Uniqueness of difference polynomials

1.   Introduction

Let $f(z)$ be a function meromorphic in the complex plane $\Bbb{C}$. We assume that the reader is familiar with the general conclussion of the Nevanlinna theory (see ^[1,2,3]). The order of $f(z)$ is denoted by $\sigma(f)$. For any $a\in \Bbb{C}$, the exponent of convergence of zeros of $f(z)-a$ is denoted by $\lambda(f, a)$. Especially, we denote $\lambda(f, 0)$ by $\lambda(f)$. Suppose that $f(z)$ is a transcendental meromorphic function of order $\sigma(f)$. If $\lambda(f, a) < \sigma(f)$, then $a$ is said to be a Borel exceptional value of $f(z)$.

Recently, some well-known facts of the Nevanlinna theory of meromorphic function and their applications were extended for the differences of meromorhic functions (see ^{[4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23]}).

For any $c\in \Bbb{C}\backslash\{0\}$ and $n\in\Bbb{N}^{+}$, we define a difference polynomial in $f(z)$ as follows (see ^[19])

where $a_{k} (k = 0, 1, 2, \cdots, n)$ are not all zero complex numbers. Following ^[4], we denote the forward difference of $f$ by $\Delta_{c}^{n} f(z)$. i.e.

Observe that

and

where $C_{n}^{k}(k = 0, 1, 2, \cdots, n)$ are the binomial coefficients. If $a_{k} = C_{n}^{k}(-1)^{n-k} (k = 0, 1, 2, \cdots, n)$ in $P(z, f)$, then $P(z, f) = \triangle_{c}^{n}f.$ Therefore, $P(z, f)$ is a more general difference polynomial than $\triangle_{c}^{n}f.$ Noting that for $\triangle_{c}^{n}f$, $\sum\limits_{k = 0}^{n}a_{k} = \sum\limits_{k = 0}^{n}(-1)^{n-k}C_{n}^{k} = 0,$ we assume that $\sum\limits_{k = 0}^{n}a_{k} = 0$ for some $a_{k}$ of $P(z, f)$ in this paper (see ^[19]). The main purpose of this paper is to study uniqueness of the difference polynomial $P(z, f)$.

Let $a\in\Bbb{C}$, $f(z)$ and $g(z)$ be two nonconstant meromorphic functions in the complex plane. If $f-a$ and $g-a$ have the same zeros counting multiplicities, then we say $f(z)$ and $g(z)$ share the value $a$ CM. We say that $f(z)$ and $g(z)$ share the value $\infty$ CM if $f(z)$ and $g(z)$ have the same poles counting multiplicities (see ^[24]). For the uniqueness of entire function $f(z)$ and its difference operator $\triangle_{c}f,$ Chen and Yi ^[15,16] had proved the following theorems.

Theorem A. ^[15] Let $f(z)$ be a transcendental entire function of finite order that is of a finite Borel exceptional value $\beta$, and let $c$ be a constant such that $f(z+c)\not\equiv f(z).$ If $\Delta_{c} f(z)$ and $f(z)$ share $a (a\neq\beta)$ CM, then,

Theorem B. ^[16] Let $f(z)$ be a transcendental entire function of finite order that is of a finite Borel exceptional value $\beta$, and let $c$ be a constant such that $f(z+c)\not\equiv f(z).$ If $\Delta_{c} f(z)$ and $f(z)$ share $\beta$ CM, then $\beta = 0$ and

for some constant $k$.

In this paper, the results on the uniqueness of entire function $f(z)$ and its difference operator $\triangle_{c}f$ established in theorems A and B are extended to meromorphic function $f(z)$ and $P(z, f)$ by using the similar method as that in ^[15,16].

Theorem 1.1. Let $f$ be a transcendental meromorphic function of finite order. Suppose that $\beta\in\Bbb{C}$ and $\infty$ are Borel exceptional values of $f$, $P(z, f)$ is defined as that in $(1.1)$ and $P(z, f)\not\equiv0$. If $\beta\neq0,$ then $P(z, f)$ and $f$ can not share the value $\beta$ CM.

Under the conditions of Theorem 1.1, there are only two possible scenarios. The first case is $P(z, f)$ and $f$ share the value $a\neq\beta$ CM for any $\beta\in\Bbb{C}$, and the second case is $\beta = 0$, $P(z, f)$ and $f$ share the value $0$ CM. For the first case, we shall prove the following Theorem.

Theorem 1.2. Let $f$ be a transcendental meromorphic function of finite order. Suppose that $\beta\in\Bbb{C}$ and $\infty$ are Borel exceptional values of $f$, $P(z, f)$ is defined as that in $(1.1)$ and $P(z, f)\not\equiv0$. If $P(z, f)$ and $f$ share the value $a\neq\beta$ CM. Then

Example 1.3. Let $f(z) = e^{z}, c = \log3$, $P(z, f) = f(z+2c)-\frac{7}{2}f(z+c)+\frac{5}{2}f(z).$ Then $P(z, f)$ and $f(z)$ share the value $2$ CM and they satisfy

where $1$ satisfies $\frac{a}{a-\beta}$, $a = 2, \beta = 0$.

Corollary 1.4. Let $f$ be a transcendental meromorphic function of finite order. Suppose that $\beta\in\Bbb{C}$ and $\infty$ are Borel exceptional values of $f$, $c\in\Bbb{C}$ is non-null and $\Delta_{c}^{n}f\not\equiv0$ and $n\in\Bbb{N}^{+}$. If $\Delta_{c}^{n}f$ and $f$ share the value $a\neq\beta$ CM. Then

For the second case, we shall prove the following Theorem.

Theorem 1.5. Let $f$ be a transcendental meromorphic function of order $\sigma(f) < 2$. $P(z, f)$ is defined as that in $(1.1)$ and $P(z, f)\not\equiv0$. If $P(z, f)$ and $f$ share the value $0$ CM. Then

where $\eta$ is a constant.

2.   Proof of Theorems

Lemma 2.1. ^[24] Suppose that $f_{1}(z), f_{2}(z), \cdots, f_{n}(z) (n\geq 2)$ are meromorphic functions and $g_{1}(z), g_{2}(z), \cdots, g_{n}(z)$ are entire functions satisfying the following conditions.

$(i) \; \sum\limits_{j = 1}^{n}f_{j}(z)e^{g_{j}(z)}\equiv0.$

$(ii) \; g_{j}(z)-g_{k}(z)$ are not constants for $1\leq j < k\leq n.$

$(iii)$ For $1\leq j\leq n, 1\leq h < k\leq n,$

where $E\subset (i, +\infty)$ is of finite linear measure or finite logarithmic measure. Then $f_{j}(z)\equiv0 (j = 1, 2, \cdots, n)$.

Lemma 2.2. Let $f$ be a transcendental meromorphic function of finite order. Suppose that $\beta\in\Bbb{C}$ and $\infty$ are Borel exceptional values of $f$, then

where $P(z)$ is a polynomial and $A(z)$ is a meromorphic function such that $\lambda(A) = \lambda(\beta, f), \lambda(\frac{1}{A}) = \lambda(\frac{1}{f})$ and

Proof. Given that $\beta$ is a Borel exceptional value of $f$, $f(z)$ can be written as

where $k\in \Bbb{Z}$, $H_{1}(z)$ and $H_{2}(z)$ are the canonical products of $f$ formed with the non-null zeros and poles of $f$, and $P(z)$ is a polynomial with $\sigma(f) = \deg P(z).$

Put

Since $\beta$ and $\infty$ are Borel exceptional values of $f$, by the Theorem 2.3 in ^[24], we have

and

Lemma 2.3. ^[17] Let $A_{0}(z), A_{1}(z), \cdots, A_{n}(z)$ be entire functions of finite order so that among those having the maximal order $\sigma: = \max\{\sigma(A_{k}(z)), 0\leq k\leq n\}$, exactly one has its type strictly greater than the others. Then for any meromorphic solution of

we have $\sigma(f)\geq \sigma+1$.

2.1.   Proof of Theorem 1.1

Suppose that $P(z, f)$ and $f(z)$ share the value $\beta$ CM, then

where $h(z)$ is a polynomial. Since $\beta$ and $\infty$ are Borel exceptional values of $f$, then by Lemma 2.2, $f(z)$ can be written as

where $A(z)$ is a meromorphic function such that

It follows from (2.1) and (2.2) that

As $\sum\limits_{i = 0}^{n}a_{i} = 0$, we get

Next, according to (2.3) and (2.4), we infer that

As $\sigma(A) < \deg P(z)$ and $\deg(P(z+ic)-P(z))\leq (\deg P(z))-1 = \sigma(f)-1, i = 0, 1, 2, \cdots, n$, then $\sum\limits_{i = 0}^{n}a_{i}\frac{A(z+ic)}{A(z)}e^{P(z+ic)-P(z)}$ is a small meromorphic function respective to $\frac{\beta}{A(z)} e^{-P(z)}.$ Applying the second fundamental theorem to $\frac{\beta}{A(z)} e^{-P(z)},$ we know that

This contradicts with $e^{h(z)}\neq0$. Thus, $P(z, f)$ and $f$ can not share the value $\beta$ CM.

2.2.   Proof of Theorem 1.2

By the conditions, we can get $a\neq0$. If $a = 0,$ then $\beta\neq0$. Since $\beta$ and $\infty$ are Borel exceptional values of $f$, then by Lemma 2.2, $f(z)$ can be written as

where $P(z)$ is a polynomial and $A(z)$ is a meromorphic function such that

Since $P(z, f)$ and $f(z)$ share the value $0$ CM, we have

where $h(z)$ is a polynomial.

It follows from (2.6) and (2.7) that

Since $\sum\limits_{i = 0}^{n}a_{i} = 0$, there is

In view of (2.8) and (2.9), it follows that

As $\deg(P(z+ic)-P(z))\leq (\deg P(z))-1 = \sigma(f)-1, i = 0, 1, 2, \cdots, n$, we see that

As $\beta\neq0$ and $\sigma(A) < \sigma(f)$, applying the second fundamental theorem to $\beta e^{-P(z)},$ we have

From (2.10)–(2.12), we can get a contradiction. Thus, $a\neq0.$ Therefore,

where $q(z)$ is a polynomial with $\deg q(z)\leq\sigma(f).$ Since $\sum\limits_{i = 0}^{n}a_{i} = 0$, we have

Hence, we can derive the following inequality by (2.13) and (2.14)

i.e.

Seeing that $q(z)$ is a polynomial with $\deg q(z)\leq\sigma(f)$, then $\deg q(z)$ only satisfies one of the following cases: $1\leq \deg q(z) < \sigma(f) = \deg P(z)$; $\deg q(z) = \sigma(f) = \deg P(z)$ and $\deg q(z) = 0.$

Case 1. $1\leq \deg q(z) < \sigma(f) = \deg P(z).$ By (2.16), we have

It follows from $\beta-a\neq0, 1\leq\deg q(z) < \deg P(z)$ that $(\beta-a)e^{q(z)}+a\not\equiv0.$ Hence, the order of $\left((\beta-a)e^{q(z)}+a\right)e^{-P(z)}$ is equal to $\sigma(f) = \deg P(z).$ As $\deg (P(z+ic)-P(z))\leq (\deg P(z))-1$, $\sigma(A(z)) < \sigma(f) = \deg P(z)$ and $\deg q(z) < \sigma(f) = \deg P(z)$, we see that the order of $\sum\limits_{i = 1}^{n}a_{i}A(z+ic)e^{P(z+ic)-P(z)}+(a_{0}-e^{q(z)})A(z)$ is less than $\sigma(f) = \deg P(z).$ We can get a contradiction from (2.17).

Case 2. $\deg q(z) = \sigma(f) = \deg P(z).$ Suppose

Thus $p_{k}$ and $q_{k}$ only satisfy one of the following cases: $p_{k} = -q_{k}$; $p_{k} = q_{k}$; $p_{k}\neq q_{k}$ and $p_{k}\neq-q_{k}.$

Subcase 2.1. $p_{k} = -q_{k}$. From (2.16), we can get

i.e.

where

Since $p_{k} = -q_{k}$, the $\deg(q(z)+P(z))\leq k-1$. Note that

By Lemma 2.1, we can get $\frac{\beta-a}{A(z)}\equiv0$. Which contradicts with $a\neq\beta.$

Subcase 2.2. $p_{k} = q_{k}$. From (2.16), we can get

i.e.

where

By Lemma 2.1, we can get a contradiction.

Subcase 2.3. $p_{k}\neq q_{k}$ and $p_{k}\neq-q_{k}.$ From (2.16), we can get

i.e.

where

By Lemma 2.1, we can get a contradiction.

Case 3. $\deg q(z) = 0.$ In this case, $e^{q(z)}$ is a constant. We denote it by $C$. Suppose that $C\neq\frac{a}{a-\beta}$, by (2.16) we can get

i.e.

Since $\deg (P(z+ic)-P(z))\leq (\deg P(z))-1$, $\sigma(A(z)) < \sigma(f) = \deg P(z)$, then

We can get a contradiction from (2.21). Hence $C = \frac{a}{a-\beta}$.

2.3.   Proof of Theorem 1.5

Since $P(z, f)$ and $f$ share the value $0$ CM, there holds

where $H(z)$ is a polynomial. If $H(z)\neq$constant, then

By Lemma 2.3, we have $\sigma(f) > \deg(H(z))+1 > 2.$ This contradicts with $\sigma(f) < 2$. Hence $H(z)$ is a constant. Denote $\eta = e^{H(z)},$ then $\eta$ is a constant and

3.   Conclusions

The main result of this paper (Theorem 1.2) shows that $P(z, f)$ is a linear function of $f$, if the following conditions are satisfied:

(1) $f$ is a transcendental meromorphic function of finite order with two Borel exceptional values $\beta$ and $\infty$;

(2) $P(z, f)$ and $f$ share the value $a(\neq\beta)$ CM.

Acknowledgments

This work was supported by the Scientific Research Project of Education Department of Hubei Province (Grant No. D20182801, B2020157).

Conflict of interest

The authors declare that they have no competing interests.