04KJD110001) and the Presidential Foundation of South China Agricultural University.

Example 1. Let k, m be positive integers satisfying m > k +1, b a nonzero constant such that b^{k} = b^{m} ≠ 1 and a = b^{k}. Set

f(z) = e^{bz} + a − 1.

Then f, f^{(k)} and f^{(m)} share a CM. But f ≢ f^{(k)} .

In Example 1, f is an entire function, and f, f^{(k)} and f^{(m)} share a CM. Although f ≢ f^{(k)} , we have f^{(k)} ≡ f^{(m)} .

Naturally, we pose the following question.

Question 2. Let f be a nonconstant meromorphic function, a a nonzero finite complex number and k, m two distinct positive integers. Suppose that f, f^{(k)} and f^{(m)} share a CM. Can we get f^{(k)} ≡ f^{(m)} ?

In this paper, we give an affirmative answer to Question 2 for entire functions. In fact, we have proved the following more general result.

Theorem 1. Let f be a nonconstant entire function, a a finite complex number, k and m two distinct positive integers, and (k, m) the greatest common divisor of k and m. If f, f^{(k)} and f^{(m)} share a CM, then

(1.1)  f(z) = (1 − \frac{1}{c})a + ∑_{j=1}^{q} C_{j}e^{λ_{j}z},

where q is a positive integer with q ≤ (k, m), c and C_{j}, 1 ≤ j ≤ q, are nonzero constants, and λ_{j}, 1 ≤ j ≤ q, are distinct nonzero constants satisfying

(1.2)  (λ_{j})^{k} = (λ_{j})^{m} = c, for a ≠ 0;

and

(1.3)  (λ_{j})^{k} = c, (λ_{j})^{m} = d, for a = 0,

where d is a nonzero constant.

By Theorem 1, we can easily obtain the following results.

Corollary 2. Let f be a nonconstant entire function, a a nonzero finite complex number, and k, m two distinct positive integers. Suppose that f, f^{(k)} and f^{(m)} share a CM. Then f^{(k)} ≡ f^{(m)} .

Corollary 2 gives an affirmative answer to Question 2 for entire functions.

Corollary 3 ([10, Theorem 1]). Let f be a nonconstant entire function, a a nonzero finite complex number and k a positive integer. If f, f^{(k)} and f^{(k+1)} share a CM, then f ≡ f′.

Corollary 4 ([10, Theorem 2]). Let f be a nonconstant entire function, a a nonzero finite complex number and k ≥ 2 a positive integer. If f, f′ and f^{(k)} share a CM, then

(1.4)  f(z) = (1 − \frac{1}{c})a + Ce^{cz},

where C and c are nonzero constants with c^{k−1} = 1.

Corollary 5 (Csillag [2], cf. [4, p. 67]). Let f be a nonconstant entire function, and k and m two distinct positive integers. If ff^{(k)}f^{(m)} ≠ 0, then f = e^{Az+B}, where A (≠ 0) and B are constants.

Let f be a nonconstant meromorphic function in the complex plane. Throughout this paper, we use the basic results and notations of Nevanlinna theory (cf. [3], [4], [11], [14]). In particular, S(r, f) denotes any function satisfying

S(r, f) = o{T (r, f)},

as r → +∞, possibly outside of a set of finite linear measure, where T (r, f) is Nevanlinna’s characteristic function.

As usual, the order ϱ(f) of f is defined as

ϱ(f) = lim sup_{r→∞} \frac{log T (r, f)}{log r}.

2. Some lemmas

We will use P_{d}[f] to denote a differential polynomial in f of degree ≤ d with constant coefficients which may be different at different occurrence. We denote the set of differential polynomials in f with constant coefficients by 𝒫[f].

Lemma 1 (Clunie [1], cf. [4, p. 68]). Let f be a nonconstant meromorphic function, n be a positive integer, P [f] and Q[f] two differential polynomials in f with constant coefficients, and P [f] ≢ 0. If the degree of P [f] is at most n and

f^{n}Q[f] = P [f],

then

m(r, Q[f]) = S(r, f).

Lemma 2 (cf. [9, p. 29–34]). Let f be a nonconstant entire function, n be a positive integer and a_{j}, 0 ≤ j ≤ n, meromorphic functions with a_{n} ≢ 0. Suppose that

(2.1)  a_{n}f^{n} + a_{n−1}f^{n−1} + ⋯ + a_{1}f + a_{0} ≡ 0.

Then

(2.2)  T (r, f) ≤ O(1 + ∑_{j=0}^{n} T (r, a_{n})).

The following result is an instant corollary of Lemma 2.

Lemma 3. Let f be a nonconstant entire function, n a positive integer and a_{j}, 0 ≤ j ≤ n, meromorphic functions satisfying T (r, a_{j}) = S(r, f). If

a_{n}f^{n} + a_{n−1}f^{n−1} + ⋯ + a_{1}f + a_{0} ≡ 0,

then a_{j} ≡ 0 for j = 0, 1, …, n.

Lemma 4 ([3, Lemma 3.12]). Let f_{j}(z) (≢ 0), j = 1, 2, …, n, be n meromorphic functions which are linearly independent such that

(2.3)  f_{1}(z) + f_{2}(z) + ⋯ + f_{n}(z) ≡ 1.

Then for every j, 1 ≤ j ≤ n,

(2.4)  T (r, f_{j}) ≤ ∑_{k=1}^{n} N(r, \frac{1}{f_{k}} ) + N(r, f_{j}) + N(r, D) + S(r),

where D = W (f_{1}, f_{2}, …, f_{n}) is the Wronskian, and S(r) is a function which satisfies

S(r) = o( max_{1≤k≤n} T (r, f_{k}))

as r → ∞, possibly outside a set of finite linear measure.

Lemma 5 ([3, Lemma 5.1]). Let a_{j}(z), j = 0, 1, …, n, be entire and of finite order ≤ ϱ (< ∞). Let g_{j}(z), j = 1, …, n, be also entire such that each of the functions g_{i} − g_{j}, i ≠ j, is a transcendental function or a polynomial of degree greater than ϱ. If

(2.5)  ∑_{j=1}^{n} a_{j}(z)e^{g_{j}(z)} ≡ a_{0}(z),

then

(2.6)  a_{j}(z) ≡ 0, j = 0, 1, …, n.

Lemma 6. Let f and α be nonconstant entire functions, a a finite complex number and k a positive integer. Suppose that

(2.7)  f^{(k)} = a + e^{α}f.

Then for any positive integer j, 1 ≤ j ≤ k − 1, we have

(2.8)  f^{(k+j)} = γ_{0,j}f + γ_{1,j}f′ + ⋯ + γ_{j,j}f^{(j)},

and γ_{i,j} are entire functions satisfying

(2.9)  ( γ_{0,j} ; ⋮ ; γ_{j,j} ) = ( A_{0,1,j}e^{α} ; ⋮ ; A_{j,1,j}e^{α} )

where

(2.10)  \begin{aligned}A_{i,1,j} & = \frac{j!}{i!(j − i)!}e^{−α}(e^{α})^{(j−i)} \\ & = \frac{j!}{i!(j − i)!}((α′)^{j−i} + P_{j−i−1}[α′]), 0 ≤ i ≤ j, \end{aligned}

are differential polynomials in α′ with constant coefficients. In particular, A_{j,1,j} ≡ 1 for 1 ≤ j ≤ k − 1. Here P_{d}[α′] ≡ 0 for d ≤ 0.

Proof. We prove this lemma by mathematical induction on j. By (2.7), we have f^{(k+1)} = α′e^{α}f + e^{α}f′, so that (2.8)–(2.10) are true for j = 1. Now suppose that (2.8)–(2.10) are true for j ≤ k − 2. Thus by (2.8), we get

(2.11)  \begin{aligned}f^{(k+j+1)} & = γ′_{0,j}f + γ′_{1,j}f′ + ⋯ + γ′_{j,j}f^{(j)} \\ & + γ_{0,j}f′ + ⋯ + γ_{j−1,j}f^{(j)} + γ_{j,j}f^{(j+1)} \\ & = γ_{0,j+1}f + γ_{1,j+1}f′ + ⋯ + γ_{j,j+1}f^{(j)} + γ_{j+1,j+1}f^{(j+1)}, \end{aligned}

where

(2.12)  γ_{0,j+1} = γ′_{0,j},
γ_{1,j+1} = γ′_{1,j} + γ_{0,j},
⋮
(2.13)  γ_{i,j+1} = γ′_{i,j} + γ_{i−1,j},
⋮
γ_{j,j+1} = γ′_{j,j} + γ_{j−1,j},
(2.14)  γ_{j+1,j+1} = γ_{j,j}.

By (2.11)–(2.14), we know that (2.8)–(2.10) are true for j + 1. Thus (2.8)–(2.10) are true for j = 1, 2, …, k − 1. Lemma 6 is proved.

Lemma 7. Let f and α be nonconstant entire functions, a a finite complex number and k a positive integer. Suppose that

(2.15)  f^{(k)} = a + e^{α}f.

Then for any positive integer j (≥ k) j = sk + l, s ≥ 1, 0 ≤ l ≤ k − 1, we have

(2.16)  f^{(k+j)} = γ_{−1,j} + γ_{0,j}f + γ_{1,j}f′ + ⋯ + γ_{k−1,j}f^{(k−1)},

and γ_{i,j} are entire functions satisfying

(2.17) ( γ_{−1,j} ; γ_{0,j} ; ⋮ ; γ_{l,j} ; γ_{l+1,j} ; ⋮ ; γ_{k−1,j} ) = ( aA_{−1,1,j}e^{α} + a ∑_{t=2}^{s−1} A_{−1,t,j}(e^{α})^{t} + aA_{−1,s,j}(e^{α})^{s} ; A_{0,1,j}e^{α} + ∑_{t=2}^{s} A_{0,t,j}(e^{α})^{t} + A_{0,s+1,j}(e^{α})^{s+1} ; ⋮ ; A_{l,1,j}e^{α} + ∑_{t=2}^{s} A_{l,t,j}(e^{α})^{t} + A_{l,s+1,j}(e^{α})^{s+1} ; A_{l+1,1,j}e^{α} + ∑_{t=2}^{s−1} A_{l+1,t,j}(e^{α})^{t} + A_{l+1,s,j}(e^{α})^{s} ; ⋮ ; A_{k−1,1,j}e^{α} + ∑_{t=2}^{s−1} A_{k−1,t,j}(e^{α})^{t} + A_{k−1,s,j}(e^{α})^{s} ) ,

where A_{i,t,j} (∈ 𝒫[α′]) satisfy

(2.18)  ( A_{−1,s,j} ; A_{0,s+1,j} ; ⋮ ; A_{l−1,s+1,j} ; A_{l,s+1,j} ; A_{l+1,s,j} ; ⋮ ; A_{k−1,s,j} ) = ( C_{−1,s,j}(α′)^{l} + P_{l−1}[α′] ; C_{0,s+1,j}(α′)^{l} + P_{l−1}[α′] ; ⋮ ; C_{l−1,s+1,j}α′ + P_{0}[α′] ; 1 ; C_{l+1,s,j}(α′)^{k−1} + P_{k−2}[α′] ; ⋮ ; C_{k−1,s,j}(α′)^{l+1} + P_{l}[α′] ) ,

and C_{i,s+1,j}, −1 ≤ i ≤ l − 1, and C_{i,s,j}, l + 1 ≤ i ≤ k − 1, are positive integers, and

(2.19)  A_{i,1,j} = \frac{j!}{i!(j − i)!}e^{−α}(e^{α})^{(j−i)} = \frac{j!}{i!(j − i)!}((α′)^{j−i} + P_{j−i−1}[α′]).

Here P_{d}[α′] ≡ 0 for d ≤ 0.

Proof. We prove this lemma by mathematical induction on j. First we prove that (2.16)–(2.19) are true for j = k. By Lemma 6, we have

(2.20)  f^{(2k−1)} = γ_{0,k−1}f + γ_{1,k−1}f′ + ⋯ + γ_{k−1,k−1}f^{(k−1)}.

This together with (2.15) yields

(2.21) \begin{aligned}f^{(2k)} & = (f^{(2k−1)})′ \\ & = γ′_{0,k−1}f + γ′_{1,k−1}f′ + ⋯ + γ′_{k−1,k−1}f^{(k−1)} \\ & + γ_{0,k−1}f′ + ⋯ + γ_{k−2,k−1}f^{(k−1)} + aγ_{k−1,k−1} + e^{α}γ_{k−1,k−1}f \\ & = γ_{−1,k} + γ_{0,k}f + γ_{1,k}f′ + ⋯ + γ_{k−1,k}f^{(k−1)}. \end{aligned}

By Lemma 6, we get

(2.22)  γ_{−1,k} = aγ_{k−1,k−1} = ae^{α},
γ_{0,k} = γ′_{0,k−1} + e^{α}γ_{k−1,k−1}

= (e^{α})^{(k)} + (e^{α})^{2}, (2.23)

γ_{i,k} = γ′_{i,k−1} + γ_{i−1,k−1}

= \frac{(k − 1)!}{i!(k − 1 − i)!}(e^{α})^{(k−i)} + \frac{(k − 1)!}{(i − 1)!(k − i)!}(e^{α})^{(k−i)}

= \frac{k!}{i!(k − i)!}(e^{α})^{(k−i)}, i = 1, …, k − 1. (2.24)

Thus (2.16)–(2.19) are true for j = k.

Now we assume that this lemma is true for a given j = sk + l with s ≥ 1 and 0 ≤ l ≤ k − 1. Next we show that this lemma is true for j + 1. First by (2.15) and (2.16), we get

\begin{aligned}f^{(k+j+1)} & = γ′_{−1,j} + γ′_{0,j}f + γ′_{1,j}f′ + ⋯ + γ′_{k−1,j}f^{(k−1)} \\ & + γ_{0,j}f′ + ⋯ + γ_{k−2,j}f^{(k−1)} + aγ_{k−1,j} + e^{α}γ_{k−1,j}f. \end{aligned}

It follows that (2.16) is true for j + 1 with

(2.25)  γ_{−1,j+1} = γ′_{−1,j} + aγ_{k−1,j},
(2.26)  γ_{0,j+1} = γ′_{0,j} + e^{α}γ_{k−1,j},
(2.27)  γ_{i,j+1} = γ′_{i,j} + γ_{i−1,j}, i = 1, 2, …, k − 1.

Thus for l ≤ k − 2, by the assumptions,

γ_{−1,j+1} = (aA_{−1,1,j}e^{α} + a ∑_{t=2}^{s−1} A_{−1,t,j}(e^{α})^{t} + aA_{−1,s,j}(e^{α})^{s})′

+ a(A_{k−1,1,j}e^{α} + ∑_{t=2}^{s−1} A_{k−1,t,j}(e^{α})^{t} + A_{k−1,s,j}(e^{α})^{s}) (2.28)

= a(A′_{−1,1,j} + α′A_{−1,1,j} + A_{k−1,1,j})e^{α}

+ a ∑_{t=2}^{s−1}(A′_{−1,t,j} + tα′A_{−1,t,j} + A_{k−1,t,j})(e^{α})^{t} + aA_{−1,s,j+1}(e^{α})^{s},

where A_{−1,s,j+1} = A′_{−1,s,j} + sα′A_{−1,s,j} + A_{k−1,s,j},

(2.29)  \begin{aligned}γ_{0,j+1} & = (A_{0,1,j}e^{α} + ∑_{t=2}^{s} A_{0,t,j}(e^{α})^{t} + A_{0,s+1,j}(e^{α})^{s+1})′ \\ & + e^{α}(A_{k−1,1,j}e^{α} + ∑_{t=2}^{s−1} A_{k−1,t,j}(e^{α})^{t} + A_{k−1,s,j}(e^{α})^{s}) \\ & = A_{0,1,j+1}e^{α} + ∑_{t=2}^{s}(A′_{0,t,j} + tα′A_{0,t,j} + A_{k−1,t−1,j})(e^{α})^{t} \\ & + A_{0,s+1,j+1}(e^{α})^{s+1}, \end{aligned}

where A_{0,1,j+1} = A′_{0,1,j} + A_{0,1,j}α′, A_{0,s+1,j+1} = A′_{0,s+1,j} + (s + 1)α′A_{0,s+1,j} + A_{k−1,s,j}, and for 1 ≤ i ≤ l,

(2.30)  \begin{aligned}γ_{i,j+1} & = γ′_{i,j} + γ_{i−1,j} \\ & = (A_{i,1,j}e^{α} + ∑_{t=2}^{s} A_{i,t,j}(e^{α})^{t} + A_{i,s+1,j}(e^{α})^{s+1})′ \\ & + A_{i−1,1,j}e^{α} + ∑_{t=2}^{s} A_{i−1,t,j}(e^{α})^{t} + A_{i−1,s+1,j}(e^{α})^{s+1} \\ & = A_{i,1,j+1}e^{α} + ∑_{t=2}^{s}(A′_{i,t,j} + tα′A_{i,t,j} + A_{i−1,t,j})(e^{α})^{t} \\ & + A_{i,s+1,j+1}(e^{α})^{s+1}, \end{aligned}

where A_{i,1,j+1} = A′_{i,1,j}+α′A_{i,1,j}+A_{i−1,1,j}, A_{i,s+1,j+1} = A′_{i,s+1,j}+(s+1)α′A_{i,s+1,j} +A_{i−1,s+1,j}, and for i = l + 1,

(2.31)  \begin{aligned}γ_{l+1,j+1} & = γ′_{l+1,j} + γ_{l,j} \\ & = (A_{l+1,1,j}e^{α} + ∑_{t=2}^{s−1} A_{l+1,t,j}(e^{α})^{t} + A_{l+1,s,j}(e^{α})^{s})′ \\ & + A_{l,1,j}e^{α} + ∑_{t=2}^{s} A_{l,t,j}(e^{α})^{t} + A_{l,s+1,j}(e^{α})^{s+1} \\ & = A_{l+1,1,j+1}e^{α} + ∑_{t=2}^{s}(A′_{l+1,t,j} + tα′A_{l+1,t,j} + A_{l,t,j})(e^{α})^{t} \\ & + A_{l+1,s+1,j+1}(e^{α})^{s+1}, \end{aligned}

where A_{l+1,1,j+1} = A′_{l+1,1,j} + α′A_{l+1,1,j} + A_{l,1,j}, A_{l+1,s+1,j+1} = A_{l,s+1,j}, and for l + 2 ≤ i ≤ k − 1,

γ_{i,j+1} = γ′_{i,j} + γ_{i−1,j}

= (A_{i,1,j}e^{α} + ∑_{t=2}^{s−1} A_{i,t,j}(e^{α})^{t} + A_{i,s,j}(e^{α})^{s})′

+ A_{i−1,1,j}e^{α} + ∑_{t=2}^{s−1} A_{i−1,t,j}(e^{α})^{t} + A_{i−1,s,j}(e^{α})^{s} (2.32)

= A_{i,1,j+1}e^{α} + ∑_{t=2}^{s−1}(A′_{i,t,j} + tα′A_{i,t,j} + A_{i−1,t,j})(e^{α})^{t} + A_{i,s,j+1}(e^{α})^{s},

where A_{i,1,j+1} = A′_{i,1,j} +α′A_{i,1,j} +A_{i−1,1,j}, A_{i,s,j+1} = A′_{i,s,j} +sα′A_{i,s,j} +A_{i−1,s,j}.

By (2.28)–(2.32), we know that (2.16)–(2.19) are true for j+1 when j = sk+l with 0 ≤ l ≤ k − 2. Similarly, we can prove (2.16)–(2.19) are true for j + 1 when j = sk + k − 1. We omit the details here. Thus Lemma 7 is proved.

Lemma 8. Let

(2.33)  Δ_{j} = ( \begin{matrix}γ_{0,j} & γ_{0,j+1} & ⋯ & γ_{0,j+k−1} \\ γ_{1,j} & γ_{1,j+1} & ⋯ & γ_{1,j+k−1} \\ ⋮ & ⋮ & ... & ⋮ \\ γ_{k−1,j} & γ_{k−1,j+1} & ⋯ & γ_{k−1,j+k−1} \end{matrix} ) ,

where γ_{i,j} are entire functions defined in Lemmas 6–7 (for 1 ≤ j ≤ k−1 and i > j set γ_{i,j} = 0). Denote the determinant of Δ_{j} by det(Δ_{j}). Then for j = sk + l (≥ 1) with s ≥ 0, 0 ≤ l ≤ k − 1, we have

(2.34)  \begin{aligned}det(Δ_{j}) & = ((α′)^{kj} + P_{kj−1}[α′])(e^{α})^{k} \\ & + ∑_{t=k+1}^{(s+1)k+l−1} A_{t,j}(e^{α})^{t} + (−1)^{l(k−l)}(e^{α})^{(s+1)k+l}, \end{aligned}

where A_{t,j} ∈ 𝒫[α′].

Proof. Obviously, by Lemmas 6–7, we have

(2.35)  det(Δ_{j}) = ∑_{t=k}^{ν} A_{t,j}(e^{α})^{t}

with ν ≥ k and A_{t,j} ∈ 𝒫[α′]. Thus we need only to show that

(2.36)  ν = (s + 1)k + l, A_{ν,j} = (−1)^{l(k−l)},

and

(2.37)  A_{k,j} = (α′)^{kj} + P_{kj−1}[α′].

First we prove (2.36). By Lemmas 6–7, we have

M_{1} = ( \begin{matrix}γ_{0,j} & γ_{0,j+1} & ⋯ & γ_{0,j+k−1−l} \\ γ_{1,j} & γ_{1,j+1} & ⋯ & γ_{1,j+k−1−l} \\ ⋮ & ⋮ & ... & ⋮ \\ γ_{l−1,j} & γ_{l−1,j+1} & ⋯ & γ_{l−1,j+k−1−l} \end{matrix} )_{l×(k−l)}
= (polynomials in e^{α} of degrees ≤ s + 1)_{l×(k−l)},
M_{2} = ( \begin{matrix}γ_{0,j+k−l} & γ_{0,j+k−l+1} & ⋯ & γ_{0,j+k−1} \\ γ_{1,j+k−l} & γ_{1,j+k−l+1} & ⋯ & γ_{1,j+k−1} \\ ⋮ & ⋮ & ... & ⋮ \\ γ_{l−1,j+k−l} & γ_{l−1,j+k−l+1} & ⋯ & γ_{l−1,j+k−1} \end{matrix} )

= ( \begin{matrix}1 & A_{0,s+2,j+k−l+1} & ⋯ & A_{0,s+2,j+k−1} \\ 0 & 1 & ⋯ & A_{1,s+2,j+k−1} \\ ⋮ & ⋮ & ... & ⋮ \\ 0 & 0 & ⋯ & 1 \end{matrix} )_{l×l} (e^{α})^{s+2}

+ (polynomials in e^{α} of degrees ≤ s + 1)_{l×l} = A_{l×l}(e^{α})^{s+2} + (polynomials in e^{α} of degrees ≤ s + 1)_{l×l},

M_{3} = ( \begin{matrix}γ_{l,j} & γ_{l,j+1} & ⋯ & γ_{l,j+k−1−l} \\ γ_{l+1,j} & γ_{l+1,j+1} & ⋯ & γ_{l+1,j+k−1−l} \\ ⋮ & ⋮ & ... & ⋮ \\ γ_{k−1,j} & γ_{k−1,j+1} & ⋯ & γ_{k−1,j+k−1−l} \end{matrix} )_{(k−l)×(k−l)}

= ( \begin{matrix}1 & A_{l,s+1,j+1} & ⋯ & A_{l,s+1,j+k−1−l} \\ 0 & 1 & ⋯ & A_{l+1,s+1,j+k−1−l} \\ ⋮ & ⋮ & ... & ⋮ \\ 0 & 0 & ⋯ & 1 \end{matrix} )_{(k−l)×(k−l)} (e^{α})^{s+1}

+ (polynomials in e^{α} of degrees ≤ s)_{(k−l)×(k−l)} = B_{(k−l)×(k−l)}(e^{α})^{s+1} + (polynomials in e^{α} of degrees ≤ s)_{(k−l)×(k−l)},

\begin{matrix}M_{4} = ( \begin{matrix}γ_{l,j+k−l} & γ_{l,j+k−l+1} & ⋯ & γ_{l,j+k−1} \\ γ_{l+1,j+k−l} & γ_{l+1,j+k−l+1} & ⋯ & γ_{l+1,j+k−1} \\ ⋮ & ⋮ & ... & ⋮ \\ γ_{k−1,j+k−l} & γ_{k−1,j+k−l+1} & ⋯ & γ_{k−1,j+k−1} \end{matrix} )_{(k−l)×l} \\ = ( \begin{matrix}A_{l,s+1,j+k−l} & A_{l,s+1,j+k−l+1} & ⋯ & A_{l,s+1,j+k−1} \\ A_{l+1,s+1,j+k−l} & A_{l+1,s+1,j+k−l+1} & ⋯ & A_{l+1,s+1,j+k−1} \\ ⋮ & ⋮ & ... & ⋮ \\ A_{k−1,s+1,j+k−l} & A_{k−1,s+1,j+k−l+1} & ⋯ & A_{k−1,s+1,j+k−1} \end{matrix} )_{(k−l)×l} (e^{α})^{s+1} \end{matrix}

+ (polynomials in e^{α} of degrees ≤ s)_{(k−l)×l}

= C_{(k−l)×l}(e^{α})^{s+1} + (polynomials in e^{α} of degrees ≤ s)_{(k−l)×l},

where A_{l×l}, B_{(k−l)×(k−l)}, C_{(k−l)×l} are matrices whose elements are differential polynomials in α′. In particular, A_{l×l} and B_{(k−l)×(k−l)} are upper triangular matrices whose principal diagonal elements equal 1. Thus by (2.33) we get

det(Δ_{j}) = | \begin{matrix}M_{1} & M_{2} \\ M_{3} & M_{4} \end{matrix} |

= | \begin{matrix}0 & A \\ B & C \end{matrix} | (e^{α})^{(s+1)k+l} + (terms of degree ≤ (s + 1)k + l − 1)

= (−1)^{l(k−l)} det(A) det(B)(e^{α})^{(s+1)k+l} + (terms of degree ≤ (s + 1)k + l − 1)

= (−1)^{l(k−l)}(e^{α})^{(s+1)k+l} + (terms of degree ≤ (s + 1)k + l − 1), where 0 = 0_{l×(k−l)} is the zero matrix, A = A_{l×l}, B = B_{(k−l)×(k−l)} , C = C_{(k−l)×l}. This proves (2.36).

Next we prove (2.37). By Lemmas 6–7, we have

(2.38)  A_{k,j} = | \begin{matrix}A_{0,1,j} & A_{0,1,j+1} & ⋯ & A_{0,1,j+k−1} \\ A_{1,1,j} & A_{1,1,j+1} & ⋯ & A_{1,1,j+k−1} \\ ⋮ & ⋮ & ... & ⋮ \\ A_{k−1,1,j} & A_{k−1,1,j+1} & ⋯ & A_{k−1,1,j+k−1} \end{matrix} |

= | \begin{matrix}1 & 1 & ⋯ & 1 \\ \binom{j}{1} & \binom{j+1}{1} & ⋯ & \binom{j+k−1}{1} \\ ⋮ & ⋮ & ... & ⋮ \\ \binom{j}{i} & \binom{j+1}{i} & ⋯ & \binom{j+k−1}{i} \\ ⋮ & ⋮ & ... & ⋮ \\ \binom{j}{k−1} & \binom{j+1}{k−1} & ⋯ & \binom{j+k−1}{k−1} \end{matrix} | (α′)^{kj} + P_{kj−1}[α′],

where

\binom{j}{i} = \frac{j!}{i!(j − i)!}

are the binomial coefficients. Since

\binom{x}{i} = \frac{x(x − 1) ⋯ (x − i + 1)}{i!}

is a polynomial in x of degree i, by the calculating properties of determinant and the well-known Vandermonde’s determinant, we see that A_{k,j} = C(α′)^{kj} + P_{kj−1}[α′], where C is a nonzero constant which is equal to

∏_{s=1}^{k−1} \frac{1}{s!} ∏_{1≤i<t≤k}(t − i) = 1.

This proves (2.37). Thus Lemma 8 is proved.

3. Proof of Theorem 1

By the assumptions, there exist two entire functions α(z) and β(z) such that

(3.1)  \frac{f^{(k)}(z) − a}{f(z) − a} = e^{α(z)},
(3.2)  \frac{f^{(m)}(z) − a}{f(z) − a} = e^{β(z)}.

Next we consider two cases.

Case 1. Either α or β is a constant. Without loss of generality, we assume that α is a constant. Set e^{α} = c. Then by (3.1), we get

(3.3)  f^{(k)} − cf = (1 − c)a.

Solving (3.3), we get

(3.4)  f(z) = (1 − \frac{1}{c})a + ∑_{j=1}^{q} C_{j}e^{λ_{j}z},

where q (≤ k) is a positive integer, and C_{j}, λ_{j} are nonzero constants satisfying (λ_{j})^{k} = c and λ_{i} ≠ λ_{j}, i ≠ j. By (3.4) and (3.2), it follows that ϱ(e^{β}) ≤ 1, where ϱ (e^{β}) is the order of e^{β}, so that e^{β} = de^{μz}, where d (≠ 0) and μ are constants. Thus by (3.2) and (3.4), we get

(3.5)  −a + ∑_{j=1}^{q}(λ_{j})^{m}C_{j}e^{λ_{j}z} = −\frac{da}{c}e^{μz} + ∑_{j=1}^{q} C_{j}de^{(λ_{j}+μ)z}.

Applying Lemma 5 to (3.5), we deduce that μ = 0 and (λ_{j})^{m} = d. Further, if a ≠ 0, then c = d.

By (λ_{j})^{k} = c, (λ_{j})^{m} = d and the fact that λ_{j}, 1 ≤ j ≤ q, are distinct, we know that q ≤ (k, m), where (k, m) is the greatest common divisor of k and m. In fact, by Euclidean division algorithm, there exist integers k_{0} and m_{0} such that (k, m) = k_{0}k + m_{0}m. Thus (λ_{j})^{(k,m)} = [(λ_{j})^{k}]^{k_{0}}[(λ_{j})^{m}]^{m_{0}} = c^{k_{0}}d^{m_{0}} . Hence by the fact that λ_{j}, 1 ≤ j ≤ q, are distinct, it follows that q ≤ (k, m).

Case 2. Both α and β are not constants.

We will prove that this case cannot occur. Without loss of generality, we assume k < m. Let

(3.6)  F (z) = f(z) − a.

Then by (3.1) and (3.2), we have

(3.7)  F^{(k)} = a + e^{α}F,
(3.8)  F^{(m)} = a + e^{β}F.

Set

(3.9)  ϕ = \frac{F^{(m)} − F^{(k)}}{F} .

Then by (3.7) and (3.8), we get

(3.10)  ϕ = e^{β} − e^{α}.

Next we consider two subcases.

Case 2.1: ϕ ≡ 0. Then by (3.10), we get

(3.11)  e^{β} = e^{α}.

Thus by (3.1), (3.2) and (3.11), we get

(3.12)  f^{(m)} − f^{(k)} = 0.

Solving (3.12), we get

(3.13)  f(z) = b(z) + ∑_{j=1}^{s} C_{j}e^{λ_{j}z},

where b is a polynomial with deg b ≤ k − 1, s ≤ m − k is a positive integer, and C_{j}, λ_{j} are nonzero constants with (λ_{j})^{m−k} = 1 and λ_{i} ≠ λ_{j}, i ≠ j. By (3.1) and (3.13), we know that ϱ(e^{α}) ≤ 1. This together with that α is nonconstant yields that e^{α} = Ce^{cz}, where C and c are nonzero constants. Thus by (3.1) and (3.13), we get

(3.14)  −a + ∑_{j=1}^{s} C_{j}(λ_{j})^{k}e^{λ_{j}z} = C[b(z) − a]e^{cz} + ∑_{j=1}^{s} CC_{j}e^{(λ_{j}+c)z}.

Applying Lemma 5 to (3.14), we get that c = 0, a contradiction.

Case 2.2: ϕ ≢ 0. Then by the logarithmic derivative lemma, it follows from (3.9) that

(3.15)  m(r, ϕ) = S(r, F).

By (3.10), ϕ is an entire function. Thus by (3.15), we get

(3.16)  T (r, ϕ) = S(r, F).

Since ϕ ≢ 0, by (3.10), we get

(3.17)  \frac{e^{β}}{ϕ} = 1 + \frac{e^{α}}{ϕ} .

Thus by (3.16), (3.17) and the second fundamental theorem we deduce that

(3.18)  \begin{aligned}T (r, \frac{e^{β}}{ϕ} ) & ≤\overline{N}(r, \frac{e^{β}}{ϕ} ) +\overline{N}(r, \frac{ϕ}{e^{β}} ) +\overline{N}(r, \frac{1}{\frac{e^{β}}{ϕ} − 1}) + S(r, \frac{e^{β}}{ϕ} ) \\ & ≤\overline{N}(r, \frac{e^{β}}{ϕ} ) +\overline{N}(r, \frac{ϕ}{e^{β}} ) +\overline{N}(r, \frac{1}{\frac{e^{α}}{ϕ}} ) + S(r, \frac{e^{β}}{ϕ} ) \\ & ≤ S(r, F) + S(r, \frac{e^{β}}{ϕ} ). \end{aligned}

This together with (3.16) yields that T (r, e^{β}) = S(r, F). It follows from (3.10) and (3.16) that T (r, e^{α}) = T (r, e^{β} − ϕ) = S(r, F). Thus we get

(3.19)  T (r, e^{α}) + T (r, e^{β}) = S(r, F).
Now, for 0 ≤ j ≤ k − 1, set
(3.20)  p_{i,j} = γ_{i,m−k+j}, i = −1, 0, 1, …, k − 1,

where γ_{i,j} are defined as in Lemmas 6–8. Then by Lemmas 6–7, we have

F^{(m+j)} = F^{(k+m−k+j)}

= p_{−1,j} + p_{0,j}F + p_{1,j}F′ + ⋯ + p_{k−1,j}F^{(k−1)}, j = 0, 1, …, k − 1. (3.21)

On the other hand, by (3.8) and Lemma 6, for 1 ≤ j ≤ k − 1, we have

(3.22)  F^{(m+j)} = q_{0,j}e^{β}F + q_{1,j}e^{β}F′ + ⋯ + q_{j,j}e^{β}F^{(j)},

where q_{i,j}, i ≤ j, are differential polynomials in β′ with constant coefficients. In particular, q_{j,j} ≡ 1 for j = 1, 2, …, k − 1. Thus by (3.8), (3.21) and (3.22), we get

(3.23)  ( F, F′, …, F^{(k−1)} ) (e^{β}Q − P) = Γ,

where

(3.24)  P = ( \begin{matrix}p_{0,0} & p_{0,1} & ⋯ & p_{0,k−1} \\ p_{1,0} & p_{1,1} & ⋯ & p_{1,k−1} \\ ⋮ & ⋮ & ... & ⋮ \\ p_{k−1,0} & p_{k−1,1} & ⋯ & p_{k−1,k−1} \end{matrix} ) ,

(3.25)  Q = ( \begin{matrix}1 & q_{0,1} & ⋯ & q_{0,k−1} \\ 0 & 1 & ⋯ & q_{1,k−1} \\ ⋮ & ⋮ & ... & ⋮ \\ 0 & 0 & ⋯ & 1 \end{matrix} ) ,

(3.26)  Γ = ( p_{−1,0} − a, p_{−1,1}, …, p_{−1,k−1} ) .

By (3.23) and the theory of linear equations, we get

(3.27)  det(e^{β}Q − P)F = det(T),

where T is a matrix whose first line is Γ and the other lines are the same as those of e^{β}Q − P.

Thus by (3.19) and (3.27), we know that

(3.28)  det(e^{β}Q − P) = 0.

This yields that

(3.29)  det(e^{β}I − R) = 0,

where I = I_{k×k} is the k th unit matrix, R = Q^{−1}P and Q^{−1} is the inverse matrix of Q. Obviously, the matrix Q^{−1} is also an upper triangular matrix whose elements are differential polynomial in β′. By (3.29), we get

(3.30)  (e^{β})^{k} − a_{1}(e^{β})^{k−1} + ⋯ + (−1)^{t}a_{t}(e^{β})^{k−t} + ⋯ + (−1)^{k}a_{k} = 0,

where a_{t} is the sum of all the principle minors of order t of R. In particular, a_{k} = det(R) = det(P). Here, for a matrix

A = (a_{i,j}) = ( \begin{matrix}a_{1,1} & a_{1,2} & ⋯ & a_{1,n} \\ a_{2,1} & a_{2,2} & ⋯ & a_{2,n} \\ ⋮ & ⋮ & ... & ⋮ \\ a_{n,1} & a_{n,2} & ⋯ & a_{n,n} \end{matrix} ) ,

and t integers 1 ≤ i_{1} < i_{2} < ⋯ < i_{t} ≤ n, we call

| \begin{matrix}a_{i_{1},i_{1}} & a_{i_{1},i_{2}} & ⋯ & a_{i_{1},i_{t}} \\ a_{i_{2},i_{1}} & a_{i_{2},i_{2}} & ⋯ & a_{i_{2},i_{t}} \\ ⋮ & ⋮ & ... & ⋮ \\ a_{i_{t},i_{1}} & a_{i_{t},i_{1}} & ⋯ & a_{i_{t},i_{t}} \end{matrix} |

a principle minor of order t of A.

Obviously, by (3.24), (3.25) and the definition of a_{t}, a_{t}, 1 ≤ t ≤ k, are polynomials in e^{α} whose coefficients are differential polynomials in α′ and β′ with constant coefficients.

Next we consider the degrees of these polynomials a_{t}. Since m > k, there exist integers s ≥ 1 and 0 ≤ l ≤ k − 1 such that

(3.31)  m = sk + l.

It is obvious that if l = 0 then s > 1. We claim that for l ≥ 1,

(3.32)  deg(a_{t}) ≤ ts + l − 1, t = 1, 2, …, k − 1,

and for l = 0,

(3.33)  deg(a_{t}) ≤ ts, t = 1, 2, …, k − 1.

and

(3.34)  deg(a_{k}) = m = ks + l.

In order to prove (3.32)–(3.34), we first consider the degree of the elements of R = (r_{i,j}) which are polynomials in e^{α}. By (3.20), we see that for 0 ≤ j ≤ k−1−l, p_{i,j} = γ_{i,(s−1)k+j+l}, while for k − l ≤ j ≤ k − 1, p_{i,j} = γ_{i,sk+j+l−k}. Thus by Lemmas 6–7, for 0 ≤ i, j ≤ k − 1,

(3.35) deg(p_{i,j}) ≤ \{ \begin{matrix}s & if 0 ≤ j ≤ k − 1 − l, 0 ≤ i ≤ j + l, \\ s − 1 & if 0 ≤ j ≤ k − 1 − l, j + l + 1 ≤ i ≤ k − 1, \\ s + 1 & if k − l ≤ j ≤ k − 1, 0 ≤ i ≤ j + l − k, \\ s & if k − l ≤ j ≤ k − 1, j + l − k + 1 ≤ i ≤ k − 1. \end{matrix}

By (3.25), we may assume that

(3.36)  Q^{−1} = ( \begin{matrix}1 & q_{0,1}^{∗} & ⋯ & q_{0,k−1}^{∗} \\ 0 & 1 & ⋯ & q_{1,k−1}^{∗} \\ ⋮ & ⋮ & ... & ⋮ \\ 0 & 0 & ⋯ & 1 \end{matrix} ) ,

where q_{i,j}^{∗}, 0 ≤ i < j ≤ k − 1, are differential polynomials in β′ with constant coefficients. Thus by (r_{i,j}) = R = Q^{−1}P, we get

(3.37)  r_{i,j} = p_{i,j} + q_{i,i+1}^{∗}p_{i+1,j} + q_{i,i+2}^{∗}p_{i+2,j} + ⋯ + q_{i,k−1}^{∗}p_{k−1,j}.

Thus by (3.35) and (3.37), we see that for 0 ≤ i, j ≤ k − 1,

(3.38) deg(r_{i,j}) ≤ \{ \begin{matrix}s & if 0 ≤ j ≤ k − 1 − l, 0 ≤ i ≤ j + l, \\ s − 1 & if 0 ≤ j ≤ k − 1 − l, j + l + 1 ≤ i ≤ k − 1, \\ s + 1 & if k − l ≤ j ≤ k − 1, 0 ≤ i ≤ j + l − k, \\ s & if k − l ≤ j ≤ k − 1, j + l − k + 1 ≤ i ≤ k − 1. \end{matrix}

Now let

L_{i_{1},i_{2},...,i_{t}} = | \begin{matrix}r_{i_{1},i_{1}} & r_{i_{1},i_{2}} & ⋯ & r_{i_{1},i_{t}} \\ r_{i_{2},i_{1}} & r_{i_{2},i_{2}} & ⋯ & r_{i_{2},i_{t}} \\ ⋮ & ⋮ & ... & ⋮ \\ r_{i_{t},i_{1}} & r_{i_{t},i_{2}} & ⋯ & r_{i_{t},i_{t}} \end{matrix} |

be a principle minor of order t ≤ k −1 of R, where 0 ≤ i_{1} < i_{2} < ⋯ < i_{t} ≤ k −1.

By (3.38), for the case of l = 0, the degrees of all r_{i,j} are at most s, so that the degree of L_{i_{1},i_{2},...,i_{t}} is at most ts. It follows that the degree of a_{t} is at most ts. This proves (3.33).

Next we consider the case of 1 ≤ l ≤ k − 1. By the definition of determinant, we have

L_{i_{1},i_{2},...,i_{t}} = ∑ δ_{j_{1},j_{2},...,j_{t}}r_{i_{1},j_{1}}r_{i_{2},j_{2}} ⋯ r_{i_{t},j_{t}},

where the sum takes over all the permutations of (i_{1}, i_{2}, …, i_{t}), and δ_{j_{1},j_{2},...,j_{t}} = ±1 according to the permutation (j_{1}, j_{2}, …, j_{t}) of (i_{1}, i_{2}, …, i_{t}) is even or odd. Let

L_{t} = r_{i_{1},j_{1}}r_{i_{2},j_{2}} ⋯ r_{i_{t},j_{t}}.

For t ≤ l − 1, by (3.38), the degree of L_{t} is at most t(s + 1) ≤ ts + l − 1. For t ≥ l, if there exist x ≤ l − 1 polynomials in r_{i_{1},j_{1}}, r_{i_{2},j_{2}}, …, r_{i_{t},j_{t}} with

degree s+1, then by (3.38), the degree of L_{t} is at most x(s+1)+(t−x)s = ts+x ≤ ts + l − 1. If there exist l polynomials in r_{i_{1},j_{1}}, r_{i_{2},j_{2}}, …, r_{i_{t},j_{t}} with degree s + 1, then by (3.38), {0, 1, …, l−1} ⊂ {i_{1}, i_{2}, …, i_{t}}. It follows that there exists at least one of r_{i_{1},j_{1}}, r_{i_{2},j_{2}}, …, r_{i_{t},j_{t}} whose degree is s − 1 (for otherwise, we must have {l, …, k−1} ⊂ {i_{1}, i_{2}, …, i_{t}}. This together with {0, 1, …, l−1} ⊂ {i_{1}, i_{2}, …, i_{t}} yields that t ≥ k, which contradicts t ≤ k − 1). Hence the degree of L_{t} is at most l(s+1)+(s−1)+(t−l−1)s = ts+l−1. It follows that deg(L_{i_{1},i_{2},...,i_{t}}) ≤ ts+l−1. Thus (3.32) is proved.

Next we prove (3.34). In fact, it can be seen from (3.20), (3.24) and Lemma 8 that

(3.39)  \begin{aligned}a_{k} & = det(P) = det(Δ_{m−k}) \\ & = ((α′)^{k(m−k)} + P_{k(m−k)−1}[α′])(e^{α})^{k} \\ & + ∑_{t=k+1}^{m−1} A_{t,m−k}(e^{α})^{t} + (−1)^{l(k−l)}(e^{α})^{m}. \end{aligned}

Thus we get (3.34).

By (3.32)–(3.34), we see that for 1 ≤ t ≤ k − 1, deg(a_{t}) < deg(a_{k}). Thus by (3.30) and Lemma 2, it follows that

\begin{aligned}T (r, e^{β}) & = O(T (r, e^{α})) + S(r, e^{β}), \\ T (r, e^{α}) & = O(T (r, e^{β})) + S(r, e^{α}). \end{aligned}

Hence we get

(3.40)  S(r, e^{α}) = S(r, e^{β}) = S(r) (say).

Next we prove the following claims.

Claim I. For any rational number θ = ν/μ with ν ∈ Z and μ ∈ N,

(3.41)  T (r, e^{β−θα}) ≠ S(r).

Suppose on the contrary that there exists a rational number θ = ν/μ such that

(3.42)  T (r, e^{β−θα}) = S(r).

Let

(3.43)  b(z) = e^{β−θα}.

Then b(z) ≠ 0 is entire and T (r, b) = S(r). By (3.43),

(3.44)  e^{β} = b(z)e^{θα} = b(z)(e^{α/μ})^{ν}.

On the other hand, by (3.20), (3.24) and Lemmas 6–7, we have

(3.45)  P = (e^{α})^{s+1}P_{0} + (e^{α})^{s}P_{1} + ⋯ + (e^{α})P_{s},

where P_{j} are k × k matrices whose elements are differential polynomials in α′. In particular, det(P_{s}) ≡ A_{k,m−k}, where A_{k,m−k} is defined by (2.38). By (3.28), (3.44) and (3.45), we get

(3.46)  det(b(e^{α/μ})^{ν}Q − (e^{α})^{s+1}P_{0} − (e^{α})^{s}P_{1} − ⋯ − (e^{α})P_{s}) = 0.

If ν > μ, then by (3.46), we get

(3.47)  det(b(e^{α/μ})^{ν−μ}Q − (e^{α/μ})^{sμ}P_{0} − ⋯ − (e^{α/μ})^{μ}P_{s−1} − P_{s}) = 0.

Since the left side of (3.47) is a polynomial in e^{α/μ} whose “constant” term is det(−P_{s}) = (−1)^{k}A_{k,m−k}, by Lemma 3, we get A_{k,m−k} = 0. Thus by (2.10), (2.19) and the fact that α is nonconstant, α′ is nonconstant. For otherwise, let α′ = c. Then c ≠ 0, and by (2.10), (2.19), we have

A_{i,1,j} = \binom{j}{i}(c)^{j−i},

so that by (2.38), it follows that A_{k,m−k} = (c)^{k(m−k)} ≠ 0, which contradicts A_{k,m−k} = 0. Hence α′ is nonconstant. Thus by (2.37) and Lemma 1, we deduce that T (r, α′) = m(r, α′) = S(r, α′), a contradiction.

If ν < μ, then by (3.46), we get

det(bQ − (e^{α/μ})^{(s+1)μ−ν}P_{0} − (e^{α/μ})^{sμ−ν}P_{1} − ⋯ − (e^{α/μ})^{μ−ν}P_{s}) = 0.

Using the same argument as that in case ν > μ, we deduce that det(bQ) = 0. Thus by det(Q) = 1, we get that b = 0, a contradiction.

If ν = μ, then e^{β} = b(z)e^{α}. Thus by (3.32)–(3.34), we see that the left side of (3.30) is a polynomial in e^{α} whose leading term is ε(e^{α})^{m}, where ε = ±1 is a constant. Thus applying Lemma 2 to (3.30), we get a contradiction: T (r, e^{α}) = S(r).

Hence Claim I is proved.

Claim II. We have

(3.48)  H = ∑_{t=1}^{k−1}(−1)^{t}a_{t}(e^{β})^{k−t} ≡ 0.

Suppose that H ≢ 0. Then by the fact that a_{t} are polynomials in e^{α}, we can rewrite H as

(3.49)  H = ∑_{(t,i)∈T ×I} a_{t,i}e^{(k−t)β+iα},

where T ⊂ {1, …, k − 1} and I are finite index sets, a_{t,i} ≢ 0 are differential polynomials in α′ and β′ such that all the functions a_{t,i}e^{(k−t)β+iα}, (t, i) ∈ T × I are linearly independent.

By (3.39), we rewrite a_{k} as

(3.50)  (−1)^{k}a_{k} = ∑_{i∈J} a_{k,i}e^{iα},

where J ⊃ {m} is a finite index set, and a_{k,i} (≢ 0), i ∈ J, are differential polynomials in α′.

Hence by (3.30), (3.48)–(3.50), we get

(3.51)  e^{kβ} + ∑_{(t,i)∈T ×I} a_{t,i}e^{(k−t)β+iα} + ∑_{i∈J} a_{k,i}e^{iα} = 0.

By (3.51), we get

(3.52)  ∑_{(t,i)∈T ×I}(−a_{t,i})e^{−tβ+iα} + ∑_{i∈J}(−a_{k,i})e^{−kβ+iα} = 1.

If the functions (−a_{t,i})e^{−tβ+iα}, (t, i) ∈ T × I and (−a_{k,i})e^{−kβ+iα}, i ∈ J are linearly independent, then by Lemma 4 and the fact that m ∈ J, we get

T (r, (−a_{k,m})e^{−kβ+mα}) = S(r),

so that

T (r, e^{−kβ+mα}) = S(r),

which contradicts Claim I.

Hence the functions (−a_{t,i})e^{−tβ+iα}, (t, i) ∈ T × I and (−a_{k,i})e^{−kβ+iα}, i ∈ J are linearly dependent. That is, there exist constants C_{t,i}, (t, i) ∈ T × I and C_{k,i}, i ∈ J, at least one of them is not equal to 0, such that

∑_{(t,i)∈T ×I} C_{t,i}a_{t,i}e^{−tβ+iα} + ∑_{i∈J} C_{k,i}a_{k,i}e^{−kβ+iα} = 0,

so that

(3.53)  ∑_{(t,i)∈T ×I} C_{t,i}a_{t,i}e^{(k−t)β+iα} + ∑_{i∈J} C_{k,i}a_{k,i}e^{iα} = 0.

By Lemma 3, at least one of C_{t,i}, (t, i) ∈ T × I is not equal to 0. Set T_{1} × I_{1} = {(t, i) ∈ T × I : C_{t,i} ≠ 0}. Then T_{1} × I_{1} ≠ ∅(empty set). By the assumption that a_{t,i}e^{(k−t)β+iα}, (t, i) ∈ T × I are linearly independent, at least one of C_{k,i}, i ∈ J is not equal to 0. Set J_{1} = {i ∈ J : C_{k,i} ≠ 0}. Then J_{1} ≠ ∅. Let i_{1} ∈ J_{1} . Then by (3.53), we get

∑_{(t,i)∈T_{1}×I_{1}} \frac{−C_{t,i}a_{t,i}}{C_{k,i_{1}}a_{k,i_{1}}} e^{(k−t)β+(i−i_{1})α} + ∑_{i∈J_{1}\\{i_{1}\}} \frac{−C_{k,i}a_{k,i}}{C_{k,i_{1}}a_{k,i_{1}}} e^{(i−i_{1})α} = 1.

If the functions

(3.54)  \begin{matrix}\frac{−C_{t,i}a_{t,i}}{C_{k,i_{1}}a_{k,i_{1}}} e^{(k−t)β+(i−i_{1})α}, (t, i) ∈ T_{1} × I_{1} and \\ \frac{−C_{k,i}a_{k,i}}{C_{k,i_{1}}a_{k,i_{1}}} e^{(i−i_{1})α}, i ∈ J_{1} \ \{i_{1}\} \end{matrix}

are linearly independent, then by Lemma 4, we get for (t_{0}, i_{0}) ∈ T_{1} × I_{1} ,

T (r, \frac{C_{t_{0},i_{0}}a_{t_{0},i_{0}}}{C_{k,i_{1}}a_{k,i_{1}}} e^{(k−t_{0})β+(i_{0}−i_{1})α}) = S(r),

so that

T (r, e^{(k−t_{0})β+(i_{0}−i_{1})α}) = S(r),

which again contradicts Claim I. Thus the functions showed in (3.54) are linearly dependent. Thus there exist constants D_{t,i}, (t, i) ∈ T_{1} × I_{1} and D_{k,i}, i ∈ J_{1} \ {i_{1}}, at least one of them is not equal to 0, such that

∑_{(t,i)∈T_{1}×I_{1}} \frac{D_{t,i}a_{t,i}}{C_{k,i_{1}}a_{k,i_{1}}} e^{(k−t)β+(i−i_{1})α} + ∑_{i∈J_{1}\\{i_{1}\}} \frac{D_{k,i}a_{k,i}}{C_{k,i_{1}}a_{k,i_{1}}} e^{(i−i_{1})α} = 0,

so that

(3.55)  ∑_{(t,i)∈T_{1}×I_{1}} D_{t,i}a_{t,i}e^{(k−t)β+iα} + ∑_{i∈J_{1}\{i_{1}}} D_{k,i}a_{k,i}e^{iα} = 0.

By Lemma 3, we see that at least one of D_{t,i}, (t, i) ∈ T ×I is not equal to 0, so that T_{2} × I_{2} = {(t, i) ∈ T_{1} × I_{1} : D_{t,i} ≠ 0} ≠ ∅. By the assumption that a_{t,i}e^{(k−t)β+iα}, (t, i) ∈ T ×I are linearly independent, at least one of D_{k,i}, i ∈ J \{i_{1}} is not equal to 0, so that J_{2} = {i ∈ J_{1} \ {i_{1}} : D_{k,i} ≠ 0} ≠ ∅. Let i_{2} ∈ J_{2} . Then using an argument similar to that in the above step, there exist constants E_{t,i}, (t, i) ∈ T_{2}×I_{2} and E_{k,i}, i ∈ J_{2} \ {i_{2}}, at least one of them is not equal to 0, such that

∑_{(t,i)∈T_{2}×I_{2}} E_{t,i}a_{t,i}e^{(k−t)β+iα} + ∑_{i∈J_{2}\{i_{2}}} E_{k,i}a_{k,i}e^{iα} = 0.

Step by step, it follows that J is an infinite set. It is impossible. Hence we have proved Claim II.

Next we continue to prove Theorem 1. By (3.30), (3.50) and Claim II, we get

e^{kβ} + ∑_{i∈J} a_{k,i}e^{iα} = 0,

so that

(3.56)  ∑_{i∈J}(−a_{k,i})e^{iα−kβ} = 1.

If the functions (−a_{k,i})e^{iα−kβ}, i ∈ J, are linearly independent, then by Lemma 3, we get

T (r, a_{k,m}e^{mα−kβ}) = S(r),

so that

T (r, e^{mα−kβ}) = S(r).

This contradicts Claim I.

286 Jianming Chang and Mingliang Fang

Hence the functions (−a_{k,i})e^{iα−kβ}, i ∈ J, are linearly dependent. Thus there exist constants C_{i}, i ∈ J, at least one of them is not equal to 0, such that

∑_{i∈J} C_{i}a_{k,i}e^{iα−kβ} = 0,

so that

∑_{i∈J} C_{i}a_{k,i}e^{iα} = 0.

This contradicts Lemma 3.

The proof of Theorem 1 is complete.

Acknowledgements. The authors are grateful to the referee for his/her many valuable suggestions.

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Received 10 September 2004