Some results on difference polynomials sharing values

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Research

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.

Keywords

Difference Equation Entire Function Meromorphic Function Open Sector Central Index 

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 [1, 2, 3]). 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 ) Open image in new window, Δ c n + 1 f ( z ) = Δ c n f ( z + c ) - Δ c n f ( z ) Open image in new window, 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 Open image in new window 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 Open image in new window

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 cC. If f(z) and f(z + c) share the values aC andCM, then
f ( z + c ) - a f ( z ) - a = τ Open image in new window

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 , Open image in new window

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 , Open image in new window

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 ) ) , Open image in new window

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 ) ) , Open image in new window

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 } , Open image in new window

holds for |z| = r ∉ [0, 1] ∪ E1, r → ∞.

Applying Lemma 2.1 to 1 f Open image in new window, 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 } , Open image in new window

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 , Open image in new window
where n is a positive integer and b n = α n e i θ n , α n > 0 , θ n [ 0 , 2 π ) Open image in new window. For any given ε ( 0 < ε < π 4 n ) Open image in new window, 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 ) . Open image in new window
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 Open image in new window
if zS j where j is even; while
R e { Q ( z ) } < - α n ( 1 - ε ) sin ( n ε ) r n Open image in new window

if zS j where j is odd.

Now for any given θ ∈ [0, 2π), if θ - θ n n + ( 2 j - 1 ) π 2 n Open image in new window, (j = 0, 1,..., 2n - 1), then we take ε sufficiently small, there is some S j , j ∈ {0, 1,...,2n - 1} such that θS 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 ) , Open image in new window
or equivalently,
f ( z + η ) f ( z ) = e η f ( z ) f ( z ) + O ( r β + ε ) , Open image in new window

holds for rE ∪ [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 Open image in new window

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 Open image in new window and z be such that|z| = r and that
| f ( z ) | > M ( r , g ) ν ( r , g ) - 1 4 + δ Open image in new window
holds. Then there exists a set FR+ of finite logarithmic measure, i.e., F d t t < Open image in new window, such that
f ( m ) ( z ) f ( z ) = ν ( r , f ) z m ( 1 + o ( 1 ) ) Open image in new window

holds for all m ≥ 0 and all rF.

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 ) Open image in new window, a point range { z k = r k e i ω k } Open image in new window 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 . Open image in new window

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 ) , Open image in new window
(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 } Open image in new window, 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 Open image in new window
(3.2)

holds for rE1 ∪ 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 , Open image in new window
(3.3)

holds for |z| = rE2 ∪ [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 ) Open image in new window
(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 } Open image in new window. By Lemma 2.6, there exist a positive number B [ 3 4 , 1 ] Open image in new window, a point range { z k = r k e i θ k } Open image in new window 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 Open image in new window
(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 ) } Open image in new window
(3.6)
Let Q ( z ) = τ e i θ 1 z + b 0 Open image in new window, 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 ) Open image in new window

For the above θ0, there are two cases: (i) θ0S0; (ii) θ0S1.

Case (i). θ0S1. 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 , Open image in new window
(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 , Open image in new window
(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 } , Open image in new window
(3.9)

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

Case (ii). θ0S0. 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 , Open image in new window
(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 . Open image in new window
(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 ) , Open image in new window
(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 , Open image in new window
(4.2)

where a is some non-zero constant.

If Rez k < -ηr k (η ∈ (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 } , Open image in new window
(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 } , Open image in new window
(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 ) , Open image in new window
(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 ) , Open image in new window
(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 } Open image in new window, 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 Open image in new window
(5.3)

holds for rE1 ∪ [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 , Open image in new window
(5.4)

holds for |z| = rE2 ∪ [0, 1], where z is chosen as in Lemma 2.5.

Set E = E1E2 and G = { - φ n n + ( 2 j - 1 ) π 2 n | j = 0 , 1 } { π 2 , 3 π 2 } Open image in new window. By Lemma 2.6, there exist a positive number B [ 3 4 , 1 ] Open image in new window, a point range { z k = r k e i θ k } Open image in new window 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 . Open image in new window
(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 ) . Open image in new window
(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 ) . Open image in new window
(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.

Notes

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).

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© Liu et al; licensee Springer. 2012

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Authors and Affiliations

  1. 1.Department of MathematicsShandong UniversityJinanP. R. China
  2. 2.Department of Physics and Mathematics, Joensuu CampusUniversity of Eastern FinlandJoensuuFinland
  3. 3.Department of MathematicsJinan UniversityJinanP. R. China

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