Open Access

Some results on difference polynomials sharing values

Advances in Difference Equations20122012:1

https://doi.org/10.1186/1687-1847-2012-1

Received: 9 June 2011

Accepted: 5 January 2012

Published: 5 January 2012

Abstract

This article is devoted to studying uniqueness of difference polynomials sharing values. The results improve those given by Liu and Yang and Heittokangas et al.

1 Introduction and main results

In this article, we shall assume that the reader is familiar with the fundamental results and the standard notations of the Nevanlinna theory (e.g., see [13]). In addition, we will use the notations λ(f) to denote the exponent of convergence of zero sequences of meromorphic function f(z); σ(f) to denote the order of f(z). We say that meromorphic functions f and g share a finite value a CM when f - a and g - a have the same zeros with the same multiplicities. For a non-zero constant c, the forward difference Δ c f ( z ) = f ( z + c ) - f ( z ) , Δ c n + 1 f ( z ) = Δ c n f ( z + c ) - Δ c n f ( z ) , n = 1, 2,.... In general, we use the notation C to denote the field of complex numbers.

Currently, there has been an increasing interest in studying difference equations in the complex plane. Halburd and Korhonen [4, 5] established a version of Nevanlinna theory based on difference operators. Ishizaki and Yanagihara [6] developed a version of Wiman-Valiron theory for difference equations of entire functions of small growth.

Recently, Liu and Yang [7] establish a counterpart result to the Brück conjecture [8] valid for transcendental entire function for which σ(f) < 1. The result is stated as follows.

Theorem A. Let f be a transcendental entire function such that σ(f) < 1. If f and Δ c n f share a finite value a CM, n is a positive integer, and c is a fixed constant, then
Δ c n f - a f - a = τ

for some non-zero constant τ.

Heittokangas et al. [9], prove the following result which is a shifted analogue of Brück conjecture valid for meromorphic functions.

Theorem B. Let f be a meromorphic function of order of growth 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 f ( z ) - a = τ

for some constant τ.

Here, we also study the shift analogue of Brück conjecture, and obtain the results as follows.

Theorem 1.1. Let f(z) be a non-constant entire function, σ(f) < 1 or 1 < σ(f) < 2 and λ(f) < σ(f) = σ. Set L1(f) = a n (z) f(z + n) + a n -1(z) f(z + n - 1) +... + a1(z) f(z + 1) + a0(z) f(z), where a j (z)(0 ≤ jn) are entire functions with a n (z)a0(z) 0. Suppose that if σ(f) < 1, then max{σ(a j )} = α < 1, and if 1 < σ(f) < 2, then max{σ(a j )} = α < σ - 1. If f and L1(f) share 0 CM, then
L 1 ( f ) = c f ,

where c is a non-zero constant.

Theorem 1.2. Let f(z) be a non-constant entire function, 2 < σ(f) <and λ(f) < σ(f). Set L2(f) = a n (z) f(z + n) + a n -1(z) f(z + n - 1) +... + a1(z) f(z + 1) + e z f(z), a j (z)(1 ≤ jn) are entire functions with σ(a j ) < 1 and a n (z) 0. If f and L2(f ) share 0 CM, then
L 2 ( f ) = h ( z ) f ,

where h(z) is an entire function of order no less than 1.

Theorem 1.3. Let f(z) be a non-constant entire function, σ(f) < 1 or 1 < σ(f) < 2, λ(f) < σ(f). Set L3(f) = a n (z) f(z + n) + a n- 1(z) f(z + n - 1) + ... + a1(z) f(z + 1) + a0(z) f(z), a j (z)(0 ≤ jn) are polynomials and a n (z) 0. If f and L3(f ) share a polynomial P(z) CM, then
L 3 ( f ) - p ( z ) = c ( f ( z ) - p ( z ) ) ,

where c is a non-zero constant.

Theorem 1.4. Let f(z) be a non-constant entire function, σ(f) < 1 or 1 < σ(f) < 2, λ(f) < σ(f). Set a(z) is an entire function with σ(a) < 1. If f and a(z)f(z + n) share a polynomial P(z) CM, then
a ( z ) f ( z + n ) - p ( z ) = c ( f ( z ) - p ( z ) ) ,

where c is a non-zero constant.

The method of the article is partly from [10].

2 Preliminary lemmas

Lemma 2.1. [11] Let f(z) be a meromorphic function with σ(f) = η < ∞. Then for any given ε > 0, there is a set E1 (1, +∞) that has finite logarithmic measure, such that
| f ( z ) | exp { r η + ε } ,

holds for |z| = r [0, 1] E1, r → ∞.

Applying Lemma 2.1 to 1 f , it is easy to see that for any given ε > 0, there is a set E2 (1, ∞) of finite logarithmic measure, such that
exp { - r η + } | f ( z ) | exp { r η + ε } ,

holds for |z| = r [0, 1] E2, r → ∞.

Lemma 2.2. [11] Let
Q ( z ) = b n z n + b n - 1 z n - 1 + + b 0 ,
where n is a positive integer and b n = α n e i θ n , α n > 0 , θ n [ 0 , 2 π ) . For any given ε ( 0 < ε < π 4 n ) , we introduce 2n open sectors
S j : - θ n + ( 2 j - 1 ) π 2 n + ε < θ < - θ n + ( 2 j + 1 ) π 2 n - ε ( j = 0 , 1 , , 2 n - 1 ) .
Then there exists a positive number R = R(ε) such that for |z| = r > R,
R e { Q ( z ) } > α n ( 1 - ε ) sin ( n ε ) r n
if z S j where j is even; while
R e { Q ( z ) } < - α n ( 1 - ε ) sin ( n ε ) r n

if z S j where j is odd.

Now for any given θ [0, 2π), if θ - θ n n + ( 2 j - 1 ) π 2 n , (j = 0, 1,..., 2n - 1), then we take ε sufficiently small, there is some S j , j {0, 1,...,2n - 1} such that θ S j .

Lemma 2.3. [12] Let f(z) be a meromorphic function of order σ = σ(f) < ∞, and let λ' and λ'' be, respectively, the exponent of convergence of the zeros and poles of f. Then for any given ε > 0, there exists a set E (1, ∞) of |z| = r of finite logarithmic measure, so that
2 π i n z , η + log f ( z + η ) f ( z ) = η f ( z ) f ( z ) + O ( r β + ε ) ,
or equivalently,
f ( z + η ) f ( z ) = e η f ( z ) f ( z ) + O ( r β + ε ) ,

holds for r E [0, 1], where n z,η is an integer depending on both z and η, β = max{σ - 2, 2λ - 2} if λ < 1 and β = max{σ - 2, λ - 1} if λ ≥ 1 and λ = max{λ', λ''} .

Lemma 2.4. [2] Let f(z) be an entire function of order σ, then
σ = lim sup r log ν ( r ) log r

where ν(r) be the central index of f.

Lemma 2.5. [2, 13, 14] Let f be a transcendental entire function, let 0 < δ < 1 4 and z be such that|z| = r and that
| f ( z ) | > M ( r , g ) ν ( r , g ) - 1 4 + δ
holds. Then there exists a set F R+ of finite logarithmic measure, i.e., F d t t < , such that
f ( m ) ( z ) f ( z ) = ν ( r , f ) z m ( 1 + o ( 1 ) )

holds for all m ≥ 0 and all r F.

Lemma 2.6. [10] Let f(z) be a transcendental entire function, σ(f) = σ <, and G = {ω1, ω2,..., ω n }, and a set E (1, ∞) having logarithmic measure lmE <. Then there is a positive number B ( 3 4 B 1 ) , a point range { z k = r k e i ω k } such that |f(z k )| ≥ BM(r k , f ), ω k [0, 2π), lim k →∞ ω k = ω0 [0, 2π), r k E [0, 1], r k → ∞, for any given ε > 0, we have
r k σ - ε < ν ( r k , f ) < r k σ + ε .

3 Proof of Theorem 1.1

Under the hypothesis of Theorem 1.1, see [3], it is easy to get that
L 1 ( f ) f = e Q ( z ) ,
(3.1)

where Q(z) is an entire function. If σ(f) < 1, we get Q(z) is a constant. Then Theorem 1.1 holds. Next, we suppose that 1 < σ(f) < 2 and λ(f) < σ(f) = σ. We divide this into two cases (Q(z) is a constant or a polynomial with deg Q = 1) to prove.

Case (1): Q(z) is a constant. Then Theorem 1.1 holds.

Case (2): deg Q = 1. By Lemma 2.3 and λ(f) < σ(f) = σ, for any given 0 < ε < min { σ - 1 2 , 1 - α 2 , σ - λ ( f ) 2 , σ - 1 - α 2 } , there exists a set E1 (1, ∞) of |z| = r of finite logarithmic measure, so that
f ( z + j ) f ( z ) = exp j f ( z ) f ( z ) + o ( r σ ( f ) - 1 - ε ) , j = 1 , 2 , , n
(3.2)

holds for r E1 0[1].

By Lemma 2.5, there exists a set E2 (0, ∞) of finite logarithmic measure, such that
f ( z ) f ( z ) = ( 1 + o ( 1 ) ) ν ( r , f ) z ,
(3.3)

holds for |z| = r E2 [0, 1], where z is chosen as in Lemma 2.5.

By Lemma 2.1, for any given ε > 0, there exists a set E3 (1, ∞) that has finite logarithmic measure such that
exp { - r α + ε } | a j ( z ) | exp { r α + ε } ( j = 0 , 1 , , n )
(3.4)

holds for |z| = r [0, 1] E3, r → ∞.

Set E = E1E2E3 and G = { - φ n n + ( 2 j - 1 ) π 2 n | j = 0 , 1 } { π 2 , 3 π 2 } . By Lemma 2.6, there exist a positive number B [ 3 4 , 1 ] , a point range { z k = r k e i θ k } such that |f(z k )| ≥ BM (r k , f], θ k [0, 2π), lim k →∞ θ k = θ0 [0, 2π) \ G, r k E [0, 1], r k → ∞, for any given ε > 0, as r k → ∞, we have
r k σ ( f ) - ε < ν ( r k , f ) < r k σ ( f ) + ε
(3.5)
By (3.1)-(3.3), we have that
a n exp { n ( 1 + o ( 1 ) ) ν ( r k , f ) z k } + + a 1 exp { ( 1 + o ( 1 ) ) ν ( r k , f ) z k } + a 0 = e Q ( z ) }
(3.6)
Let Q ( z ) = τ e i θ 1 z + b 0 , τ > 0, θ1 [0, 2π). By Lemma 2.4, there are two opened angles for above ε,
S j : - θ 1 + ( 2 j - 1 ) π 2 + ε < θ < - θ 1 + ( 2 j + 1 ) π 2 + ε ( j = 0 , 1 )

For the above θ0, there are two cases: (i) θ0 S0; (ii) θ0 S1.

Case (i). θ0 S1. Since S j is an opened set and lim k →∞ θ k = θ0, there is a K > 0 such that θ k S j when k > K. By Lemma 2.2, we have
R e { Q ( r k e i θ k ) } < - η r k ,
(3.7)
where η = η(1 - ε) sin(ε) > 0. By Lemma 2.2, if Rez k > ζr k (0 < ζ ≤ 1). By (3.4)-(3.7), we have
exp { r k σ ( f ) - 1 - ε - r k α + ε } a n exp n ( 1 + o ( 1 ) ) ν ( r k , f ) z k 3 a n exp n ( 1 + o ( 1 ) ) ν ( r k , f ) z k + + a 1 exp { ( 1 + o ( 1 ) ) ν ( r k , f ) z k } + a 0 = 3 e Q ( z ) 3 e - η r k ,
(3.8)

which contradicts that 0 < σ(f) - 1 - α - ε.

If Rez k < - ζr k (0 < ζ ≤ 1), By (3.4)-(3.7), we have
1 a n a 0 exp n ( 1 + o ( 1 ) ) ν ( r k , f ) z k + + a 1 a 0 exp ( 1 + o ( 1 ) ) ν ( r k , f ) z k + e Q ( z ) a 0 2 n exp - η r k σ ( f ) - 1 + ε + 2 r k α + ε + e - η r k exp { r k α + ε } ,
(3.9)

which implies that 1 < 0, r → ∞, a contradiction.

Case (ii). θ0 S0. Since S0 is an opened set and lim k →∞ θ k = θ0, there is K > 0 such that θ k S j when k > K. By Lemma 2.2, we have
R e { Q ( r k e i θ k ) } > η r k ,
(3.10)
where η = τ(1 - ε) sin(ε) > 0. By (3.4)-(3.6), (3.9), we obtain
( n + 1 ) exp { n r k σ ( f ) - 1 + ε + r k α + ε } | a n exp { n ( 1 + o ( 1 ) ) ν ( r k , f ) z k } + + a 1 exp { ( 1 + o ( 1 ) ) ν ( r k , f ) z k } + a 0 | = | e Q ( z ) | e η r k .
(3.11)

From (3.11), we get that σ(f) ≥ 2, a contradiction. Theorem 1.1 is thus proved.

4 Proof of Theorem 1.2

Under the hypothesis of Theorem 1.2, see [3], it is easy to get that
L 2 ( f ) f = e Q ( z ) ,
(4.1)

where Q(z) is an entire function. For Q(z), we discuss the following two cases.

Case (1): Q(z) is a polynomial with deg Q = n ≥ 1. Then Theorem 1.2 is proved.

Case (2): Q(z) is a constant. Using the similar reasoning as in the proof of Theorem 1.1, we get that
a n exp n ( 1 + o ( 1 ) ) ν ( r k , f ) z k + + a 1 exp ( 1 + o ( 1 ) ) ν ( r k , f ) z k + a = - e z k ,
(4.2)

where a is some non-zero constant.

If Rez k < -ηr k (0, 1]), By (3.4), (3.5), (4.2), we have
| a | a n exp n ( 1 + o ( 1 ) ) ν ( r k , f ) z k + + a 1 exp ( 1 + o ( 1 ) ) ν ( r k , f ) z k + | exp { z k } | exp { - η r k } + n exp { - η r k σ ( f ) - 1 + ε + 2 r k α + ε } ,
(4.3)

which is impossible.

If Rez k > ηr k (η (0, 1]), By (3.4), (3.5) and (4.2), we get
exp η r k σ ( f ) - 1 - ε < exp n ν ( r k , f ) z k - r k α + ε 2 a n exp n ( 1 + o ( 1 ) ) ν ( r k , f ) z k + + a 1 exp ( 1 + o ( 1 ) ) ν ( r k , f ) z k + a = 2 | - exp { z k } | 2 exp { r k } ,
(4.4)

which contradicts that σ(f) > 2. This completes the proof of Theorem 1.2.

5 Proof of Theorem 1.3

Since f and L3(f) share P CM, we get
L 3 ( f ) f = e Q ( z ) ,
(5.1)
where Q(z) is an entire function. If σ(f) < 1, we get Q(z) is a constant. Then Theorem 1.3 holds. Next, we suppose that 1 < σ(f) < 2 and λ(f) < σ(f) = σ. Set F(z) = f(z) - P(z), then σ(F) = σ(f). Substituting F(z) = f(z) - p(z) into (5.1), we obtain
a n ( z ) F ( z + n ) + a n - 1 ( z ) F ( z + n - 1 ) + + a 1 ( z ) F ( z + 1 ) F ( z ) + a 0 ( z ) + b ( z ) F ( z ) = e Q ( z ) ,
(5.2)

where b(z) = a n (z)P(z + n) + ... + a1(z)P (z + 1) + a0(z)p(z) is a polynomial. We discuss the following two cases.

Case 1. Q(z) is a complex constant. Then Theorem 1.3 holds.

Case 2. Q(z) is a polynomial with deg Q = 1. By Lemma 2.3 and λ(f) < σ(f) = σ, for any given 0 < ε < min { σ - 1 2 , 1 - α 2 , σ - λ ( f ) 2 , σ - 1 - α 2 } , there exists a set E1 (1, ∞) of |z| = r of finite logarithmic measure, so that
f ( z + j ) f ( z ) = exp { j f ( z ) f ( z ) + o ( r σ ( f ) - 1 - ε ) } , j = 1 , 2 , , n
(5.3)

holds for r E1 [0, 1].

By Lemma 2.5, there exists a set E2 (0, ∞) of finite logarithmic measure, such that
f ( z ) f ( z ) = ( 1 + o ( 1 ) ) ν ( r , f ) z ,
(5.4)

holds for |z| = r E2 [0, 1], where z is chosen as in Lemma 2.5.

Set E = E1 E2 and G = { - φ n n + ( 2 j - 1 ) π 2 n | j = 0 , 1 } { π 2 , 3 π 2 } . By Lemma 2.6, there exist a positive number B [ 3 4 , 1 ] , a point range { z k = r k e i θ k } such that | f (z k )| ≥ BM(r k , f), θ k [0, 2π), lim k →∞θ k = θ0 [0, 2π) \ G, r k E 0[1], r k → ∞, for any given ε > 0, as r k → ∞, we have
r k σ ( f ) - ε < ν ( r k , f ) < r k σ ( f ) + ε .
(5.5)
Since F is a transcendental entire function and |f(z k )| ≥ BM (r k , f), we obtain
b ( z k ) F ( z k ) 0 , ( r k ) .
(5.6)
By (5.2)-(5.6), we have that
a n exp n ( 1 + o ( 1 ) ) ν ( r k , f ) z k + + a 1 exp ( 1 + o ( 1 ) ) ν ( r k , f ) z k + a 0 + o ( 1 ) = e Q ( z ) .
(5.7)

Using similar proof as in proof of Theorem 1.1, we can get a contradiction. Hence, Theorem 1.3 holds.

6 Proof of Theorem 1.4

Using similar proof as in proof of Theorem 1.1, we can get Theorem 1.4 holds.

Author's contributions

YL completed the main part of this article, YL, XQ and HX corrected the main theorems. All authors read and approved the final manuscript.

Declarations

Acknowledgements

The authors thank the referee for his/her valuable suggestions to improve the present article. This research was partly supported by the NNSF of China (No. 11171184), the NSF of Shangdong Province, China (No. Z2008A01) and Shandong University graduate student independent innovation fund (yzc11024).

Authors’ Affiliations

(1)
Department of Mathematics, Shandong University
(2)
Department of Physics and Mathematics, Joensuu Campus, University of Eastern Finland
(3)
Department of Mathematics, Jinan University

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© Liu et al; licensee Springer. 2012

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