Unicity of transcendental meromorphic functions concerning differential-difference polynomials

1.   Introduction and main results

In this paper, we assume that the reader is familiar with the basic notions of Nevanlinna's value distribution theory, see ^[9,11,16,17]. In the following, a meromorphic function always means meromorphic in the whole complex plane. By $S(r, f)$, we denote any quantity satisfying $S(r, f) = o(T(r, f))$ as $r\rightarrow \infty$ possible outside of an exceptional set $E$ with finite logarithmic measure $\int_{E}dr/r < \infty$. A meromorphic function $\alpha$ is said to be a small function of $f$ if it satisfies $T(r, \alpha) = S(r, f)$.

Let $f$ be a nonconstant meromorphic function. The order of $f$ is defined by

The exponent of convergence of poles of $f$ is defined by

Let $a$ be a complex number, and let $f$ be a nonconstant meromorphic function. If

for $\rho(f) > 0$; and $N\left(r, \frac{1}{f-a}\right) = O(\log r)$ for $\rho(f) = 0$, then $a$ is called a Borel exceptional value of $f$.

Let $f$ and $g$ be two meromorphic functions, and let $\alpha$ be a small functions of both $f$ and $g$. We say that $f$ and $g$ share $\alpha$ CM(IM) if $f-\alpha$ and $g-\alpha$ have the same zeros counting multiplicities (ignoring multiplicities).

$N(r, \alpha)$ is the counting function of common zeros of $f-\alpha$ and $g-\alpha$ with the same multiplicities and the multiplicity is counted. If $N\left(r, \frac{1}{f-\alpha}\right)+N\left(r, \frac{1}{g-\alpha}\right)-2N(r, \alpha) \leq S(r, f)+S(r, g)$, then we call that $f$ and $g$ share $\alpha$ CM almost.

Let $\overline N_{(k}(r, f)$ be the counting function for poles of $f$ with multiplicity $\geq k$ where multiplicity is not counted. Set $N_{k}(r, f) = \overline N(r, f)+\overline N_{(2}(r, f)+ \cdots +\overline N_{(k}(r, f).$

Nevanlinna ^[17] proved the following famous five-value theorem.

Theorem A. Let $f$ and $g$ be two nonconstant meromorphic functions, and let $a_j$ ($j = 1, 2, 3, 4, 5$) be five distinct values in the extended complex plane. If $f$ and $g$ share $a_j$ ($j = 1, 2, 3, 4, 5$) IM, then $f\equiv g$.

In 2000, Li and Qiao ^[13] improved Theorem A as follows.

Theorem B. Let $f$ and $g$ be two nonconstant meromorphic functions, and let $\alpha_j$ ($j = 1, 2, 3, 4, 5$) be five distinct small functions of both $f$ and $g$. If $f$ and $g$ share $\alpha_j$ ($j = 1, 2, 3, 4, 5$) IM, then $f\equiv g$.

Recently, value distribution in difference analogue of meromorphic functions has become a subject of some interests, see ^{[1,2,3,4,6,7,8,10,12,15,18,19]}.

In 2010, Zhang ^[18] proved the following result.

Theorem C. Let $f$ and $g$ be two transcendental entire functions of finite order, let $\alpha$ $(\not\equiv 0)$ be a small function of both $f$ and $g$, and let $c$ be a nonzero finite complex constant and $n\ge 7$ an integer. If $f^n(z)(f(z)-1)f(z+c)$ and $g^n(z)(g(z)-1)g(z+c)$ share $\alpha$ CM, then $f \equiv g.$

In 2012, Chen and Chen ^[2] extended Theorem C as follows.

Theorem D. Let $f$ and $g$ be two transcendental entire functions of finite order, let $\alpha$ $(\not\equiv 0)$ be a small function of both $f$ and $g$, let $d, n, m$ and $v_j (j = 1, 2, \cdots, d)$ be positive integers, and let $c_j (j = 1, 2, \cdots, d)$ be distinct nonzero finite values. If $n\ge m+8\sigma$, where $\sigma = v_1+v_2+ \cdots +v_d$, and $f^n(z)(f^m(z)-1)\prod _{j = 1}^{d}f^{v_j}(z+c_j)$ and $g^n(z)(g^m(z)-1)\prod _{j = 1}^{d}g^{v_j}(z+c_j)$ share $\alpha$ CM, then $f \equiv tg$, where $t^m = t^{n+\sigma } = 1.$

Zhang and Yi ^[19], Banerjee and Majumder ^[1], Husna et al. ^[10], Sahoo and Biswas ^[15] continued to study this problem and proved:

Theorem E. ^[1]Let $f$ and $g$ be two transcendental entire functions of finite order, let $\alpha$ $(\not\equiv 0)$ be a small function of both $f$ and $g$ with finitely many zeros, let $d, k, n, m$ and $v_j (j = 1, 2, \cdots, d)$ be positive integers, and let $c_j (j = 1, 2, \cdots, d)$ be distinct nonzero finite values. If $n\ge \max \{2k+m+\sigma +5, \sigma +2d+3\}$, where $\sigma = v_1+v_2+\cdots +v_d$, and $(f^n(z)(f^m(z)-1)\prod_{j = 1}^{d}f^{v_j}(z+c_j))^{(k)}$ and $(g^n(z)(g^m(z)-1)\prod_{j = 1}^{d}g^{v_j}(z+c_j))^{(k)}$ share $\alpha$ CM, then $f \equiv tg$, where $t^m = t^{n+\sigma } = 1.$

In ^[1], Banerjee and Majumder posed a problem as follows.

Problem 1. Whether Theorem E is valid or not for any small function?

In 2021, Majumder and Saha ^[14] gave a positive answer to Problem 1 and proved:

Theorem F. Let $f$ and $g$ be two transcendental entire functions of finite order, let $\alpha$ $(\not\equiv 0)$ be a small function of both $f$ and $g$, let $d, k, n, m$ and $v_j (j = 1, 2, \cdots, d)$ be positive integers, and let $c_j (j = 1, 2, \cdots, d)$ be distinct nonzero finite values. If $n\ge \max \{2k+m+\sigma +5, \sigma +2d+3\}$, where $\sigma = v_1+v_2+\cdots +v_d$, and $(f^n(z)(f^m(z)-1)\prod_{j = 1}^{d}f^{v_j}(z+c_j))^{(k)}$ and $(g^n(z)(g^m(z)-1)\prod_{j = 1}^{d}g^{v_j}(z+c_j))^{(k)}$ share $\alpha$ CM, then $f \equiv tg$, where $t^m = t^{n+\sigma } = 1.$

In this paper, we consider the case of meromophic functions and obtain:

Theorem 1. Let $f$ and $g$ be two transcendental meromorphic functions of finite order with a Borel exceptional value $\infty$, let $\alpha$ $(\not\equiv 0)$ be a small function of both $f$ and $g$, let $d, k, n, m$ and $v_j (j = 1, 2, \cdots, d)$ be positive integers, and let $c_j (j = 1, 2, \cdots, d)$ be distinct nonzero finite values. If $n\ge \max \{2k+m+\sigma +5, \sigma +2d+3\}$, where $\sigma = v_1+v_2+\cdots +v_d$, and $(f^n(z)(f^m(z)-1)\prod_{j = 1}^{d}f^{v_j}(z+c_j))^{(k)}$ and $(g^n(z)(g^m(z)-1)\prod_{j = 1}^{d}g^{v_j}(z+c_j))^{(k)}$ share $\alpha$ CM, then $f \equiv tg$, where $t^m = t^{n+\sigma } = 1.$

Remark. By Theorem 1, we get Theorem F.

2.   Some lemmas

Lemma 1. ^[9,17]Let $f$ be a nonconstant meromorphic function, and let $k$ be a positive integer. Then

Lemma 2. ^[3,6]Let $f$ be a nonconstant meromorphic function of finite order, and let $c$ be a nonzero finite complex number. Then

and for any $\varepsilon > 0$, we have

Lemma 3. ^[16]Let $k$ be a positive integer, and let $f$ be a nonconstant meromorphic function satisfying $f^{(k)} \not\equiv 0$. Then

Lemma 4. ^[9,17]Let $f$ be a nonconstant meromorphic function, and let $\alpha, \beta$ be two distinct small functions of $f$. Then

Lemma 5. ^[8]Let $f$ be a nonconstant meromorphic function of finite order, and let $c$ be a nonzero finite complex number. Then

Lemma 6. Let $f$ be a nonconstant meromorphic function of finite order, and let $F(z) = f^{n}(z)(f^{m}(z)-1)\prod_{j = 1} ^{d}f^{v_{j}}(z+c_{j}),$ where $d, n, m, c_{j}$ and $v_j (j = 1, 2, \cdots, d)$ are positive integers and $\sigma = v_{1}+ \cdots +v_{d}$. Then

Proof. By Lemma 2 and Lemma 5, we have

By Lemma 2 and Lemma 5, we have

Lemma 7. ^[5]Let $f$ and $g$ be two nonconstant meromorphic functions. If $f$ and $g$ share 1 CM almost, then one of the following cases must occur

(1) $T(r, f)+T(r, g)\leq 2\left\{N_{2}(r, f)+N_{2}(r, g)+N_{2}\left(r, \frac{1}{f}\right)+N_{2}\left(r, \frac{1}{g}\right)\right\}+S(r, f)+S(r, g);$

(2) $f = \frac{(b+1)g+(a-b-1)}{bg+(a-b)}$, where $a(\neq 0)$ and $b$ are two constants.

3.   Proof of Theorem 1

Since $\infty$ is a Borel exceptional value of both $f$ and $g$, we have

Thus for $\frac{\rho(f)-\lambda\left(\frac{1}{f}\right)}{2}$, then exist a positive number $R$ such that $r\geq R$, we have

Similarly, we get

Set

Since $(f^n(z)(f^m(z)-1)\prod\nolimits _{j = 1}^{d}f^{v_j}(z+c_j))^{(k)}$ and $(g^n(z)(g^m(z)-1)\prod\nolimits _{j = 1}^{d}g^{v_j}(z+c_j))^{(k)}$ share $\alpha(z)$ CM, then $F(z)$ and $G(z)$ share 1 CM almost. By Lemma 7, we consider two cases.

Case 1.

Set

By Lemma 1 and Lemma 3, we obtain

Similarly,

It follows from Lemma 6, (3.3)–(3.7) that

Thus, we have

Without loss of generality we assume that $\rho(f) \leq \rho(g)$. By (3.1), (3.2) and (3.8), we have

Since $n\geq 2k+m+\sigma+5$, then by (3.9), we get $\rho(g)\leq \max\left\{\frac{\rho(f)+\lambda\left(\frac{1}{f}\right)}{2}, \frac{\rho(g)+\lambda\left(\frac{1}{g}\right)}{2}\right\}$ that is $\rho(g) < \rho(g)$, a contradiction.

Case 2.

where $a(\neq 0)$ and $b$ are two constants.

Obviously,

Next, we consider three subcases.

Case 2.1. $b\neq 0, -1.$ In the following, we consider two subcases.

Case 2.1.1. $a-b-1\neq 0.$ From (3.10), we have $\overline N\left(r, \frac{1}{G+\frac{a-b-1}{b+1}}\right) = \overline N\left(r, \frac{1}{F}\right).$

By Nevanlinna's second fundamental theorem, we get

By (3.11) and (3.12), we have

It follows from Lemma 3 that

Similarly,

By Lemma 6, (3.13)–(3.17), we get

Since $n\geq 2k+m+\sigma+5$, then by (3.1), (3.2), (3.18) and using the same argument as used in Case 1, we get $\rho(g) < \rho(g)$, a contradiction.

Case 2.1.2. $a-b-1 = 0.$ Then by (3.10), we have $\overline N\left(r, \frac{1}{G+ \frac{1}{b}}\right) = \overline N(r, F).$

By Nevanlinna's second fundamental theorem, we get

Similarly,

It follows from (3.19) and (3.20) that

By Lemma 6, (3.14)–(3.17), (3.21), we get

Since $n\geq 2k+m+\sigma+5$, then by (3.1), (3.2), (3.22), and using the same argument as used in Case 1, we get $\rho(g) < \rho(g)$, a contradiction.

Case 2.2. $b = -1.$ From (3.10), we have

Next, we consider two subcases.

Case 2.2.1. $a+1\neq 0$. Then by (3.23), we have $\overline N\left(r, \frac{1}{G-(a+1)}\right) = \overline N(r, F).$ Next, by using the same argument as used in Case 2.1.2, we get $\rho(g) < \rho(g)$, a contradiction.

Case 2.2.2. $a+1 = 0$. Then by (3.23), we get $FG\equiv 1.$ That is

Since $\infty$ is a Borel exceptional value of both $f$ and $g$, then by Hadamard's factorization theorem and (3.24), we have

where $\beta(\not\equiv 0, \infty), \gamma(\not\equiv 0, \infty)$ are two meromophic functions and $p_{1}$, $p_{2}$ are two nonconstant polynomials with $\deg{p_{1}} = \deg{p_{2}}$.

Hence, by the simple analysis, we get

By (3.25), we have

Further, we have

Obviously,

where $j = 1, 2, \cdots, d$.

By (3.26), (3.30) and Lemma 5, we get

Set

Then it follows from (3.31) and (3.32) that $\beta_{1}(z)(\not\equiv 0)$ is a small function of $e^{p_{1}(z)}$.

By (3.29) and (3.32), we get

Similarly, we get

where $\beta_{2}(z) = -\beta^{n}(z)\prod\nolimits_{j = 1}^{d}\beta^{v_{j}}(z+c_{j})e^{v_{j}[p_{1}(z+c_{j})-p_{1}(z)]}$. By using the same argument as used in (3.29)–(3.32), we get $\beta_{2}(z)$ is a small function of $e^{p_{1}(z)}$.

By (3.28), (3.33) and (3.34), we have

It follows from (3.35) that

It is easy to show that

Set

By (3.36)–(3.40) we have

where $\beta_{3}(z), \beta_{4}(z)$ are two nonzero small functions of $e^{p_{1}(z)}$.

By mathematical induction, we obtain

where $\beta_{2k+1}(z), \beta_{2k+2}(z)$ are two nonzero small functions of $e^{p_{1}(z)}$.

Similarly, we get

where $\gamma_{2k+1}(z), \gamma_{2k+2}(z)$ are two nonzero small functions of $e^{p_{2}(z)}$.

By (3.24), (3.42) and (3.43), we have

Thus, we deduce that the zeros of $\beta_{2k+1}e^{mp_{1}}+\beta_{2k+2}$ are either the zeros of $\alpha$ or the poles of $\gamma_{2k+1}e^{mp_{2}}+\gamma_{2k+2}$. Hence, by Lemma 4 we get

It follows $T(r, e^{p_{1}}) \leq S(r, e^{p_{1}})$, a contradiction.

Case 2.3. $b = 0.$ Then by (3.10), we obtain

Next, we consider two subcases.

Case 2.3.1. $a-1\neq 0.$ Then by (3.46), we have $\overline N\left(r, \frac{1}{G+(a-1)}\right) = \overline N(r, \frac{1}{F}).$ Next, by using the same argument as used in Case 2.1.1, we get $\rho(g) < \rho(g)$, a contradiction.

Case 2.3.2. $a-1 = 0$. Then by (3.46), we get $F\equiv G.$

It follows

where $p$ is a polynomial with $\deg p\leq k-1.$

Now, we prove $p\equiv 0.$ Suppose on the contrary that $p\not\equiv 0.$ Then by Lemma 4, Lemma 6 and (3.47), we have

Likewise,

By (3.48) and (3.49), we get

Since $n\geq 2k+m+\sigma+5$, then by (3.1), (3.2), (3.50), and using the same argument as used in Case 1, we get $\rho(g) < \rho(g)$, a contradiction.

Hence,

Set $h = \frac {f}{g}.$ From (3.51), we get

We claim that $h$ is a constant. Suppose on the contrary that $h$ is a nonconstant.

We assert that both $h^{m+n}(z)\prod\nolimits _{j = 1}^{d}h^{v_j}(z+c_j)$ and $h^{n}(z)\prod\nolimits _{j = 1}^{d}h^{v_j}(z+c_j)$ are nonconstant. Without loss of generality, we suppose that $h^{m+n}(z)\prod\nolimits _{j = 1}^{d}h^{v_j}(z+c_j)\equiv c,$ where $c$ is a nonzero complex number. Then

By Lemma 5 and Nevanlinna's first fundamental theorem, we get

From (3.53), we have

It follows from $n\geq 2k+m+\sigma+5$ that $T(r, h)\leq S(r, h)$, a contradiction.

Hence, from (3.52), we get

Thus, the zeros of $h^{m+n}(z)\prod\nolimits _{j = 1}^{d}h^{v_j}(z+c_j)-1$ are either the poles of $g(z)$ or the zeros of $h^{n}(z)\prod\nolimits _{j = 1}^{d}h^{v_j}(z+c_j)-1$, that is the zeros of $(h^{m}(z)-1)$.

By (3.54), we have

It follows from Lemma 6, (3.55) and Nevanlinna's first fundamental theorem that

By Lemma 6, we obtain

Since $n\geq 2k+m+\sigma+5$, then by (3.56) and (3.57), we get

It follows from Lemma 5 and Nevanlinna's second fundamental theorem that

Since $n \geq \sigma+2d+3$, then by (3.58), we have

It follows from (3.2), (3.58) and (3.59) that $\rho(g) < \lambda\left(\frac{1}{g}\right)$, a contradiction. Therefore, $h$ is a nonzero constant.

From (3.51), we get

Then

Therefore, $f \equiv tg$, where $t$ is a constant such that $t^{m} = t^{n+\sigma} = 1.$

The proof of Theorem 1 is complete.

4.   Conclusions

In this paper, by using Nevanlinna theory, we study the uniqueness problem of certain type of differential-difference polynomials sharing a small function and extend the existing results to the case of meromorphic functions with a Borel exceptional value $\infty$.

Acknowledgments

This research was funded by the National Natural Science Foundation of China (Grant No 12171127) and the Natural Science Foundation of Zhejiang Province (Grant No LY21A010012).

Conflict of interest

The authors declare that none of the authors have any competing interests in the manuscript.