# Value distribution of difference and q-difference polynomials

## Abstract

In this paper, we investigate the value distribution of difference polynomial and obtain the following result, which improves a recent result of K. Liu and L.Z. Yang: Let f be a transcendental meromorphic function of finite order σ, c be a nonzero constant, and $α(z)≢0$ be a small function of f, and let

$P(z)= a n z n + a n − 1 z n − 1 +⋯+ a 1 z+ a 0$

be a polynomial with a multiple zero. If $λ(1/f)<σ$, then $P(f)f(z+c)−α(z)$ has infinitely many zeros. We also obtain a result concerning the value distribution of q-difference polynomial.

MSC:30D35, 39A05.

## 1 Introduction and main results

Throughout the paper, we assume that the reader is familiar with the standard symbols and fundamental results of Nevanlinna theory as found in [13]. A function $f(z)$ is called the meromorphic function, if it is analytic in the complex plane except at isolated poles. For any non-constant meromorphic function f, we denote by $S(r,f)$ any quantity satisfying

$lim r → ∞ S ( r , f ) T ( r , f ) =0,$

possibly outside of a set of finite linear measure in $R +$. A meromorphic function $a(z)$ is called a small function of $f(z)$ provided that $T(r,a)=S(r,f)$. As usual, we denote by $σ(f)$ the order of a meromorphic function $f(z)$, and denote by $λ(f)$ ($λ(1/f)$) the exponent of convergence of the zeros (poles) of $f(z)$.

Recently, a number of papers concerning the complex difference products and the differences analogues of Nevanlinna’s theory have been published (see [412] for example), and many excellent results have been obtained. In 2007, Laine and Yang [10] investigated the value distribution of difference products of entire functions, and obtained the following result.

Theorem A Let $f(z)$ be a transcendental entire function of finite order, and c be a non-zero complex constant. Then for $n≥2$, $f ( z ) n f(z+c)$ assumes every non-zero value $a∈C$ infinitely often.

Liu and Yang [11] improved Theorem A, and proved the next result.

Theorem B Let $f(z)$ be a transcendental entire function of finite order, and c be a non-zero complex constant. Then for $n≥2$, $f ( z ) n f(z+c)−p(z)$ has infinitely many zeros, where $p(z)≢0$ is a polynomial in z.

The purpose of this paper is to investigate the value distribution of difference polynomial $P(f)f(z+c)−α(z)$ and q-difference polynomial $P(f)f(qz)−α(z)$, where $P(z)= a n z n + a n − 1 z n − 1 +⋯+ a 1 z+ a 0$ with constant coefficients $a n (≠0), a n − 1 ,…, a 0$, and $α(z)$ is a mall function of $f(z)$.

For the sake of simplicity, we denote by $s(P)$ and $m(P)$ the number of the simple zeros and the number of multiple zeros of a polynomial

$P(z)= a n z n + a n − 1 z n − 1 +⋯+ a 1 z+ a 0$

respectively.

We obtain the following result which improves Theorem A and Theorem B.

Theorem 1.1 Let f be a transcendental meromorphic function of finite order $σ(f)=σ$, and c be a non-zero constant, and let

$P(z)= a n z n + a n − 1 z n − 1 +⋯+ a 1 z+ a 0$

be a polynomial with constant coefficients $a n (≠0), a n − 1 ,…, a 0$ and $m(P)>0$. If $λ( 1 f )<σ$, then $P(f)f(z+c)−α(z)$ has infinitely many zeros, where $α(z)≢0$ is a small function of f.

Remark 1 The result of Theorem 1.1 may be false if $α(z)≡0$, for example, $f(z)= e z 2 z$, it is obvious that $f 2 f(z+1)$ has no zeros. The following example shows that the assumption $λ( 1 f )<σ$ in Theorem 1.1 cannot be deleted. In fact, let $f(z)= 1 − e z 1 + e z$, $c=πi$, $α(z)=−1$, and $P(z)= z 2$. Then $λ( 1 f )=σ(f)=1$ and $P(f)f(z+c)−α(z)= 2 1 + e z$ has no zeros. Also, let $f(z)=i+ e z$, $c=πi$, $α(z)=1$, and $P(z)=z(z−i+1)(z−i−1)$. Then $P(f)f(z+c)−α(z)=− e 4 z$ has no zeros. This shows that the restriction in Theorem 1.1 to the multiple zero case is essential.

Considering the value distribution of q-differences polynomials, we obtain the following result.

Theorem 1.2 Let $f(z)$ be a transcendental entire function of zero order, and $α(z)∈S(r,f)$. Suppose that q is a non-zero complex constant and n is an integer. If $m(P)>0$, then $P(f)f(qz)−α(z)$ has infinitely many zeros.

## 2 Some lemmas

Lemma 2.1 [6]

Given two distinct complex constants $η 1$, $η 2$, let f be a meromorphic function of finite order σ. Then, for each $ε>0$, we have

$m ( r , f ( z + η 1 ) f ( z + η 2 ) ) =O ( r σ − 1 + ε ) .$

Lemma 2.2 [6]

Let f be a transcendental meromorphic function of finite order σ, c be a complex number. Then, for each $ε>0$, we have

$T ( r , f ( z + c ) ) =T ( r , f ( z ) ) +O ( r σ − 1 + ε ) +O(logr).$

The following lemma is a revised form of Lemma 2.4.2 in [2].

Lemma 2.3 Let $f(z)$ be a transcendental meromorphic solution of

$f n A(z,f)=B(z,f),$

where $A(z,f)$, $B(z,f)$ are differential polynomials in f and its derivatives with meromorphic coefficients, say ${ a λ ∣λ∈I}$, n be a positive integer. If the total degree of $B(z,f)$ as a polynomial in f and its derivatives is less than or equal to n, then

$m ( r , A ( z , f ) ) ≤ ∑ λ ∈ I m(r, a λ )+S(r,f).$

Lemma 2.4 [12]

Let $f(z)$ be a non-constant meromorphic function of finite order, $c∈C$. Then

$N ( r , 1 f ( z + c ) ) ≤ N ( r , 1 f ( z ) ) + S ( r , f ) , N ( r , f ( z + c ) ) ≤ N ( r , f ) + S ( r , f ) , N ¯ ( r , 1 f ( z + c ) ) ≤ N ¯ ( r , 1 f ( z ) ) + S ( r , f ) , N ¯ ( r , f ( z + c ) ) ≤ N ¯ ( r , f ) + S ( r , f ) ,$

outside of a possible exceptional set E with finite logarithmic measure.

Lemma 2.5 [4]

Let f be a non-constant zero-order meromorphic function, and $q∈C∖{0}$. Then

$m ( r , f ( q z ) f ( z ) ) =o ( T ( r , f ) )$

on a set of logarithmic density 1.

Remark 2 For the similar reason in Theorem 1.1 in [4], we can easily deduce that

$m ( r , f ( z ) f ( q z ) ) =o ( T ( r , f ) )$

also holds on a set of logarithmic density 1.

Proof

Using the identity

$ρ 2 − r 2 ρ 2 − 2 ρ r cos ( φ − θ ) + r 2 =Re ( ρ e i θ + z ρ e i θ − z ) ,$

and let Poisson-Jensen formula with $R=ρ$, we see

$log | f ( z ) f ( q z ) | = ∫ 0 2 π log | f ( ρ e i θ ) | Re ( ρ e i θ + z ρ e i θ − z − ρ e i θ + q z ρ e i θ − q z ) d θ 2 π + ∑ | a n | < ρ log | ( z − a n ) ( ρ 2 − a ¯ n q z ) ( q z − a n ) ( ρ 2 − a ¯ n z ) | − ∑ | b m | < ρ log | ( z − b m ) ( ρ 2 − b ¯ m q z ) ( q z − b m ) ( ρ 2 − b ¯ m z ) | = S 1 ′ ( z ) + S 1 ′ ( z ) − S 3 ′ ( z ) ,$

where ${ a n }$ and ${ b m }$ are the zeros and poles of f, respectively. Integration on the set $E:={φ∈[0,2π]:| f ( r e i φ ) f ( q r e i φ ) |≥1}$ gives us the proximity function,

$m ( r , f ( z ) f ( q z ) ) = ∫ E log | f ( z ) f ( q z ) | d ψ 2 π = ∫ E ( S 1 ′ ( r e i ψ ) + S 2 ′ ( r e i ψ ) − S 3 ′ ( r e i ψ ) ) d ψ 2 π ≤ ∫ 0 2 π ( | S 1 ′ ( r e i ψ ) | + | S 2 ′ ( r e i ψ ) | + | S 3 ′ ( r e i ψ ) | ) d ψ 2 π .$

Since $S i ′ =− S i$ ($i=1,2,3$) in [4], we get $| S i ′ |=| S i |$ ($i=1,2,3$).

Following the similar method in the proof of Theorem 1.1 in [4], we get the result. □

Lemma 2.6 Let f be a non-constant zero-order entire function, and $q∈C∖{0}$. Then

$T ( r , P ( f ) f ( q z ) ) =T ( r , P ( f ) f ( z ) ) +S(r,f)$

on a set of logarithmic density 1.

Proof Since f is an entire function of zero-order, we deduce from Lemma 2.5 that

$T ( r , P ( f ) f ( q z ) ) = m ( r , P ( f ) f ( q z ) ) ≤ m ( r , P ( f ) f ( z ) ) + m ( r , f ( q z ) f ( z ) ) ≤ m ( r , P ( f ) f ( z ) ) + S ( r , f ) = T ( r , P ( f ) f ( z ) ) + S ( r , f ) ,$

that is

$T ( r , P ( f ) f ( q z ) ) ≤T ( r , P ( f ) f ( z ) ) +S(r,f).$
(2.1)

On the other hand, using Remark 2, we get

$T ( r , P ( f ) f ( z ) ) = m ( r , P ( f ) f ( z ) ) ≤ m ( r , P ( f ) f ( q z ) ) + m ( r , f ( z ) f ( q z ) ) ≤ m ( r , P ( f ) f ( q z ) ) + S ( r , f ) = T ( r , P ( f ) f ( q z ) ) + S ( r , f ) ,$

that is

$T ( r , P ( f ) f ( z ) ) ≤T ( r , P ( f ) f ( q z ) ) +S(r,f).$
(2.2)

The assertion follows from (2.1) and (2.2). □

## 3 Proof of Theorem 1.1

Let $β(z)$ be the canonical products of the nonzero poles of $P(f)f(z+c)−α(z)$. Since $λ(1/f)<σ$ and $α(z)$ is a small function of $f(z)$, we know that $σ(β)=λ(β)<σ(f)$. Suppose on contrary to the assertion that $P(f)f(z+c)−α(z)$ has finitely many zeros. Then we have

$P(f)f(z+c)−α(z)=R(z) e Q ( z ) /β(z),$

where $Q(z)$ is a polynomial, and $R(z)≢0$ is a rational function. Set $H(z)=R(z)/β(z)$. Then

$σ(H)<σ(f)=σ,$
(3.1)

and

$P(f)f(z+c)−α(z)=H(z) e Q ( z ) .$
(3.2)

Differentiating (3.2) and eliminating $e Q ( z )$, we obtain

$P ′ ( f ) f ′ ( z ) f ( z + c ) H ( z ) + P ( f ) f ′ ( z + c ) H ( z ) − P ( f ) f ( z + c ) H ′ ( z ) − P ( f ) f ( z + c ) Q ′ ( z ) H ( z ) = α ′ ( z ) H ( z ) − α ( z ) H ′ ( z ) − α ( z ) Q ′ ( z ) H ( z ) .$
(3.3)

Let $α 1 , α 2 ,…, α t$ be the distinct zeros of $P(z)$. Then

$P(f)= a n ( f − α 1 ) n 1 ( f − α 2 ) n 2 ⋯ ( f − α t ) n t .$

Substituting this into (3.3), we have

$a n ∏ j = 1 t ( f − α j ) n j − 1 { ( n 1 ∏ j ≠ 1 ( f − α j ) + n 2 ∏ j ≠ 2 ( f − α j ) + ⋯ + n t ∏ j ≠ t ( f − α j ) ) × f ( z + c ) H ( z ) f ′ ( z ) + f ′ ( z + c ) H ( z ) × ∏ j = 1 t ( f − α j ) − f ( z + c ) ( H ′ ( z ) + Q ′ ( z ) H ( z ) ) ∏ j = 1 t ( f − α j ) } = α ′ ( z ) H ( z ) − α ( z ) H ′ ( z ) − α ( z ) Q ′ ( z ) H ( z ) .$

Note that $P(z)$ has at least one multiple zero, we may assume that $n 1 >1$ without loss of generality, and we have

$a n ( f − α 1 ) n 1 − 1 F(z,f)= α ′ (z)H(z)−α(z) H ′ (z)−α(z) Q ′ (z)H(z),$
(3.4)

where

$F ( z , f ) = ∏ j = 2 t ( f − α j ) n j − 1 { ( n 1 ∏ j ≠ 1 ( f − α j ) + n 2 ∏ j ≠ 2 ( f − α j ) + ⋯ + n t ∏ j ≠ t ( f − α j ) ) × f ( z + c ) H ( z ) f ′ ( z ) + f ′ ( z + c ) H ( z ) ∏ j = 1 t ( f − α j ) − f ( z + c ) ( H ′ ( z ) + Q ′ ( z ) H ( z ) ) ∏ j = 1 t ( f − α j ) } .$

Now we distinguish two cases.

Case 1. $F(z,f)≡0$. In this case, we obtain from (3.4) that

$α ′ (z)H(z)−α(z) H ′ (z)−α(z) Q ′ (z)H(z)≡0.$

Since $α(z)≢0$ and $H(z)≢0$, by integrating, we have

$α ( z ) H ( z ) =k e Q ( z ) ,$
(3.5)

where k is a non-zero constant. From (3.2) and (3.5), we have

$P(f)f(z+c)= ( 1 k + 1 ) α(z).$

By Lemma 2.2, we have

$n T ( r , f ( z ) ) = T ( r , P ( f ) ) + O ( 1 ) ≤ T ( r , f ( z + c ) ) + T ( r , α ( z ) ) + O ( 1 ) = T ( r , f ( z ) ) + O ( r σ − 1 + ε ) + S ( r , f ) .$

Since $n≥ n 1 ≥2$, and $f(z)$ is a transcendental, this is impossible.

Case 2. $F(z,f)≢0$. In this case, we set

$F ∗ ( z , f ) = F ( z , f ) f − α 1 = ∏ j = 2 t ( f − α j ) n j − 1 { ( n 1 ∏ j ≠ 1 ( f − α j ) + n 2 ∏ j ≠ 2 ( f − α j ) + ⋯ + n t ∏ j ≠ t ( f − α j ) ) × f ( z + c ) f ( z ) f ( z ) H ( z ) f ′ ( z ) f − α 1 + f ′ ( z + c ) f ( z + c ) f ( z + c ) f ( z ) f ( z ) H ( z ) ∏ j = 2 t ( f − α j ) − f ( z + c ) f ( z ) f ( z ) ( H ′ ( z ) + Q ′ ( z ) H ( z ) ) ∏ j = 2 t ( f − α j ) } .$

Since $f(z)=(f(z)− α 1 )+ α 1$ and $f ( k ) = ( f − α 1 ) ( k )$, we know that $F ∗ (z,f)$ is a differential polynomial of $f(z)− α 1$ with meromorphic coefficients, and

$a n ( f − α 1 ) n 1 F ∗ (z,f)= α ′ (z)H(z)−α(z) H ′ (z)−α(z) Q ′ (z)H(z).$
(3.6)

By Lemma 2.3, we have

$m ( r , ( f − α 1 ) k F ∗ ( z , f ) ) ≤ 3 m ( r , f ( z + c ) f ( z ) ) + m ( r , f ′ ( z + c ) f ( z + c ) ) + m ( r , f ′ ( z ) f − α 1 ) + 5 T ( r , H ) + S ( r , f )$
(3.7)

for $k=0$ and $k=1$.

Now for any given ε ($0<ε<1$), we obtain from Lemma 2.1, Lemma 2.2 and (3.1) that

$m ( r , f ( z + c ) f ( z ) ) =O ( r σ − ε ) ,T(r,H)=O ( r σ − ε ) ,$
(3.8)
$m ( r , f ′ ( z + c ) f ( z + c ) ) =O ( r σ − ε ) +S(r,f).$
(3.9)

The lemma of logarithmic derivative implies that

$m ( r , f ′ ( z ) f − α 1 ) =S(r,f).$
(3.10)

It follows from (3.7) to (3.10) that

$m ( r , F ∗ ( z , f ) ) =O ( r σ − ε ) +S(r,f),$
(3.11)
$m ( r , ( f − α 1 ) F ∗ ( z , f ) ) =O ( r σ − ε ) +S(r,f).$
(3.12)

Since $(f− α 1 ) F ∗ (z,f)=F(z,f)$, we obtain from the definition of $F(z,f)$ that

$N ( r , F ( z , f ) ) =O ( N ( r , H ( z ) ) + N ( r , f ) ) =O ( r σ − ε ) +S(r,f).$

Thus,

$T ( r , ( f − α 1 ) F ∗ ( z , f ) ) =O ( r σ − ε ) +S(r,f).$
(3.13)

Note that, a zero of $f(z)− α 1$ which is not a pole of $f(z+c)$ and $H(z)$, is a pole of $F ∗ (z,f)$ with the multiplicity at most 1, we know from (3.6), (3.1), Lemma 2.4 and $λ(1/f)<σ$ that

$( n 1 − 1 ) N ( r , 1 f ( z ) − α 1 ) ≤ N ( r , 1 α ′ ( z ) H ( z ) − α ( z ) H ′ ( z ) − α ( z ) Q ′ ( z ) H ( z ) ) + O ( N ( r , f ( z + c ) ) ) + O ( N ( r , H ) ) = O ( r σ − ε )$
(3.14)

for the positive ε sufficiently small. Hence (see the definition of $F ∗ (z,f)$),

$N ( r , F ∗ ( z , f ) ) = O ( N ( r , 1 f − α 1 ) + N ( r , f ) + N ( r , H ) ) = O ( r σ − ε ) + S ( r , f ) .$
(3.15)

It follows from (3.15) and (3.11) that

$T ( r , F ∗ ( z , f ) ) =O ( r σ − ε ) +S(r,f).$
(3.16)

Thus, we deduce from (3.16) and (3.13) that

$T ( r , f ( z ) ) = T ( r , f ( z ) − α 1 ) + O ( 1 ) = T ( r , ( f − α 1 ) F ∗ ( z , f ) F ∗ ( z , f ) ) = O ( r σ − ε ) + S ( r , f ) .$

This contradicts that f is of order σ. Theorem 1.1 is proved.

## 4 Proof of Theorem 1.2

Denote $F(z)=P(f)f(qz)$. From Lemma 2.6 and the standard Valiron-Mohon’ko theorem, we deduce

$T ( r , F ( z ) ) = T ( r , P ( f ) f ( z ) ) + S ( r , f ) = ( n + 1 ) T ( r , f ( z ) ) + S ( r , f ) .$

Since f is a entire function, then by the second main theorem and Lemma 2.5, we have

$T ( r , F ( z ) ) ≤ N ¯ ( r , F ( z ) ) + N ¯ ( r , 1 F ( z ) ) + N ¯ ( r , 1 F ( z ) − α ( z ) ) + S ( r , f ) ≤ N ¯ ( r , 1 P ( f ) ) + N ¯ ( r , 1 f ( q z ) ) + N ¯ ( r , 1 F ( z ) − α ( z ) ) + S ( r , f ) ≤ ( s ( P ) + m ( P ) ) T ( r , f ( z ) ) + T ( r , f ( q z ) ) + N ¯ ( r , 1 F ( z ) − α ( z ) ) + S ( r , f ) ≤ ( s ( P ) + m ( P ) ) T ( r , f ( z ) ) + m ( r , f ( q z ) f ( z ) ) + m ( r , f ( z ) ) + N ¯ ( r , 1 F ( z ) − α ( z ) ) + S ( r , f ) ≤ ( s ( P ) + m ( P ) + 1 ) T ( r , f ( z ) ) + N ¯ ( r , 1 F ( z ) − α ( z ) ) + S ( r , f ) ,$

that is,

$N ¯ ( r , 1 F ( z ) − α ( z ) ) ≥ ( n − s ( P ) − m ( P ) ) T ( r , f ( z ) ) +S ( r , f ( z ) ) .$

Since f is a transcendental entire function with $m(P)>0$, we deduce that $P(f)f(qz)−α(z)$ has infinitely many zeros.

## References

1. 1.

Hayman WK: Meromorphic Functions. Clarendon, Oxford; 1964.

2. 2.

Laine I: Nevanlinna Theory and Complex Differential Equations. de Gruyter, Berlin; 1993.

3. 3.

Yi HX, Yang CC: Uniqueness Theory of Meromorphic Functions. Kluwer Academic, Dordrecht; 2003.

4. 4.

Barnett DC, Halburd RG, Korhonen RJ, Morgan W: Nevanlinna theory for the q -difference operator and meromorphic solutions of q -difference equations. Proc. R. Soc. Edinb. A 2007, 137: 457–474.

5. 5.

Bergweiler W, Langley JK: Zeros of difference of meromorphic functions. Math. Proc. Camb. Philos. Soc. 2007, 142: 133–147. 10.1017/S0305004106009777

6. 6.

Chiang YM, Feng SJ:On the Nevanlinna characteristic $f(z+η)$ and difference equations in the complex plane. Ramanujan J. 2008, 16: 105–129. 10.1007/s11139-007-9101-1

7. 7.

Chiang YM, Feng SJ: On the growth of logarithmic differences, difference quotients and logarithmic derivatices of meromorphic functions. Trans. Am. Math. Soc. 2009, 361(7):3767–3791. 10.1090/S0002-9947-09-04663-7

8. 8.

Halburd RG, Korhonen RJ: Nevanlinna theory for the difference operator. Ann. Acad. Sci. Fenn. Math. 2006, 31: 463–478.

9. 9.

Halburd RG, Korhonen RJ: Difference analogue of the lemma on the logarithmic derivative with applications to difference equations. J. Math. Anal. Appl. 2006, 314: 477–487. 10.1016/j.jmaa.2005.04.010

10. 10.

Laine I, Yang CC: Value distribution of difference polynomials. Proc. Jpn. Acad., Ser. A, Math. Sci. 2007, 83: 148–151. 10.3792/pjaa.83.148

11. 11.

Liu K, Yang LZ: Value distribution of the difference operator. Arch. Math. 2009, 92: 270–278. 10.1007/s00013-009-2895-x

12. 12.

Qi XG, Yang LZ, Liu K: Uniqueness and periodicity of meromorphic functions concerning difference operator. Comput. Math. Appl. 2010, 60(6):1739–1746. 10.1016/j.camwa.2010.07.004

## Acknowledgements

This work was supported by the NSF of Shandong Province, P.R. China (No. ZR2010AM030) and the NNSF of China (No. 11171013 and No. 11041005).

## Author information

Authors

### Corresponding author

Correspondence to Lianzhong Yang.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors drafted the manuscript, read and approved the final manuscript.

## Rights and permissions

Reprints and Permissions

Li, N., Yang, L. Value distribution of difference and q-difference polynomials. Adv Differ Equ 2013, 98 (2013). https://doi.org/10.1186/1687-1847-2013-98