Open Access

A note on entire functions and their differences

Journal of Inequalities and Applications20132013:587

https://doi.org/10.1186/1029-242X-2013-587

Received: 17 April 2013

Accepted: 19 November 2013

Published: 17 December 2013

Abstract

In this paper, we prove that for a transcendental entire function f ( z ) of finite order such that λ ( f a ( z ) ) < σ ( f ) , where a ( z ) is an entire function and satisfies σ ( a ( z ) ) < 1 , n is a positive integer and if Δ η n f ( z ) and f ( z ) share the function a ( z ) CM, where η ( C ) satisfies Δ η n f ( z ) 0 , then

a ( z ) 0 and f ( z ) = c e c 1 z ,

where c, c 1 are two nonzero constants.

MSC:39A10, 30D35.

Keywords

complex difference meromorphic function Borel exceptional value sharing value

1 Introduction and results

In this paper, we assume that the reader is familiar with the fundamental results and the standard notations of Nevanlinna’s value distribution theory of meromorphic functions (see [13]). In addition, we use the notation λ ( f ) for the exponent of convergence of the sequence of zeros of a meromorphic function f, and σ ( f ) to denote the order growth of f. For a nonzero constant η, the forward differences Δ η n f ( z ) are defined (see [4, 5]) by
Δ η f ( z ) = Δ η 1 f ( z ) = f ( z + η ) f ( z ) and Δ η n + 1 f ( z ) = Δ η n f ( z + η ) Δ η n f ( z ) , n = 1 , 2 , .

Throughout this paper, we denote by S ( r , f ) any function satisfying S ( r , f ) = o ( T ( r , f ) ) as r , possibly outside a set of r of finite logarithmic measure. A meromorphic function α ( z ) is said to be a small function of f ( z ) if T ( r , α ( z ) ) = S ( r , f ) , and we denote by S ( f ) the set of functions which are small compared to f ( z ) .

Let f and g be two nonconstant meromorphic functions, and let a C . We say that f and g share the value a CM (IM) provided that f a and g a have the same zeros counting multiplicities (ignoring multiplicities), that f and g share the value ∞ CM (IM) provided that f and g have the same poles counting multiplicities (ignoring multiplicities). Using the same method, we can define that f and g share the function a ( z ) CM (IM), where a ( z ) S ( f ) S ( g ) . Nevanlinna’s four values theorem [6] says that if two nonconstant meromorphic functions f and g share four values CM, then f g or f is a Möbius transformation of g. The condition ‘f and g share four values CM’ has been weakened to ‘f and g share two values CM and two values IM’ by Gundersen [7, 8], as well as by Mues [9]. But whether the condition can be weakened to ‘f and g share three values IM and another value CM’ is still an open question.

In the special case, we recall a well-known conjecture by Brück [10].

Conjecture Let f be a nonconstant entire function such that hyper order σ 2 ( f ) < and σ 2 ( f ) is not a positive integer. If f and f share the finite value a CM, then
f a = c ( f a ) ,

where c is a nonzero constant.

The notation σ 2 ( f ) denotes hyper-order (see [11]) of f ( z ) which is defined by
σ 2 ( f ) = lim ¯ r log log T ( r , f ) log r .

The conjecture has been verified in the special cases when a = 0 [10], or when f is of finite order [12], or when σ 2 ( f ) < 1 2 [13].

Recently, many authors [1417] started to consider sharing values of meromorphic functions with their shifts. Heittokangas et al. proved the following theorems.

Theorem A (See [15])

Let f be a meromorphic function with σ ( f ) < 2 , and let c C . If f ( z ) and f ( z + c ) share the values a ( C ) andCM, then
f ( z + c ) a = τ ( f ( z ) a )

for some constant τ.

In [15], Heittokangas et al. give the example f ( z ) = e z 2 + 1 which shows that σ ( f ) < 2 cannot be relaxed to σ ( f ) 2 .

Theorem B (See [16])

Let f be a meromorphic function of finite order, let c C . If f ( z ) and f ( z + c ) share three distinct periodic functions a 1 , a 2 , a 3 S ˆ ( f ) with period c CM (where S ˆ ( f ) = S ( f ) { } ), then f ( z ) = f ( z + c ) for all z C .

Recently, many results of complex difference equations have been rapidly obtained (see [1825]). In the present paper, we utilize a complex difference equation to consider uniqueness problems.

The main purpose of this paper is to utilize a complex difference equation to study problems concerning sharing values of meromorphic functions and their differences. It is well known that Δ η f ( z ) = f ( z + η ) f ( z ) (where η ( C ) is a constant satisfying f ( z + η ) f ( z ) 0 ) is regarded as the difference counterpart of f . So, Chen and Yi [20] considered the problem that Δ η f ( z ) and f ( z ) share one value a CM and proved the following theorem.

Theorem C (See [20])

Let f be a finite order transcendental entire function which has a finite Borel exceptional value a, and let η ( C ) be a constant such that f ( z + η ) f ( z ) . If Δ η f ( z ) = f ( z + η ) f ( z ) and f ( z ) share the value a CM, then
a = 0 and f ( z + η ) f ( z ) f ( z ) = A ,

where A is a nonzero constant.

Question 1 What can be said if we consider the forward difference Δ η n f ( z ) and f ( z ) share one value or one small function?

In this paper, we answer Question 1 and prove the following theorem.

Theorem 1.1 Let f ( z ) be a finite order transcendental entire function such that λ ( f a ( z ) ) < σ ( f ) , where a ( z ) is an entire function and satisfies σ ( a ) < 1 . Let n be a positive integer. If Δ η n f ( z ) and f ( z ) share a ( z ) CM, where η ( C ) satisfies Δ η n f ( z ) 0 , then
a ( z ) 0 and f ( z ) = c e c 1 z ,

where c, c 1 are two nonzero constants.

In the special case, if we take a ( z ) a in Theorem 1.1, we can get the following corollary.

Corollary 1.1 Let f ( z ) be a finite order transcendental entire function which has a finite Borel exceptional value a. Let n be a positive integer. If Δ η n f ( z ) and f ( z ) share value a CM, where η ( C ) satisfies Δ η n f ( z ) 0 , then
a = 0 and f ( z ) = c e c 1 z ,

where c, c 1 are two nonzero constants.

Remark 1.1 From Corollary 1.1, we know that Δ η n f ( z ) f ( z ) = ( e c 1 η 1 ) n and it shows that the quotient of Δ η n f ( z ) and f ( z ) is related to η, n and c 1 , but not related to c. On the other hand, Corollary 1.1 shows that if f has a nonzero finite Borel exceptional value b , then, for any constant η satisfying Δ η n f ( z ) 0 , the value b is not shared CM by Δ η n f ( z ) and f ( z ) . See the following example.

Example 1.1 Suppose that f ( z ) = e z + b , where b is a nonzero finite value. Then f has a nonzero finite Borel exceptional value b . For any η 2 k π i , k Z , the value b is not shared CM by Δ η n f ( z ) and f ( z ) . Observe that Δ η n f ( z ) = j = 0 n ( 1 ) j C n j f ( z + ( n j ) η ) , where C n j are the binomial coefficients. Thus, for any η 2 k π i , k Z , we have Δ η n f ( z ) = ( e η 1 ) n e z . Thus, we can see that f ( z ) b = e z has no zero, but Δ η n f ( z ) b = ( e η 1 ) n e z b has infinitely many zeros. Hence, the value b is not shared CM by Δ η n f ( z ) and f ( z ) .

In the special case, if we take n = 1 in Theorem 1.1 and n = 1 in Corollary 1.1, we can obtain the following corollaries.

Corollary 1.2 Let f ( z ) be a finite order transcendental entire function such that λ ( f a ( z ) ) < σ ( f ) , where a ( z ) is an entire function and satisfies σ ( a ) < 1 . If Δ η f ( z ) = f ( z + η ) f ( z ) and f ( z ) share a ( z ) CM, where η ( C ) satisfies f ( z + η ) f ( z ) , then
a ( z ) 0 and f ( z ) = c e c 1 z ,

where c, c 1 are two nonzero constants.

Corollary 1.3 Let f ( z ) be a finite order transcendental entire function which has a finite Borel exceptional value a. If Δ η f ( z ) = f ( z + η ) f ( z ) and f ( z ) share value a CM, where η ( C ) satisfies f ( z + η ) f ( z ) , then
a = 0 and f ( z ) = c e c 1 z ,

where c, c 1 are two nonzero constants.

Remark 1.2 The Corollary 1.2 shows that if a nonzero polynomial a ( z ) satisfies λ ( f a ) < σ ( f ) , then a ( z ) is not shared CM by Δ f ( z ) and f ( z ) . For example, if we take a ( z ) z , and λ ( f z ) < σ ( f ) holds, then Δ f ( z ) and f ( z ) do not have any common fixed point (counting multiplicities). See the following example.

Example 1.2 Suppose that f ( z ) = e z + z . Then f ( z ) satisfies λ ( f ( z ) z ) = 0 < 1 = σ ( f ) and has no fixed point. But for any η 2 k π i , k Z , the function Δ η f ( z ) = f ( z + η ) f ( z ) = ( e η 1 ) e z + η has infinitely many fixed points by Milloux’s theorem (see [1, 3]). Hence, the nonzero polynomial a ( z ) z is not shared CM by Δ η f ( z ) and f ( z ) .

Remark 1.3 From Corollary 1.3, we can see that under the hypothesis of Theorem C, we can get the expression of f ( z ) , that is, f ( z ) = c e c 1 z . Thus, we know that the constant A in Theorem C is related to η and c 1 , but not related to c. Actually, from the proof of Lemma 2.9, we have A = e c 1 η 1 (obviously, we can obtain A 1 ). Hence, Corollary 1.3 contains and improves Theorem C. Obviously, Theorem 1.1 generalizes Theorem C.

2 Lemmas for the proof of theorems

Lemma 2.1 (See [21])

Let f be a meromorphic function with a finite order σ, η be a nonzero constant. Let ε > 0 be given, then there exists a subset E ( 1 , ) with finite logarithmic measure such that for all z satisfying | z | = r E [ 0 , 1 ] , we have
exp { r σ 1 + ε } | f ( z + η ) f ( z ) | exp { r σ 1 + ε } .

Lemma 2.2 (See [11, 26])

Suppose that n 2 and let f 1 ( z ) , , f n ( z ) be meromorphic functions and g 1 ( z ) , , g n ( z ) be entire functions such that
  1. (i)

    j = 1 n f j ( z ) exp { g j ( z ) } = 0 ;

     
  2. (ii)

    when 1 j < k n , g j ( z ) g k ( z ) is not constant;

     
  3. (iii)
    when 1 j n , 1 h < k n ,
    T ( r , f j ) = o { T ( r , exp { g h g k } ) } ( r , r E ) ,
     

where E ( 1 , ) has finite linear measure or logarithmic measure.

Then f j ( z ) 0 , j = 1 , , n .

ε-set Following Hayman [1], we define an ε-set to be a countable union of open discs not containing the origin and subtending angles at the origin whose sum is finite. If E is an ε-set, then the set of r 1 , for which the circle S ( 0 , r ) meets E, has finite logarithmic measure, and for almost all real θ, the intersection of E with the ray arg z = θ is bounded.

Lemma 2.3 (See [4])

Let f be a function transcendental and meromorphic in the plane of order <1. Let h > 0 . Then there exists an ε-set E such that
f ( z + c ) f ( z ) = c f ( z ) ( 1 + o ( 1 ) ) as  z  in  C E ,

uniformly in c for | c | h .

Lemma 2.4 (See [25])

Let f be a transcendental meromorphic solution of finite order ρ of a difference equation of the form
U ( z , f ) P ( z , f ) = Q ( z , f ) ,
where U ( z , f ) , P ( z , f ) , Q ( z , f ) are difference polynomials such that the total degree deg U ( z , f ) = n in f ( z ) and its shifts, and deg Q ( z , f ) n . Moreover, we assume that U ( z , f ) contains just one term of maximal total degree in f ( z ) and its shifts. Then, for each ε > 0 ,
m ( r , P ( z , f ) ) = O ( r ρ 1 + ε ) + S ( r , f ) ,

possibly outside of an exceptional set of finite logarithmic measure.

Remark 2.1 From the proof of Lemma 2.4 in [25], we can see that if the coefficients of U ( z , f ) , P ( z , f ) , Q ( z , f ) , namely α λ ( z ) , satisfy m ( r , α λ ) = S ( r , f ) , then the same conclusion still holds.

Lemma 2.5 (See [27])

Let P n ( z ) , , P 0 ( z ) be polynomials such that P n P 0 0 and satisfy
P n ( z ) + + P 0 ( z ) 0 .
(2.1)
Then every finite order transcendental meromorphic solution f ( z ) (0) of the equation
P n ( z ) f ( z + n ) + P n 1 ( z ) f ( z + n 1 ) + + P 0 ( z ) f ( z ) = 0
(2.2)

satisfies σ ( f ) 1 , and f ( z ) assumes every nonzero value a C infinitely often and λ ( f a ) = σ ( f ) .

Remark 2.2 If equation (2.2) satisfies condition (2.1) and all P j ( z ) are constants, we can easily see that equation (2.2) does not possess any nonzero polynomial solution.

Lemma 2.6 (See [27])

Let F ( z ) , P n ( z ) , , P 0 ( z ) be polynomials such that F P n P 0 0 . Then every finite order transcendental meromorphic solution f ( z ) (0) of the equation
P n ( z ) f ( z + n ) + P n 1 ( z ) f ( z + n 1 ) + + P 0 ( z ) f ( z ) = F
(2.3)

satisfies λ ( f ) = σ ( f ) 1 .

Remark 2.3 From the proof of Lemma 2.5 in [27], we can see that if we replace f ( z + j ) by f ( z + j η ) ( j = 1 , , n ) in equation (2.2) or (2.3), then the corresponding conclusion still holds.

Lemma 2.7 (See [4])

Let f be a function transcendental and meromorphic in the plane which satisfies lim ̲ r T ( r , f ) r = 0 . Then g ( z ) = f ( z + 1 ) f ( z ) and G ( z ) = f ( z + 1 ) f ( z ) f ( z ) are both transcendental.

Remark 2.4 From the proof of Lemma 2.7 in [4], we can see that, under the same hypotheses of Lemma 2.7, we can obtain the following conclusion: Δ η f ( z ) = f ( z + η ) f ( z ) and G ( z ) = Δ η f ( z ) f ( z ) = f ( z + η ) f ( z ) f ( z ) are both transcendental.

Lemma 2.8 Let f ( z ) = H ( z ) e c 1 z , where H ( z ) (0) is an entire function such that σ ( H ) < 1 and c 1 is a nonzero constant. If Δ η n f ( z ) 0 for some constant η, and
Δ η n f ( z ) f ( z ) = A
(2.4)

holds, where A is a constant, then H ( z ) is a constant.

Proof From Δ η n f ( z ) 0 , we can see that A 0 . In order to prove that H ( z ) is a constant, we only need to prove H ( z ) 0 . Substituting f ( z ) = H ( z ) e c 1 z into (2.4), we can obtain
j = 0 n 1 ( 1 ) j C n j e ( n j ) c 1 η H ( z + ( n j ) η ) + ( ( 1 ) n A ) H ( z ) = 0 .
(2.5)
First, we assert that the sum of all coefficients of equation (2.5) is equal to zero, that is,
e n c 1 η C n 1 e ( n 1 ) c 1 η + + ( 1 ) n 1 C n n 1 e c 1 η + ( ( 1 ) n A ) = 0 .
(2.6)
On the contrary, we suppose that
e n c 1 η C n 1 e ( n 1 ) c 1 η + + ( 1 ) n 1 C n n 1 e c 1 η + ( ( 1 ) n A ) 0 .
Thus, applying Lemma 2.5 and Remarks 2.2-2.3 to (2.5), we have σ ( H ) 1 , a contradiction. Hence, (2.6) holds. Thus, by (2.6) and (2.5), we have
j = 0 n 1 ( 1 ) j C n j e ( n j ) c 1 η ( H ( z + ( n j ) η ) H ( z ) ) = 0 .
(2.7)
By Lemma 2.3, we see that there exists an ε-set E such that for j = 1 , 2 , , n ,
H ( z + j η ) H ( z ) = j η H ( z ) ( 1 + o ( 1 ) ) as  z  in  C E .
(2.8)
Substituting (2.8) into (2.7), we can get
η H ( z ) K + η H ( z ) K o ( 1 ) = 0 as  z  in  C E ,
(2.9)
where K is a constant and satisfies
K = n e n c 1 η C n 1 ( n 1 ) e ( n 1 ) c 1 η + + ( 1 ) n 2 C n n 2 2 e 2 c 1 η + ( 1 ) n 1 C n n 1 e c 1 η .
Secondly, we assert that K 0 . If n = 1 , then K = e c 1 η 0 ; if n 2 , on the contrary, we suppose that K = 0 . Then, for j = 0 , 1 , , n 1 , noting that
C n j ( n j ) = n ! ( n j ) ( n j ) ! j ! = ( n 1 ) ! n ( n 1 j ) ! j ! = n C n 1 j ,
we have
j = 0 n 1 ( 1 ) j C n j ( n j ) e ( n j ) c 1 η = n e c 1 η ( e c 1 η 1 ) n 1 = 0 .

Thus, we can obtain from the equality above that e c 1 η = 1 since n 1 1 . By (2.6) we have A = ( e c 1 η 1 ) n = 0 , which contradicts A 0 . Hence K 0 and (2.9) implies H ( z ) 0 . Thus, we can know that H ( z ) is a nonzero constant.

Thus, Lemma 2.8 is proved. □

Lemma 2.9 Suppose that f ( z ) is a finite order transcendental entire function such that λ ( f a ( z ) ) < σ ( f ) , where a ( z ) is an entire function and satisfies σ ( a ) < 1 . Let n be a positive integer. If Δ η n f ( z ) 0 for some constant η ( C ), and
Δ η n f ( z ) a ( z ) f ( z ) a ( z ) = A
(2.10)
holds, where A is a constant, then
a ( z ) 0 , A 0 and f ( z ) = c e c 1 z ,

where c, c 1 are two nonzero constants.

Proof Since f ( z ) is a transcendental entire function of finite order and satisfies λ ( f a ( z ) ) < σ ( f ) , we can write f ( z ) in the form
f ( z ) = a ( z ) + H ( z ) e h ( z ) ,
(2.11)
where H (0) is an entire function, h is a polynomial with deg h = k ( k 1 ), H and h satisfy
λ ( H ) = σ ( H ) = λ ( f a ( z ) ) < σ ( f ) = deg h .
(2.12)
First, we assert that a ( z ) 0 . Substituting (2.11) into (2.10), we can get that
Δ η n f ( z ) a ( z ) f ( z ) a ( z ) = j = 0 n ( 1 ) j C n j H ( z + ( n j ) η ) e h ( z + ( n j ) η ) + b ( z ) H ( z ) e h ( z ) = A ,
(2.13)
where b ( z ) = Δ η n a ( z ) a ( z ) . Rewrite (2.13) in the form
j = 0 n 1 ( 1 ) j C n j H ( z + ( n j ) η ) e h ( z + ( n j ) η ) h ( z ) + ( ( 1 ) n A ) H ( z ) = b ( z ) e h ( z ) .
(2.14)
Suppose that b ( z ) 0 . Then, from σ ( H ( z + ( n j ) η ) ) = σ ( H ( z ) ) < deg h ( z ) = k ( j = 0 , 1 , , n 1 ), deg ( h ( z + ( n j ) η ) h ( z ) ) = k 1 and σ ( b ( z ) ) σ ( a ( z ) ) < 1 k , we can see that the order of growth of the left-hand side of (2.14) is less than k, and the order of growth of the right-hand side of (2.14) is equal to k. This is a contradiction. Hence, b ( z ) Δ η n a ( z ) a ( z ) 0 . Namely,
a ( z + n η ) C n 1 a ( z + ( n 1 ) η ) + + ( 1 ) n 1 C n n 1 a ( z + η ) + ( ( 1 ) n 1 ) a ( z ) = 0 .
(2.15)
Suppose that a ( z ) 0 . Note that the sum of all coefficients of (2.15) does not vanish. Then we can apply Lemma 2.5 and Remarks 2.2-2.3 to (2.15) and obtain σ ( a ( z ) ) 1 , which contradicts our hypothesis. Hence, a ( z ) 0 . Thus, (2.13) can be rewritten as
Δ η n f ( z ) f ( z ) = j = 0 n ( 1 ) j C n j H ( z + ( n j ) η ) e h ( z + ( n j ) η ) h ( z ) H ( z ) = A .
(2.16)

Secondly, we prove that A 0 . In fact, if A = 0 , we obtain from (2.16) that Δ η n f ( z ) 0 , which contradicts our hypothesis.

Thirdly, we prove that σ ( f ) = k = 1 . On the contrary, we suppose that σ ( f ) = k 2 . Thus, we will deduce a contradiction for cases A = ( 1 ) n and A ( 1 ) n , respectively.

Case 1. Suppose that A = ( 1 ) n . Thus, for a positive integer n, there are three subcases: (1) n = 1 ; (2) n = 2 ; (3) n 3 .

Subcase 1.1. Suppose that n = 1 . Then, by A = 1 , we can obtain from (2.16) that
e h ( z + η ) h ( z ) = ( 1 + A ) H ( z ) H ( z + η ) 0 ,

a contradiction.

Subcase 1.2. Suppose that n = 2 . Then, by A = ( 1 ) 2 = 1 and (2.16), we have
e h ( z + 2 η ) h ( z + η ) = 2 H ( z + η ) H ( z + 2 η ) .
(2.17)
Set Q 1 ( z ) = 2 H ( z + 2 η ) H ( z + η ) . Then, from (2.17), we can know that Q 1 ( z ) is a nonconstant entire function. Set σ ( H ) = σ 1 . Then σ 1 < σ ( f ) = k . By Lemma 2.1, we see that for any given ε 1 ( 0 < 3 ε 1 < k σ 1 ), there exists a set E 1 ( 1 , ) of finite logarithmic measure such that for all z satisfying | z | = r [ 0 , 1 ] E 1 , we have
exp { r σ 1 1 + ε 1 } | 2 H ( z + η ) H ( z + 2 η ) | exp { r σ 1 1 + ε 1 } .
(2.18)
Since Q 1 ( z ) is an entire function, by (2.18), we have
T ( r , Q 1 ( z ) ) = m ( r , Q 1 ( z ) ) m ( r , 2 H ( z + η ) H ( z + 2 η ) ) + O ( 1 ) r σ 1 1 + ε 1 ,

so that σ ( Q 1 ( z ) ) σ 1 1 + ε 1 < k 1 . Thus, by deg ( h ( z + η ) h ( z ) ) = k 1 and σ ( Q 1 ) < k 1 , we can see that the order of growth of the left-hand side of (2.17) is equal to k 1 , and the order of growth of the right-hand side of (2.17) is less than k 1 . This is a contradiction.

Subcase 1.3. Suppose that n 3 . Then we can obtain from (2.16) that
j = 0 n 2 ( 1 ) j C n j H ( z + ( n j ) η ) H ( z + η ) e h ( z + ( n j ) η ) h ( z + η ) + ( 1 ) n 1 C n n 1 = 0 .
(2.19)
Set Q 2 ( z ) = e h ( z + 2 η ) h ( z + η ) . Then Q 2 ( z ) is a transcendental entire function since σ ( Q 2 ( z ) ) = k 1 1 . For j = 3 , 4 , , n , we have
e h ( z + j η ) h ( z + η ) = Q 2 ( z + ( j 2 ) η ) Q 2 ( z + ( j 3 ) η ) Q 2 ( z ) .
Thus, (2.19) can be rewritten as
U 2 ( z , Q 2 ( z ) ) Q 2 ( z ) = ( 1 ) n C n n 1 ,
(2.20)
where
U 2 ( z , Q 2 ( z ) ) = H ( z + n η ) H ( z + η ) Q 2 ( z + ( n 2 ) η ) Q 2 ( z + ( n 3 ) η ) Q 2 ( z + η ) C n 1 H ( z + ( n 1 ) η ) H ( z + η ) Q 2 ( z + ( n 3 ) η ) Q 2 ( z + ( n 4 ) η ) Q 2 ( z + η ) + + ( 1 ) n 2 C n n 2 H ( z + 2 η ) H ( z + η ) .
Noting that ( 1 ) n C n n 1 0 , we can see that U 2 ( z , Q 2 ( z ) ) 0 . Set σ ( H ) = σ 2 . Then σ 2 < k . Since Q 2 ( z ) is of regular growth and σ ( Q 2 ( z ) ) = k 1 , for any given ε 2 ( 0 < 3 ε 2 < k σ 2 ) and all r > r 0 (>0), we have
T ( r , Q 2 ( z ) ) > r k 1 ε 2 .
(2.21)
By Lemma 2.1, we see that for ε 2 , there exists a set E 2 ( 1 , ) of finite logarithmic measure such that for all z satisfying | z | = r [ 0 , 1 ] E 2 , we have
exp { r σ 2 1 + ε 2 } | H ( z + j η ) H ( z + η ) | exp { r σ 2 1 + ε 2 } ( j = 2 , 3 , , n ) .
(2.22)
Thus, from (2.21) and (2.22), we can get that for j = 2 , 3 , , n ,
m ( r , H ( z + j η ) H ( z + η ) ) T ( r , Q 2 ( z ) ) r σ 2 1 + ε 2 r k 1 ε 2 0 ( r  and  r [ 0 , 1 ] E 2 ) ,
that is,
m ( r , H ( z + j η ) H ( z + η ) ) = S ( r , Q 2 ) ( j = 2 , 3 , , n ) .
(2.23)
Noting that deg Q 2 U ( z , Q 2 ) = n 2 1 and by Lemma 2.4 and Remark 2.1, we have
T ( r , Q 2 ) = m ( r , Q 2 ) = S ( r , Q 2 ) ,

a contradiction.

Case 2. Suppose that A ( 1 ) n . Thus, for a positive integer n, there are two subcases: (1)  n = 1 ; (2) n 2 .

Subcase 2.1. Suppose that n = 1 . Thus, (2.16) can be rewritten as
H ( z + η ) H ( z ) = ( A ( 1 ) n ) e h ( z ) h ( z + η ) = ( A + 1 ) e h ( z ) h ( z + η ) .

Noting the A + 1 0 , we can use the same method as in the proof of Subcase 1.2 and deduce a contradiction.

Subcase 2.2. Suppose that n 2 . Then we can obtain from (2.16) that
j = 0 n 1 ( 1 ) j C n j H ( z + ( n j ) η ) H ( z ) e h ( z + ( n j ) η ) h ( z ) + ( 1 ) n A = 0 .
(2.24)
Set Q 3 ( z ) = e h ( z + η ) h ( z ) . Then Q 3 ( z ) is a transcendental entire function since σ ( Q 3 ( z ) ) = k 1 1 . For j = 1 , 2 , , n , we have
e h ( z + j η ) h ( z ) = Q 3 ( z + ( j 1 ) η ) Q 3 ( z + ( j 2 ) η ) Q 3 ( z ) .
Thus, (2.24) can be rewritten as
U 3 ( z , Q 3 ( z ) ) Q 3 ( z ) = A ( 1 ) n ,
(2.25)
where
U 3 ( z , Q 3 ( z ) ) = H ( z + n η ) H ( z ) Q 3 ( z + ( n 1 ) η ) Q 3 ( z + ( n 2 ) η ) Q 3 ( z + η ) C n 1 H ( z + ( n 1 ) η ) H ( z ) Q 3 ( z + ( n 2 ) η ) Q 3 ( z + ( n 3 ) η ) Q 3 ( z + η ) + + ( 1 ) n 1 C n n 1 H ( z + η ) H ( z ) .

We can see that U 3 ( z , Q 3 ( z ) ) 0 since A ( 1 ) n 0 . Noting that deg Q 3 U 3 ( z , Q 3 ( z ) ) = n 1 1 , we can use the same method as in the proof of Subcase 1.3 and deduce a contradiction.

Thus, we have proved that σ ( f ) = k = 1 . And f ( z ) can be written as
f ( z ) = H ( z ) e c 1 z + c 0 = H ( z ) e c 1 z ,
(2.26)
where c 0 , c 1 (≠0) are two constants and H ( z ) = e c 0 H ( z ) (0) is an entire function and satisfies
σ ( H ( z ) ) = λ ( H ( z ) ) = λ ( f ) < σ ( f ) = 1 .
(2.27)
Thus, by (2.26), (2.27), (2.16) and Lemma 2.8, we can get that H ( z ) is a nonzero constant, and so, f ( z ) can be written as
f ( z ) = c e c 1 z ,

where c, c 1 are two nonzero constants.

Thus, Lemma 2.9 is proved. □

Remark 2.5 From the proof of Lemma 2.9 or Remark 1.3, we can see that A 1 in Lemma 2.9 when n = 1 and Theorem C. Unfortunately, we cannot obtain A ( 1 ) n when n 2 in Lemma 2.9. This is because we can get a contradiction from the equality e c 1 η 1 = 1 , but we cannot obtain a contradiction from the equality ( e c 1 η 1 ) n = ( 1 ) n when n 2 .

3 Proof of Theorem 1.1

By the hypotheses of Theorem 1.1, we can write f ( z ) in the form (2.11), and (2.12) holds. Since Δ η n f ( z ) and f ( z ) share an entire function a ( z ) CM, then
Δ η n f ( z ) a ( z ) f ( z ) a ( z ) = j = 0 n ( 1 ) n j C n j H ( z + j η ) e h ( z + j η ) + b ( z ) H ( z ) e h ( z ) = e P ( z ) ,
(3.1)

where P ( z ) is a polynomial and b ( z ) = Δ η n a ( z ) a ( z ) . Obviously, σ ( b ( z ) ) σ ( a ( z ) ) < 1 .

First step. We prove
Δ η n f ( z ) a ( z ) f ( z ) a ( z ) = A ,
(3.2)

where A (≠0) is a constant. If P ( z ) 0 , then, by (3.1), we see that (3.2) holds and A = 1 .

Now suppose that P ( z ) 0 and deg P ( z ) = s . Set
h ( z ) = a k z k + a k 1 z k 1 + + a 0 , P ( z ) = b s z s + b s 1 z s 1 + + b 0 ,
(3.3)

where k = σ ( f ) 1 , a k ( 0 ) , a k 1 , , a 0 , b s ( 0 ) , b s 1 , , b 0 are constants. By (3.1), we can see that 0 deg P = s deg h = k .

In this case, we prove that P ( z ) is a constant, that is, s = 0 . To this end, we will deduce a contradiction for the cases s = k and 1 s < k , respectively.

Case 1. Suppose that 1 s = k . Thus, there are two subcases: (1)  a ( z ) 0 ; (2) a ( z ) 0 .

Subcase 1.1. Suppose that a ( z ) 0 . First we suppose that b k a k . Then (3.1) is rewritten as
g 11 ( z ) e P ( z ) + g 12 e h ( z ) + g 13 e h 0 ( z ) = 0 ,
(3.4)
where h 0 ( z ) 0 and
g 11 ( z ) = H ( z ) ; g 12 ( z ) = b ( z ) ; g 13 ( z ) = j = 0 n ( 1 ) n j C n j H ( z + j η ) e h ( z + j η ) h ( z ) .
Since σ ( H ) < k , σ ( b ) < 1 k and deg ( h ( z + j η ) h ( z ) ) = k 1 ( j = 1 , 2 , , n ), we can see that σ ( g 1 m ( z ) ) < k ( m = 1 , 2 , 3 ). On the other hand, by b k a k , we can see that deg ( P ( h ) ) = deg ( P h 0 ) = deg ( h h 0 ) = k . Since e P ( h ) , e P h 0 and e h h 0 are of regular growth, and σ ( g 1 m ) < k ( m = 1 , 2 , 3 ), we can see that for m = 1 , 2 , 3 ,
T ( r , g 1 m ) = o ( T ( r , e P ( h ) ) ) = o ( T ( r , e P h 0 ) ) = o ( T ( r , e h h 0 ) ) .
(3.5)

Thus, applying Lemma 2.2 to (3.4), by (3.5), we can obtain g 1 m ( z ) 0 ( m = 1 , 2 , 3 ). Clearly, this is a contradiction.

Now we suppose that b k = a k . Then (3.1) is rewritten as
[ H ( z ) e P ( z ) + h ( z ) b ( z ) ] e h ( z ) = j = 0 n ( 1 ) n j C n j H ( z + j η ) e h ( z + j η ) h ( z ) .
(3.6)
We affirm that H ( z ) e P ( z ) + h ( z ) b ( z ) 0 . In fact, if H ( z ) e P ( z ) + h ( z ) b ( z ) 0 , then, by (3.6), we can obtain
j = 1 n ( 1 ) n j C n j H ( z + j η ) e h ( z + j η ) h ( z ) + ( 1 ) n H ( z ) 0 ,
(3.7)
this is the special case of (2.14) when b ( z ) 0 and A = 0 . Hence, using the same method as in the proof of Case 2 in the proof of Lemma 2.9, we can get that σ ( f ) = k = 1 . Hence, substituting h ( z ) = c 1 z + c 0 into (3.7), we have
j = 0 n ( 1 ) j C n j e ( n j ) c 1 η H ( z + ( n j ) η ) = 0 .
(3.8)
On this occasion, we assert that ( e c 1 η 1 ) n = 0 . On the contrary, we suppose that ( e c 1 η 1 ) n 0 . Then the sum of all coefficients of (3.8) is ( e η 1 ) n , which does not vanish. By Lemma 2.5 and Remarks 2.2-2.3, we have σ ( H ) 1 , a contradiction. Hence, ( e c 1 η 1 ) n = 0 . Thus, e c 1 η = 1 . Substituting it into (3.8), we have
j = 0 n ( 1 ) j C n j H ( z + ( n j ) η ) = 0 .
(3.9)

First, we suppose that H ( z ) is transcendental. Then, noting that σ ( H ) < 1 implies lim ̲ r T ( r , H ) r = 0 , by Lemma 2.7 and Remark 2.4, we know that Δ η H ( z ) = H ( z + η ) H ( z ) is transcendental. Moreover, σ ( Δ η H ( z ) ) σ ( H ( z ) ) < 1 implies lim ̲ r T ( r , Δ η H ) r = 0 . Repeating the process above n 1 times, we can see that Δ η n H ( z ) is transcendental. That is, the left-hand side of (3.9) is a transcendental function. Hence (3.9) is impossible.

Secondly, we suppose that H ( z ) is a nonzero polynomial. Then, noting that b k = a k , we can see that e p ( z ) + h ( z ) is a nonzero constant. Thus, from b ( z ) = H ( z ) e p ( z ) + h ( z ) , we can know that b ( z ) is a nonzero polynomial. Thus, applying Lemma 2.6 to the equation Δ η n a ( z ) a ( z ) = b ( z ) and by Remark 2.3, we have σ ( a ) 1 , a contradiction. Hence, H ( z ) e P ( z ) + h ( z ) b ( z ) 0 . Thus, since deg ( P + h ) k 1 , deg ( h ) = k , deg ( h ( z + j η ) h ( z ) ) = k 1 ( j = 1 , 2 , , n ) and σ ( H ) < k , we see that the order of growth of the left-hand side of (3.6) is equal to k, and the order of growth of the right-hand side of (3.6) is less than k. This is a contradiction.

Subcase 1.2. Suppose that a ( z ) 0 . Then (3.1) is rewritten as
H ( z ) e P ( z ) = j = 0 n ( 1 ) n j C n j H ( z + j η ) e h ( z + j η ) h ( z ) .
(3.10)

Since H ( z ) 0 , σ ( H ) < k , deg P = s = k and deg ( h ( z + j η ) h ( z ) ) = k 1 ( j = 1 , 2 , , n ), we can see that the order of growth of the left-hand side of (3.10) is equal to k, and the order of growth of the right-hand side of (3.10) is less than k. This is a contradiction.

Case 2. Suppose that 1 s < k . Thus, there are two subcases: (1)  a ( z ) 0 ; (2) a ( z ) 0 .

Subcase 2.1. Suppose that a ( z ) 0 . Then, by (3.1), we can obtain
j = 0 n ( 1 ) n j C n j H ( z + j η ) e h ( z + j η ) h ( z ) H ( z ) e P ( z ) = b ( z ) e h ( z ) .
(3.11)

We assert that b ( z ) 0 . In fact, if b ( z ) 0 , then (2.15) obviously holds. Hence, using the same method as in the proof of Lemma 2.9, by Lemma 2.5 and Remarks 2.2-2.3, we can get that σ ( a ) 1 , a contradiction. Hence, b ( z ) 0 . Since deg h = k , deg ( h ( z + j η ) h ( z ) ) = k 1 ( j = 1 , 2 , , n ), deg P = s < k and σ ( H ) < k , we see that the order of growth of the left-hand side of (3.11) is less than k, and the order of growth of the right-hand side of (3.11) is equal to k. This is a contradiction.

Subcase 2.2. Suppose that a ( z ) 0 . Then, by (3.1), we obtain
j = 1 n ( 1 ) n j C n j H ( z + j η ) H ( z ) e h ( z + j η ) h ( z ) + ( 1 ) n = e P ( z ) .
(3.12)

Thus, there are two subcases: (1) n = 1 ; (2)  n 2 .

Subcase 2.2.1. Suppose that n = 1 . Then (3.12) can be rewritten as
H ( z + η ) H ( z ) e h ( z + η ) h ( z ) 1 = e P ( z ) .
(3.13)
By (3.13), we see that H ( z + η ) H ( z ) is a nonzero entire function. Set σ ( H ) = σ 4 . Then σ 4 < σ ( f ) = k . By Lemma 2.1, we see that for any given ε 4 ( 0 < 3 ε 4 < k σ 4 ), there exists a set E 4 ( 1 , ) of finite logarithmic measure such that for all z satisfying | z | = r [ 0 , 1 ] E 4 , we have
exp { r σ 4 1 + ε 4 } | H ( z + η ) H ( z ) | exp { r σ 4 1 + ε 4 } .
(3.14)
Since H ( z + η ) H ( z ) is an entire function, by (3.13), we have
T ( r , H ( z + η ) H ( z ) ) = m ( r , H ( z + η ) H ( z ) ) r σ 4 1 + ε 4 ,
so that
σ ( H ( z + η ) H ( z ) ) σ 4 1 + ε 4 < k 1 .
(3.15)

Since s < k , we can see that deg P k 1 . If deg P < k 1 , then, by (3.15) and deg ( h ( z + η ) h ( z ) ) = k 1 , we can see that the order of growth of the left-hand side of (3.13) is equal to k 1 , and the order of growth of the right-hand side of (3.13) is equal to degP which is less than k 1 . This is a contradiction.

If deg P = k 1 , then since H ( z + η ) H ( z ) is an entire function and deg ( h ( z + η ) h ( z ) ) = k 1 1 , by (3.15), we can see that the entire function H ( z + η ) H ( z ) e h ( z + η ) h ( z ) has a Borel exceptional value 0, thus the value 1 must be not its Borel exceptional value. Hence, the left-hand side of (3.13), H ( z + η ) H ( z ) e h ( z + η ) h ( z ) 1 , has infinitely many zeros, but the right-hand side of (3.13), e P ( z ) , has no zero. This is a contradiction.

Subcase 2.2.2. Now we suppose that n 2 . Thus, for s ( = deg P ), there are two subcases: (1)  s < k 1 ; (2)  s = k 1 .

Subcase 2.2.2.1. Now we suppose that s < k 1 . Set Q 5 ( z ) = e h ( z + η ) h ( z ) . Since σ ( Q 5 ) = k 1 1 , Q 5 ( z ) is a transcendental entire function. Thus, (3.12) can be rewritten as
U 5 ( z , Q 5 ( z ) ) Q 5 ( z ) = e P ( z ) ( 1 ) n ,
(3.16)
where
U 5 ( z , Q 5 ( z ) ) = H ( z + n η ) H ( z ) Q 5 ( z + ( n 1 ) η ) Q 5 ( z + ( n 2 ) η ) Q 5 ( z + η ) C n 1 H ( z + ( n 1 ) η ) H ( z ) Q 5 ( z + ( n 2 ) η ) Q 5 ( z + ( n 3 ) η ) Q 5 ( z + η ) + + ( 1 ) n 1 C n n 1 H ( z + η ) H ( z ) .
(3.17)
Thus, using the same method as in the proof of Subcase 1.3 in the proof of Lemma 2.9 and noting that σ ( e P ( z ) ( 1 ) n ) = deg P < k 1 , we have
m ( r , e P ( z ) ( 1 ) n ) = S ( r , Q 5 ) .
Noting that n 2 and so deg U 5 ( z , Q 5 ) = n 1 1 . Using the same method as in the proof of Subcase 1.3 in the proof of Lemma 2.9, we can obtain
T ( r , Q 5 ) = m ( r , Q 5 ) = S ( r , Q 5 ) .

Clearly, this is a contradiction.

Subcase 2.2.2.2. Now we suppose that s = k 1 . Thus, (3.12) is written as
j = 1 n ( 1 ) n j C n j H ( z + j η ) H ( z ) e T j ( z ) + ( 1 ) n e P ( z ) = 0 ,
(3.18)
where T j ( z ) = h ( z + j η ) h ( z ) ( j = 1 , 2 , , n ). Thus, by (3.3), we have
T j ( z ) = j k η a k z k 1 + P k 2 , j ( z ) ,
(3.19)
where P k 2 , j ( z ) is a polynomial with degree at most k 2 . Thus, we have
T j ( z ) T t ( z ) = ( j t ) k η a k z k 1 + P j , t ( z ) ( 1 j t n ) ,

where P j , t ( z ) is a polynomial with degree at most k 2 .

First, we suppose that there is some j 0 ( 1 j 0 n ) such that j 0 k η a k = b k 1 , that is, deg ( T j 0 ( z ) P ( z ) ) k 2 . Thus, (3.18) can be written as
1 j n , j j 0 ( 1 ) n j C n j H ( z + j η ) H ( z ) e h ( z + j η ) h ( z ) + B j 0 ( z ) e h ( z + j 0 η ) h ( z ) = ( 1 ) n + 1 ,
(3.20)
where
B j 0 ( z ) = ( 1 ) n j 0 C n n j 0 H ( z + j 0 η ) H ( z ) e P ( z ) + h ( z ) h ( z + j 0 η ) .
Set Q 6 ( z ) = e h ( z + η ) h ( z ) and σ ( H ) = σ 6 . Then (3.20) can be rewritten as
U 6 ( z , Q 6 ( z ) ) Q 6 ( z ) = ( 1 ) n + 1 ,
(3.21)
where
U 6 ( z , Q 6 ( z ) ) = 1 j n , j j 0 ( 1 ) n j C n n j H ( z + j η ) H ( z ) Q 6 ( z + ( j 1 ) η ) Q 6 ( z + ( j 2 ) η ) Q 6 ( z + η ) + B j 0 ( z ) Q 6 ( z + ( j 0 1 ) η ) Q 6 ( z + ( j 0 2 ) η ) Q 6 ( z + η ) ( j 0 2 ) ,
(3.22)
or
U 6 ( z , Q 6 ( z ) ) = 2 j n ( 1 ) n j C n n j H ( z + j η ) H ( z ) Q 6 ( z + ( j 1 ) η ) Q 6 ( z + ( j 2 ) η ) Q 6 ( z + η ) + B j 0 ( z ) ( j 0 = 1 ) .
(3.23)
Noting that ( 1 ) n + 1 0 , we can see that U 6 ( z , Q 6 ( z ) ) 0 . Since σ ( H ) < k and σ ( e P ( z ) + h ( z ) h ( z + j 0 η ) ) k 2 < k 1 , using the same method as in the proof of Subcase 1.3 in the proof of Lemma 2.9, we have
m ( r , B j 0 ( z ) ) = S ( r , Q 6 ) .
(3.24)
Noting that n 2 and so deg U 6 ( z , Q 6 ) = n 1 1 . Combining (3.21)-(3.24), using the same method as in the proof of Subcase 1.3 in the proof of Lemma 2.9, we can obtain
T ( r , Q 6 ) = m ( r , Q 6 ) = S ( r , Q 6 ) .

Clearly, this is a contradiction.

Secondly, we suppose that j k η a k b k 1 for any 1 j n . Thus, equation (3.18) can be rewritten as
e P ( z ) = e b k 1 z k 1 e P k 2 ( z ) = j = 0 n ( 1 ) n j C n j H ( z + j η ) H ( z ) e h ( z + j η ) h ( z ) ,
(3.25)
where P k 2 ( z ) = P ( z ) b k 1 z k 1 = b k 2 z k 2 + b k 3 z k 3 + + b 0 . For dealing with equation (3.25), we just compare | b k 1 | with n k | η a k | since n k | η a k | > ( n 1 ) k | η a k | > > k | η a k | . Without loss of generality, we suppose that n k | η a k | | b k 1 | . Let arg b k 1 = θ 1 , arg ( η a k ) = θ 2 and σ ( H ) = σ 7 < k . Take θ 0 such that cos ( ( k 1 ) θ 0 + θ 1 ) = 1 . By Lemma 2.1, we see that for any given ε 7 ( 0 < 3 ε 7 < k σ 7 ), there exists a set E 7 ( 1 , ) of finite logarithmic measure such that for all z = r e i θ 0 satisfying | z | = r [ 0 , 1 ] E 7 , we have
exp { r σ 7 1 + ε 7 } | H ( z + j η ) H ( z ) | exp { r σ 7 1 + ε 7 } ( j = 1 , , n ) .
(3.26)
Thus, noting that e P k 2 ( z ) is of regular growth, we can deduce from (3.25) and (3.26) that
| e b k 1 z k 1 | = | e P ( z ) e P k 2 ( z ) | | j = 0 n ( 1 ) j C n j H ( z + ( n j ) η ) H ( z ) e h ( z + ( n j ) η ) h ( z ) | | e b k 2 z k 2 + b k 3 z k 3 + + b 0 | ( n + 1 ) n ! exp { r σ 7 1 + ε 7 } exp { n k | η a k | cos ( ( k 1 ) θ 0 + θ 2 ) r k 1 + O ( r k 2 ) } exp { | b k 2 | 2 r k 2 } ,
that is,
exp { | b k 1 | r k 1 } exp { n k | η a k | cos ( ( k 1 ) θ 0 + θ 2 ) r k 1 + r σ 7 1 + ε 7 + O ( r k 2 ) | b k 2 | 2 r k 2 } exp { n k | η a k | cos ( ( k 1 ) θ 0 + θ 2 ) r k 1 + o ( r k 1 ) } .
(3.27)
We assert that
n k | η a k | cos ( ( k 1 ) θ 0 + θ 2 ) < | b k 1 | .

In fact, if n k | η a k | = | b k 1 | , then, by b k 1 n k η a k , we know that cos ( ( k 1 ) θ 0 + θ 2 ) 1 , that is, cos ( ( k 1 ) θ 0 + θ 2 ) < 1 , and hence n k | η a k | cos ( ( k 1 ) θ 0 + θ 2 ) < n k | η a k | = | b k 1 | . If n k | η a k | < | b k 1 | , then we have n k | η a k | cos ( ( k 1 ) θ 0 + θ 2 ) n k | η a k | < | b k 1 | .

Thus, taking a positive constant ε 8 ( 0 < ε 8 < | b k 1 | n k | η a k | cos ( ( k 1 ) θ 0 + θ 2 ) 3 ), we can deduce from (3.27) that
exp { | b k 1 | r k 1 } exp { n k | η a k | cos ( ( k 1 ) θ 0 + θ 2 ) r k 1 + o ( r k 1 ) } exp { ( | b k 1 | ε 8 ) r k 1 } ,

a contradiction. Thus, we have proved that P is only a constant and (3.2) holds.

Second step. Applying Lemma 2.9 to (3.2), we can obtain the conclusion.

Thus, Theorem 1.1 is proved.

Declarations

Acknowledgements

This research was supported by the National Natural Science Foundation of China (Nos. 11171119, 11226090) and supported by the Natural Science Foundation of Guangdong Province, China (No. S2013040014347).

Authors’ Affiliations

(1)
School of Mathematical Sciences, South China Normal University

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© Chen and Chen; licensee Springer. 2013

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