# 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 . 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  for example), and many excellent results have been obtained. In 2007, Laine and Yang  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  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 

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 

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 .

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 

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 

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 , 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 , we get $| S i ′ |=| S i |$ ($i=1,2,3$).

Following the similar method in the proof of Theorem 1.1 in , 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.

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## 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).

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Correspondence to Lianzhong Yang.

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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 