Open Access

Uniqueness of difference operators of meromorphic functions

Journal of Inequalities and Applications20122012:48

https://doi.org/10.1186/1029-242X-2012-48

Received: 16 November 2011

Accepted: 29 February 2012

Published: 29 February 2012

Abstract

In this article, we investigate the uniqueness problems of difference operators of two meromorphic functions. Uniqueness of a meromorphic function f n and its difference operator with the same 1-points and poles is also proved.

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

Keywords

uniqueness difference operator meromorphic function

1. Introduction and main results

In this article, a meromorphic function always means meromorphic in the whole complex plane, and c always means a non-zero constant. For a meromorphic function f(z), we define its shift by f(z + c), and define its difference operators by
Δ c f ( z ) = f ( z + c ) - f ( z ) and Δ c n f ( z ) = Δ c n - 1 ( Δ c f ( z ) ) , n , n 2 .

We adopt the standard notations of the Nevanlinna theory of meromorphic functions such as T(r, f), m(r, f), N(r, f) and N ¯ ( r , f ) as explained in [13]. In addition, we use N 2 r , 1 f to denote the counting function of the zeros of f(z) such that simple zeros are counted once and multiple zeros twice.

Let f(z) and a(z) be two meromorphic functions. We say that a(z) is a small function with respect to f(z) if T(r, a) = S(r, f), where S(r, f) = o(T(r, f)), as r → ∞ outside of a possible exceptional set of finite logarithmic measure. For a small function a(z) related to f(z), we define
δ ( a , f ) = lim inf r m r , 1 f - a T ( r , f ) .

Let f(z) and g(z) be two meromorphic functions, and let a(z) be a small function with respect to f(z) and g(z). We say that f(z) and g(z) share a(z) IM, provided that f(z) - a(z) and g(z) - a(z) have the same zeros (ignoring multiplicities), and we say that f(z) and g(z) share a(z) CM, provided that f(z) - a(z) and g(z) - a(z) have the same zeros with the same multiplicities.

Recently, a number of articles including [410] have focused on value distribution in difference analogues of meromorphic functions. In particular, there has been an increasing interest in studying the uniqueness problems related to meromorphic functions and their shifts or their difference operators (see, e.g., [813]). Some of them (see, e.g., [9, 10]) dealt with the uniqueness problems on the case that shifts or difference polynomials of two entire functions share a small function. Our aim in this article is to investigate the uniqueness problems of difference operators of meromorphic functions.

To begin the statement of our results, we recall the following Theorem A (see [3]), which is an improvement of an original result of Shibazaki [14].

Theorem A ([[3], Theorem 9.16]). Let f and g be non-constant entire functions. If f' and g' share 1 CM, and δ(0, f) + δ(0, g) > 1, then fg or f'g' ≡ 1.

Qi and Liu [9] considered the case that shifts of two entire functions share a small function. They proved

Theorem B ([[9], Theorem 5]). Suppose that f and g are two entire functions of finite order, and let a and b be distinct small functions related to f and g such that δ(a) = δ(a, f) + δ(a, g) > 1. If f(z + c1) and g(z + c2) share b CM, then exactly one of the following assertions holds
  1. (i)

    f(z) ≡ g(z + c), where c = c 2 - c 1;

     
  2. (ii)

    f(z + c 1) ≡ (a - b)e h + a, g(z + c 2) ≡ (a - b)e -h+ a, where h(z) is an entire function.

     

As a difference analogue of Theorem A, we prove the following Theorem 1.1, whose proof is omitted as it is similar as the proof of Theorem 1.2.

Theorem 1.1. Let f(z) and g(z) be entire functions of finite order, and let a ( z ) ( 0 ) be a small entire function with respect to f(z) and g(z). Suppose that c1, c2 \{0} such that Δ c 1 f ( z ) Δ c 2 g ( z ) 0 . If Δ c 1 f ( z ) and Δ c 2 g ( z ) share a CM, and δ(0) = δ(0, f) + δ(0, g) > 1, then one of the following assertions holds:
  1. (i)

    Δ c 1 f ( z ) Δ c 2 g ( z ) ;

     
  2. (ii)

    Δ c 1 f ( z ) - a ( z ) e h ( z ) , Δ c 2 g ( z ) - a ( z ) e - h ( z ) , where h(z) is a polynomial.

     

We consider the case that c1 = c2 and obtain the following Theorem 1.2, which is an extension of Theorem 1.1 for this case.

Theorem 1.2. Let f(z) and g(z) be entire functions of finite order, and let a(z) and b(z) be small entire functions with respect to f(z) and g(z). Suppose that c \{0} such that Δ c ( f - b ) Δ c ( g - b ) ( a - Δ c b ) 0 . If Δ c f(z) and Δ c g(z) share a CM, and δ(b) = δ(b, f) + δ(b, g) > 1, then one of the following assertions holds:
  1. (i)

    Δ c f(z) ≡ Δ c g(z);

     
  2. (ii)

    Δ c f ≡ (Δ c b - a)e h(z)+ Δ c b, Δ c g ≡ (Δ c b - a)e -h(z)+ Δ c b, where h(z) is a polynomial.

     
Example 1. We list three examples to show that there exist entire functions satisfying the cases in Theorems 1.1 and 1.2.
  1. (1)

    Let f(z) = (z + 1)e z , g(z) = ze z , and c 1 = c 2 = 2πi. Then δ(0) = δ(0,f) + δ(0,g) > 1, and for any a , Δ c 1 f ( z ) and Δ c 2 g ( z ) share a CM, and we have Δ c 1 f ( z ) Δ c 2 g ( z ) . This example satisfies Theorems 1.1(i) and 1.2(i).

     
  2. (2)

    Let f(z) = e z , g(z) = -e(e + 1)e -z, c 1 = 2, and c 2 = 1. Then δ(0) = δ(0, f) + δ(0, g) > 1, and for a = 1 - e 2, Δ c 1 f ( z ) and Δ c 2 g ( z ) share a CM, and we have Δ c 1 f ( z ) - a e z and Δ c 2 g ( z ) - a e - z . This example satisfies Theorem 1.1(ii).

     
  3. (3)

    Let f(z) = -e z + 2, g(z) = e · e -z+ 2, and c = 1. Then δ(2) = δ(2, f) + δ(2, g) > 1, and for a = e - 1, Δ c f(z) and Δ c g(z) share a CM, and we have Δ c f(z) ≡ -ae z and Δ c g(z) = -ae -z. This example satisfies Theorem 1.2(ii).

     

Now an interesting question is whether the deficiency condition in Theorem 1.1 can be replaced by other condition or not. Considering this question, we prove the following result, which is an improvement of Theorem 1.4 in [12].

Theorem 1.3. Let c1, c2 \ {0}. Let f(z) and g(z) be entire functions of finite order ρ(f) and ρ(g), respectively. Suppose that a and b are distinct complex constants. If Δ c 1 f ( z ) and Δ c 2 g ( z ) share a CM, and Δ c 1 f ( z ) - b and Δ c 2 g ( z ) - b have at least max{[ρ(f)], [ρ(g)],1} distinct common zeros of multiplicity ≥ 2, then Δ c 1 f ( z ) Δ c 2 g ( z ) .

Example 2. We give examples (1) and (2) to show that there exist entire functions satisfying Theorem 1.3. Moreover, Example (3) shows that the condition related to common zeros cannot be omitted in Theorem 1.3.
  1. (1)

    Let f(z) = e z log 2sin2(2πz), g(z) = e z log 2sin2(2πz) + e 2πiz, and c = 1. Then Δ c f(z) ≡ Δ c g(z) ≡ e z log 2sin2(2πz). In this example, Δ c f(z) and Δ c g(z) have infinitely many distinct common zeros of multiplicity 2.

     
  2. (2)

    Let f(z) = e z log 2sin2(2πz), g ( z ) = 1 3 e z log 2 sin 2 ( 2 π z ) , and c 1 = 1, c 2 = 2. Then Δ c 1 f ( z ) Δ c 2 g ( z ) e z log 2 sin 2 ( 2 π z ) . In this example, Δ c 1 f ( z ) and Δ c 2 g ( z ) have infinitely many distinct common zeros of multiplicity 2.

     
  3. (3)

    Let f(z) = ze z - z, g(z) = ze -z- z, and c = 2πi. Then Δ c f(z) = 2πi(e z - 1) and Δ c g(z) = 2πi(1 - e z )e -zshare -2πi CM. However, max{[ρ(f)],[ρ(g)],1} = 1, while Δ c f(z) and Δ c g(z) have only simple common zeros and Δ c f ( z ) Δ c g ( z ) .

     

In 2008, Yang and Zhang [15] considered the uniqueness problems on the meromorphic function f n sharing values with its first derivative. One of their results can be stated as follows.

Theorem C ([[15], Theorem 4.3]). Let f(z) be a non-constant meromorphic function and n ≥ 12 be an integer. Let F = f n . If F and F' share 1 CM, then F = F', and f assumes the form
f ( z ) = c e 1 n z ,

where c is a non-zero constant.

To replace F' by Δ c F in Theorem C, we prove the following Theorem 1.4.

Theorem 1.4. Let f(z) be a non-constant meromorphic function of finite order and n ≥9 be an integer. Let F(z) = f(z) n . If F(z) and Δ c F share 1, ∞ CM, then F(z) = Δ c F.

For the entire functions case, using the same method as in the proof of Theorem 1.4, we get the following result.

Theorem 1.5. Let f(z) be a non-constant entire function of finite order and n ≥ 6 be an integer. Let F(z) = f(z) n . If F(z) and Δ c F share 1 CM, then F(z) = Δ c F.

Example 3. The following example (1) satisfies Theorem 1.4, while example (2) satisfies both Theorems 1.4 and 1.5.
  1. (1)

    Let f ( z ) = e z 9 log 2 / sin ( 2 π z ) , n = 9 and c = 1. Then we see that F(z) = f(z) n = e z log 2 / sin9(2πz) and Δ c F(z) = F(z + c) - F(z) = e z log 2/sin9(2πz) share 1, ∞ CM, and F(z) ≡ Δ c F.

     
  2. (2)

    Let f ( z ) = 2 1 9 e z , n = 9 and c = 1 9 log 2 . Then we see that F(z) = f(z) n = 2e 9zand Δ c F(z) = F(z + c) - F(z) = 2e 9zshare 1 CM, and F(z) ≡ Δ c F.

     

But we still wonder whether the lower bound of n in our results is sharp or not.

2. Proof of Theorem 1.2

Lemma 2.1 ([[7], Lemma 2.3]). Let c , n , and let f(z) be a meromorphic function of finite order. Then for any small periodic function a(z) with period c, with respect to f(z),
m r , Δ c n f f - a = S ( r , f ) ,

where the exceptional set associated with S(r, f) is of at most finite logarithmic measure.

The proof of Theorem 1.2 is based on a result in [15], which can be read as follows:

Lemma 2.2 ([[15], Theorem 3.1]). Let f j (z)(j = 1,2,3) be meromorphic functions that satisfy
j = 1 3 f j ( z ) 1 .
If f1(z) is not a constant, and
j = 1 3 N 2 r , 1 f j + j = 1 3 N ¯ ( r , f j ) < ( λ + o ( 1 ) ) T ( r ) , r I ,

where 0 ≤ λ < 1, T(r) = max1≤j≤3{T(r, f j )}, and I has infinite linear measure, then either f2(z) ≡ 1 or f3(z) ≡ 1.

Proof of Theorem 1.2. By the condition that δ(b) = δ(b, f) + δ (b, g) > 1, we see easily that δ(b, f) > 0 and δ(b, g) > 0. Then for any given ε such that 0 < ε < min δ ( b , f ) 2 , δ ( b , g ) 2 , δ ( b ) - 1 2 , we have
( δ ( b , f ) - ε ) T ( r , f ) m r , 1 f - b ,
(2.1)
and
( δ ( b , g ) - ε ) T ( r , g ) m r , 1 g - b .
(2.2)
From Lemma 2.1, we see that
m ( r , Δ c f ) m r , Δ c f f + m ( r , f ) = m ( r , f ) + S ( r , f ) .
(2.3)
Thus,
T ( r , Δ c f ) = m ( r , Δ c f ) m ( r , f ) + S ( r , f ) = T ( r , f ) + S ( r , f ) .
(2.4)
On the other hand, by Lemma 2.1, we have
m r , 1 f - b = m r , Δ c ( f - b ) f - b 1 Δ c f - Δ c b m r , 1 Δ c f - Δ c b + S ( r , f ) .
(2.5)
Combining (2.1) and (2.5) gives
( δ ( b , f ) - ε ) T ( r , f ) m r , 1 f - b m r , 1 Δ c f - Δ c b + S ( r , f ) T ( r , Δ c f - Δ c b ) + S ( r , f ) T ( r , Δ c f ) + S ( r , f ) .
(2.6)

Hence, by (2.4) and (2.6), we have S(r, Δ c f) = S(r, f). Similarly, S(r, Δ c g) = S(r, g).

From (2.1), (2.4) and (2.5), we have
( δ ( b , f ) - ε ) T ( r , Δ c f ) ( δ ( b , f ) - ε ) T ( r , f ) + S ( r , f ) m r , 1 f - b + S ( r , f ) m r , 1 Δ c f - Δ c b + S ( r , f ) T ( r , Δ c f ) - N r , 1 Δ c f - Δ c b + S ( r , f ) .
That is
N r , 1 Δ c f - Δ c b ( 1 - δ ( b , f ) + ε ) T ( r , Δ c f ) + S ( r , f ) .
(2.7)
By the same reasoning, we have
N r , 1 Δ c g - Δ c b ( 1 - δ ( b , g ) + ε ) T ( r , Δ c g ) + S ( r , g ) .
(2.8)

Set I1 = {r: T(r, Δ c f) ≥ T(r, Δ c g)} (0, ∞), and I2 = (0, ∞) \I1. Then there exists at least one of I i (i = 1, 2), which has infinite logarithmic measure. Without loss of generality, we may suppose I1 has infinite logarithmic measure.

Since Δ c f(z) and Δ c g(z) share a CM, we have
Δ c f ( z ) - a Δ c g ( z ) - a = e h ( z ) ,
(2.9)

where h(z) is a polynomial.

We rewrite (2.9) as
F 1 ( z ) + F 2 ( z ) + F 3 ( z ) 1 ,
where
F 1 ( z ) = Δ c f - Δ c b a - Δ c b , F 2 ( z ) = - Δ c g - Δ c b a - Δ c b e h ( z ) , F 3 ( z ) = e h ( z ) .
Set T(r) = max1≤j≤3{T(r, F j )} and S(r) = o(T(r)). Then
T ( r ) T ( r , F 1 ) = T ( r , Δ c f ) + S ( r , f ) .
We deduce from the definition of F j (z) (j = 1, 2, 3) that
N ¯ ( r , F 1 ) = N ¯ ( r , F 2 ) = N ¯ r , 1 a - Δ c b = S ( r , f ) , N ¯ ( r , F 3 ) = 0 , N 2 r , 1 F 3 = 0 .
(2.10)
From (2.7) and (2.8), we have
N 2 r , 1 F 1 + N 2 r , 1 F 2 N r , 1 Δ c f - Δ c b + N r , 1 Δ c g - Δ c b ( 2 - δ ( b ) + 2 ε ) T ( r , Δ c f ) + S ( r , Δ c f ) , r I 1 ,
(2.11)

where I1 has infinite logarithmic measure.

Then we get from (2.10) and (2.11) that
j = 1 3 N 2 r , 1 F j + j = 1 3 N ¯ ( r , F j ) ( 2 - δ ( b ) + 2 ε ) T ( r , Δ c f ) + S ( r , f ) ( 2 - δ ( b ) + 2 ε ) T ( r ) + S ( r ) , r I 1 ,

where I1 has infinite logarithmic measure, and also infinite linear measure.

It is obvious that F1 is not a constant. By Lemma 2.2, we have F2(z) ≡ 1 or F3(z) ≡ 1.

If F2(z) = 1, we have
Δ c g ( Δ c b - a ) e - h ( z ) + Δ c b , Δ c f ( Δ c b - a ) e h ( z ) + Δ c b .

If F3(z) ≡ 1, we have eh(z)≡ 1. By (2.9), the conclusion holds.

3. Proof of Theorem 1.3

Lemma 3.1 ([[6], Theorem 2.1]). Let f(z) be a meromorphic function of finite order ρ and let c be a non-zero complex constant. Then, for each ε > 0,
T ( r , f ( z + c ) ) = T ( r , f ) + O ( r ρ - 1 + ε ) + O ( log r ) .
Proof of Theorem 1.3. Since Δ c 1 f ( z ) and Δ c 2 g ( z ) share a CM, we have
Δ c 1 f ( z ) - a Δ c 2 g ( z ) - a = e h ( z ) ,
(3.1)

where h(z) is a polynomial.

Then by (3.1) and Lemma 3.1, we deduce that
T ( r , e h ( z ) ) = T r , Δ c 1 f ( z ) - a Δ c 2 g ( z ) - a T ( r , Δ c 1 f ( z ) ) + T ( r , Δ c 2 g ( z ) ) + O ( 1 ) T ( r , f ( z + c 1 ) ) + T ( r , f ( z ) ) + T ( r , g ( z + c 2 ) ) + T ( r , g ( z ) ) + O ( 1 ) 2 T ( r , f ) + 2 T ( r , g ) + S ( r , f ) + S ( r , g ) .
(3.2)

It follows from (3.2) that ρ(eh(z)) ≤ max{ρ(f), ρ(g)}. Clearly, ρ(eh(z)) = deg h(z) is an integer. Thus, we deduce that deg h(z) ≤ max{[ρ(f)], [ρ(g)]}.

Suppose that h(z) is a constant. Notice that Δ c 1 f ( z ) - b and Δ c 2 g ( z ) - b have at least a common zero. Suppose that a point z0 satisfies Δ c 1 f ( z 0 ) = Δ c 2 g ( z 0 ) = b . It follows from (3.1) that eh(z)e h ( z 0)= 1. Thus, Δ c 1 f ( z ) Δ c 2 g ( z ) .

Suppose that h(z) is not a constant. By taking the derivative in (3.1), we get
Δ c 1 f ( z ) - Δ c 2 g ( z ) e h ( z ) - h ( z ) ( Δ c 2 g ( z ) - a ) e h ( z ) 0 .
(3.3)
Set k = max{[ρ(f)],[ρ(g)],1}. Then k ≥ max{[ρ(f)],[ρ(g)]} ≥ deg h(z). Notice that Δ c 1 f ( z ) - b and Δ c 2 g ( z ) - b have at least k distinct common zeros of multiplicity ≥ 2. Suppose that z j (j = 1,2,..., k) satisfies
Δ c 1 f ( z j ) = Δ c 2 g ( z j ) = b , Δ c 1 f ( z j ) = Δ c 2 g ( z j ) = 0 .
(3.4)

Then we get from (3.3) and (3.4) that h'(z j ) = 0, j = 1,2,..., k. It implies that h(z) is a polynomial with deg h(z) ≥ k + 1, which is a contradiction since k ≥ deg h(z).

4. Proof of Theorem 1.4

Since f(z) is a non-constant meromorphic function of finite order, then F(z) = f(z) n is a non-constant meromorphic function of finite order. By Lemma 3.1, we have
T ( r , F ( z + c ) ) = n T ( r , f ( z + c ) ) = n T ( r , f ) + S ( r , f ) .
(4.1)

From (4.1), we see that F(z + c) and Δ c F are meromorphic functions of finite order, and obviously S(r, F(z)) = S(r, F(z + c)) = S(r, f).

Set ω = e 2 π i n . Then by the second main theorem, we have
m r , 1 F ( z ) - 1 = m r , 1 ( f ( z ) - ω 0 ) ( f ( z ) - ω 1 ) ( f ( z ) - ω n - 1 ) j = 0 n - 1 m r , 1 f ( z ) - ω j 2 T ( r , f ) - m ( r , f ) - 2 N ( r , f ) - N ( r , f ) + N r , 1 f + S ( r , f ) T ( r , f ) + N ( r , f ) - N ( r , f ) - N r , 1 f + S ( r , f ) T ( r , f ) + N ¯ ( r , f ) - N r , 1 f + S ( r , f ) 2 T ( r , f ) + S ( r , f ) .
(4.2)
Since F(z) and Δ c F share 1, ∞ CM, we have
F ( z ) - 1 Δ c F - 1 = e h ( z ) ,
(4.3)

where h(z) is a polynomial.

By (4.2) and (4.3), we get
T ( r , e h ( z ) ) = T ( r , e - h ( z ) ) + O ( 1 ) = m ( r , e - h ( z ) ) + O ( 1 ) = m r , Δ c F - 1 F ( z ) - 1 + O ( 1 ) m r , Δ c F F ( z ) - 1 + m r , 1 F ( z ) - 1 + O ( 1 ) S ( r , F ( z ) ) + 2 T ( r , f ) + S ( r , f ) = 2 T ( r , f ) + S ( r , f ) .
(4.4)

Thus, S(r, eh(z)) = o(T(r, f)).

Now we rewrite (4.3) as
F ( z + c ) - ( e h ( z ) + 1 ) e - h ( z ) F ( z ) + e - h ( z ) 1 .
(4.5)

Set F1(z) = F(z + c), F2(z) = -(eh(z)+ 1)e-h(z)F(z) and F3(z) = e-h(z).

Then
F 1 ( z ) + F 2 ( z ) + F 3 ( z ) 1
and
T ( r ) = max 1 j 3 { T ( r , F j ) } T ( r , F ( z + c ) ) = n T ( r , f ) + S ( r , f ) , S ( r ) = o ( T ( r ) ) .
We easily get
N ¯ ( r , F 2 ) = N ¯ ( r , f ) T ( r , f ) + S ( r , f ) , N ¯ ( r , F 3 ) = 0 , N 2 r , 1 F 3 = 0 .
(4.6)
From Lemma 3.1, we get
N ¯ ( r , F 1 ) = N ¯ ( r , f ( z + c ) ) T ( r , f ( z + c ) ) + S ( r , f ( z + c ) ) = T ( r , f ) + S ( r , f ) ,
(4.7)
and
N 2 r , 1 F 1 = 2 N ¯ r , 1 F 1 = 2 N ¯ r , 1 f ( z + c ) 2 T r , 1 f ( z + c ) = 2 T ( r , f ) + S ( r , f ) .
(4.8)
By (4.4), we see that
N 2 r , 1 F 2 2 N ¯ r , 1 F + N r , 1 e h ( z ) + 1 2 N ¯ r , 1 f + T ( r , e h ( z ) ) + S ( r , e h ( z ) ) + O ( 1 ) 4 T ( r , f ) + S ( r , f ) .
(4.9)
By (4.6)-(4.9), we obtain
j = 1 3 N 2 r , 1 F j + j = 1 3 N ¯ ( r , F j ) 8 T ( r , f ) + S ( r , f ) 8 n T ( r ) + S ( r ) .

Noting that n ≥ 9, we get from Lemma 2.2 that F2(z) ≡ 1 or F3(z) ≡ 1.

If F3(z) ≡ 1, we have eh(z)≡ 1. By (4.3), the conclusion holds.

If F2(z) ≡ 1, we have F ( z ) - e h ( z ) e h ( z ) + 1 . Then by (44), we see that
n T ( r , f ) = T ( r , F ) = T r , - e h ( z ) e h ( z ) + 1 = T ( r , e h ( z ) ) + O ( 1 ) 2 T ( r , f ) + S ( r , f ) ,

which is a contradiction since n ≥ 9.

Declarations

Acknowledgements

This study was supported by the National Natural Science Foundation of China (No. 11171119). The authors would like to thank referees for their valuable suggestions which greatly improve the article.

Authors’ Affiliations

(1)
School of Mathematical Sciences, South China Normal University
(2)
College of Science, Guangdong Ocean University

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© Chen et al.; licensee Springer. 2012

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