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A note on entire functions and their differences

Abstract

In this paper, we prove that for a transcendental entire function f(z) of finite order such that λ(fa(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)0andf(z)=c e c 1 z ,

where c, c 1 are two nonzero constants.

MSC:39A10, 30D35.

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 aC. We say that f and g share the value a CM (IM) provided that fa and ga 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 fg 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(fa),

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 cC. 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 cC. 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 zC.

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=0and 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(fa(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)0andf(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=0andf(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 η2kπi, kZ, 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+(nj)η), where C n j are the binomial coefficients. Thus, for any η2kπi, kZ, 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(fa(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)0andf(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=0andf(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(fa)<σ(f), then a(z) is not shared CM by Δf(z) and f(z). For example, if we take a(z)z, and λ(fz)<σ(f) holds, then Δf(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 η2kπi, kZ, 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 A1). 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|=rE[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 n2 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 1j<kn, g j (z) g k (z) is not constant;

  3. (iii)

    when 1jn, 1h<kn,

    T(r, f j )=o { T ( r , exp { g h g k } ) } (r,rE),

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 r1, 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 argz=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 CE,

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 degU(z,f)=n in f(z) and its shifts, and degQ(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+n1)++ P 0 (z)f(z)=0
(2.2)

satisfies σ(f)1, and f(z) assumes every nonzero value aC infinitely often and λ(fa)=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+n1)++ 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+jη) (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 A0. 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+jη)H(z)=jη H (z) ( 1 + o ( 1 ) ) as z in CE.
(2.8)

Substituting (2.8) into (2.7), we can get

η H (z)K+η H (z)Ko(1)=0as z in CE,
(2.9)

where K is a constant and satisfies

K=n e n c 1 η C n 1 (n1) 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 K0. If n=1, then K= e c 1 η 0; if n2, on the contrary, we suppose that K=0. Then, for j=0,1,,n1, noting that

C n j (nj)= 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 (nj) 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 n11. By (2.6) we have A= ( e c 1 η 1 ) n =0, which contradicts A0. Hence K0 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(fa(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,A0andf(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(fa(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 degh=k (k1), H and h satisfy

λ(H)=σ(H)=λ ( f a ( z ) ) <σ(f)=degh.
(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+(nj)η))=σ(H(z))<degh(z)=k (j=0,1,,n1), deg(h(z+(nj)η)h(z))=k1 and σ(b(z))σ(a(z))<1k, 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+nη) 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 A0. 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)=k2. 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) n3.

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 <k1. Thus, by deg(h(z+η)h(z))=k1 and σ( Q 1 )<k1, we can see that the order of growth of the left-hand side of (2.17) is equal to k1, and the order of growth of the right-hand side of (2.17) is less than k1. This is a contradiction.

Subcase 1.3. Suppose that n3. 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))=k11. 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))=k1, 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 )=n21 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) n2.

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+10, we can use the same method as in the proof of Subcase 1.2 and deduce a contradiction.

Subcase 2.2. Suppose that n2. 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))=k11. 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))=n11, 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 A1 in Lemma 2.9 when n=1 and Theorem C. Unfortunately, we cannot obtain A ( 1 ) n when n2 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 n2.

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 degP(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 0degP=sdegh=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 1s<k, respectively.

Case 1. Suppose that 1s=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+jη) e h ( z + j η ) h ( z ) .

Since σ(H)<k, σ(b)<1k and deg(h(z+jη)h(z))=k1 (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+jη) 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+jη) 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 n1 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)k1, deg(h)=k, deg(h(z+jη)h(z))=k1 (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+jη) e h ( z + j η ) h ( z ) .
(3.10)

Since H(z)0, σ(H)<k, degP=s=k and deg(h(z+jη)h(z))=k1 (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 1s<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+jη) 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 degh=k, deg(h(z+jη)h(z))=k1 (j=1,2,,n), degP=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) n2.

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 <k1.
(3.15)

Since s<k, we can see that degPk1. If degP<k1, then, by (3.15) and deg(h(z+η)h(z))=k1, we can see that the order of growth of the left-hand side of (3.13) is equal to k1, and the order of growth of the right-hand side of (3.13) is equal to degP which is less than k1. This is a contradiction.

If degP=k1, then since H ( z + η ) H ( z ) is an entire function and deg(h(z+η)h(z))=k11, 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 n2. Thus, for s (=degP), there are two subcases: (1) s<k1; (2) s=k1.

Subcase 2.2.2.1. Now we suppose that s<k1. Set Q 5 (z)= e h ( z + η ) h ( z ) . Since σ( Q 5 )=k11, 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 )=degP<k1, we have

m ( r , e P ( z ) ( 1 ) n ) =S(r, Q 5 ).

Noting that n2 and so deg U 5 (z, Q 5 )=n11. 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=k1. 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+jη)h(z) (j=1,2,,n). Thus, by (3.3), we have

T j (z)=jkη 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 k2. Thus, we have

T j (z) T t (z)=(jt)kη a k z k 1 + P j , t (z)(1jtn),

where P j , t (z) is a polynomial with degree at most k2.

First, we suppose that there is some j 0 (1 j 0 n) such that j 0 kη a k = b k 1 , that is, deg( T j 0 (z)P(z))k2. 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 ) )k2<k1, 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 n2 and so deg U 6 (z, Q 6 )=n11. 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 jkη a k b k 1 for any 1jn. 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 nk|η a k | since nk|η a k |>(n1)k|η a k |>>k|η a k |. Without loss of generality, we suppose that nk|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((k1) 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

nk|η a k |cos ( ( k 1 ) θ 0 + θ 2 ) <| b k 1 |.

In fact, if nk|η a k |=| b k 1 |, then, by b k 1 nkη a k , we know that cos((k1) θ 0 + θ 2 )1, that is, cos((k1) θ 0 + θ 2 )<1, and hence nk|η a k |cos((k1) θ 0 + θ 2 )<nk|η a k |=| b k 1 |. If nk|η a k |<| b k 1 |, then we have nk|η a k |cos((k1) θ 0 + θ 2 )nk|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.

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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).

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Correspondence to Zong-Xuan Chen.

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Keywords

  • complex difference
  • meromorphic function
  • Borel exceptional value
  • sharing value