## 1 Introduction and main results

Let $$f(z)$$ be a meromorphic function in the complex plane C. We assume that the reader is familiar with the standard notations and results in Nevanlinna’s value distribution theory of meromorphic functions (see e.g. [13]). We use $$\rho(f)$$ to denote the growth order of a meromorphic function $$f(z)$$. The notation $$S(r,f)$$ denotes any quantity that satisfies the condition $$S(r,f)= o(T(r,f))$$ as $$r \rightarrow\infty$$ possibly outside an exceptional set of r of finite logarithmic measure. A meromorphic function $$a(z)$$ is called a small function of $$f(z)$$ if and only if $$T(r,a(z))=S(r,f)$$.

In the last ten years, there has been a great deal of interest in studying the properties of complex difference equations (see e.g. [420]). Especially, a number of papers (see e.g. [4, 6, 10, 11, 15, 1719]) focusing on a Malmquist type theorem of the complex difference equations emerged. In 2000, Ablowitz et al. [4] proved some results on the Malmquist theorem of the complex difference equations by utilizing Nevanlinna theory. They obtained the following two results.

### Theorem A

If the second-order difference equation

$$f(z+1)+f(z-1)=\frac{a_{0}(z)+a_{1}(z)f+ \cdots +a_{p}(z)f^{p}}{b_{0}(z)+b_{1}(z)f+ \cdots+b_{q}(z)f^{q}},$$

with polynomial coefficients $$a_{i}$$ ($$i=1,2,\ldots,p$$) and $$b_{j}$$ ($$j=1,2,\ldots,q$$), admits a transcendental meromorphic solution of finite order, then $$d=\max\{p,q\}\leq2$$.

### Theorem B

If the second-order difference equation

$$f(z+1)f(z-1)=\frac{a_{0}(z)+a_{1}(z)f+ \cdots +a_{p}(z)f^{p}}{b_{0}(z)+b_{1}(z)f+ \cdots+b_{q}(z)f^{q}},$$

with polynomial coefficients $$a_{i}$$ ($$i=1,2,\ldots,p$$) and $$b_{j}$$ ($$j=1,2,\ldots,q$$), admits a transcendental meromorphic solution of finite order, then $$d=\max\{p,q\}\leq2$$.

Subsequently, Heittokangas et al. [10], Laine et al. [15] and Huang et al. [11], respectively, gave some generalizations of the above two results. In 2010, the first author in this paper and Liao [18] obtained the following more general result.

### Theorem C

Let $$c_{1}, c_{2}, \ldots, c_{n}$$ be distinct, nonzero complex numbers, and suppose that $$f(z)$$ is a transcendental meromorphic solution of the difference equation

$$\sum_{\lambda\in I}\alpha_{\lambda}(z) \Biggl(\prod_{\nu=1}^{n}f(z+c_{\nu})^{l_{\lambda, \nu}} \Biggr) = \frac{a_{0}(z) + a_{1}(z)f(z) + \cdots+ a_{p}(z)f(z)^{p}}{b_{0}(z) + b_{1}(z)f(z) + \cdots+ b_{q}(z)f(z)^{q}},$$
(1)

with coefficients $$\alpha_{\lambda}(z)$$ ($$\lambda\in I$$), $$a_{i}(z)$$ ($$i=0,1,\ldots, p$$), and $$b_{j}(z)$$ ($$j=0,1,\ldots, q$$), which are small functions relative to $$f(z)$$, where $$I = \{\lambda=(l_{\lambda, 1}, l_{\lambda, 2}, \ldots, l_{\lambda,n})| l_{\lambda, \nu}\in{N} \cup\{0\}, \nu=1,2, \ldots,n\}$$ is a finite index set, and denote

$$\sigma_{\nu}= \max_{\lambda}\{l_{\lambda, \nu}\}\quad ( \nu=1,2,\ldots,n), \qquad \sigma=\sum_{\nu=1}^{n} \sigma_{\nu}.$$

If the order $$\rho(f)$$ is finite, then $$d=\max\{p,q\}\leq \sigma$$.

If all the coefficients in the complex difference equation (1) are rational functions, then in [19], we have the following Malmquist type result, which is reminiscent of the classical Malmquist theorem in complex differential equations.

### Theorem D

Let $$c_{1}, c_{2}, \ldots, c_{n}$$ be distinct, nonzero complex numbers and suppose that $$f(z)$$ is a transcendental meromorphic solution of the equation

$$P[z,f]:=\sum_{\lambda\in I}\alpha_{\lambda}(z) \Biggl( \prod_{\nu=1}^{n}f(z+c_{\nu})^{l_{\lambda, \nu}} \Biggr)=R\bigl(z,f(z)\bigr)= \frac{P(z,f(z))}{Q(z,f(z))},$$

where $$I = \{\lambda=(l_{\lambda, 1}, l_{\lambda, 2}, \ldots, l_{\lambda,n})| l_{\lambda, \nu}\in{N} \cup\{0\}, \nu=1,2, \ldots,n\}$$ is a finite index set, P and Q are relatively prime polynomials in f over the field of rational functions, the coefficients $$\alpha_{\lambda }$$ ($$\lambda\in I$$) are rational functions. Denoting the degree of $$P[z,f]$$ by

$$\gamma_{P}:= \max_{\lambda\in I}\bigl\{ l_{\lambda, 1}+l_{\lambda, 2}+ \cdots +l_{\lambda, n}|\lambda=(l_{\lambda, 1}, l_{\lambda, 2}, \ldots, l_{\lambda,n})\bigr\} .$$

If $$f(z)$$ is finite order and has at most finitely many poles, then $$R(z,f)$$ reduces to a polynomial in f of degree $$d \leq\gamma_{P}$$.

More recently, people began to study the properties of meromorphic solutions of systems of complex difference equations. In [21], Gao discussed the proximity function and counting function of meromorphic solutions of some classes of systems of complex difference equations. In 2013, Wang et al. [16] investigated the growth of meromorphic solutions of systems of complex difference equations.

Now, we give the Malmquist type result of a system of complex difference equations as follows.

### Theorem 1

Let $$c_{1}, c_{2}, \ldots, c_{n}$$ be distinct, nonzero complex numbers and suppose that $$(f(z), g(z))$$ is a transcendental meromorphic solution of a system of complex difference equations of the form

$$\left \{ \textstyle\begin{array}{l} \sum_{\lambda_{1} \in I_{1}, \mu_{1}\in J_{1}}\alpha_{\lambda_{1}, \mu_{1}}(z) (\prod_{\nu=1}^{n}f(z+c_{\nu})^{l_{\lambda_{1}, \nu}}\prod_{\nu=1}^{n}g(z+c_{\nu})^{m_{\mu_{1}, \nu}} ) = \frac{\sum_{i=0}^{p}a_{i}(z)g(z)^{i} }{\sum_{j=0}^{q}b_{j}(z)g(z)^{j}} , \\ \sum_{\lambda_{2} \in I_{2}, \mu_{2}\in J_{2}}\beta_{\lambda_{2}, \mu_{2}}(z) (\prod_{\nu =1}^{n}f(z+c_{\nu})^{l_{\lambda_{2}, \nu}}\prod_{\nu=1}^{n}g(z+c_{\nu})^{m_{\mu_{2}, \nu}} ) = \frac{\sum_{k=0}^{s}d_{k}(z)f(z)^{k} }{\sum_{l=0}^{t}e_{l}(z)f(z)^{l}}, \end{array} \right .$$
(2)

with coefficients $$\alpha_{\lambda_{1}, \mu_{1}}(z)$$ ($$\lambda_{1} \in I_{1}$$, $$\mu_{1}\in J_{1}$$), $$\beta_{\lambda_{2}, \mu_{2}}(z)$$ ($$\lambda_{2} \in I_{2}$$, $$\mu_{2}\in J_{2}$$), $$a_{i}(z)$$ ($$i=0,1,\ldots, p$$), $$b_{j}(z)$$ ($$j=0,1,\ldots, q$$), $$d_{k}(z)$$ ($$k=0,1,\ldots, s$$), and $$e_{l}(z)$$ ($$l=0,1,\ldots, t$$) are small functions relative to $$f(z)$$ and $$g(z)$$, $$a_{p}(z), b_{q}(z), d_{s}(z), e_{t}(z)\not\equiv0$$, where $$I_{i} = \{\lambda_{i}=(l_{\lambda_{i}, 1}, l_{\lambda_{i}, 2}, \ldots, l_{\lambda_{i},n})| l_{\lambda_{i}, \nu} \in{N} \cup\{0\}, \nu=1,2, \ldots,n\}$$ ($$i=1,2$$), and $$J_{j} = \{\mu_{j}=(m_{\mu_{j}, 1}, m_{\mu _{j}, 2}, \ldots, m_{\mu_{j},n})| m_{\mu_{j}, \nu}\in{N} \cup\{0\},\nu=1,2, \ldots,n\}$$ ($$j=1, 2$$) are finite index sets, and denote

\begin{aligned}& \xi_{1,\nu}= \max_{\lambda_{1}\in I_{1}}\{l_{\lambda_{1}, \nu}\},\qquad \eta_{1,\nu}= \max_{\mu_{1}\in J_{1}}\{m_{\mu_{1}, \nu}\}, \\& \xi_{2,\nu}= \max_{\lambda_{2}\in I_{2}}\{l_{\lambda_{2}, \nu}\},\qquad \eta_{2,\nu}= \max_{\mu_{2}\in J_{2}}\{m_{\mu_{2}, \nu}\} \end{aligned}

($$\nu=1,2,\ldots,n$$), and

$$\sigma_{11} =\sum_{\nu=1}^{n} \xi_{1,\nu},\qquad \sigma_{12} =\sum _{\nu =1}^{n}\eta_{1,\nu},\qquad \sigma_{21} =\sum_{\nu=1}^{n} \xi_{2,\nu},\qquad \sigma_{22} =\sum _{\nu =1}^{n}\eta_{2,\nu}.$$

If $$\max\{p,q\}>\sigma_{12}$$, $$\max\{s,t\}>\sigma_{21}$$, and $$\max\{\rho (f), \rho(g)\}<+\infty$$, then $$\rho(f)= \rho(g)$$ and $$(\max\{p,q\} -\sigma_{12})\cdot(\max\{s,t\}-\sigma_{21})\leq \sigma_{11}\sigma_{22}$$.

### Example 1

It is easy to check that $$(f(z), g(z))=(\tan z, \cot z)$$ satisfies the following system of difference equations:

$$\left \{ \begin{array}{l} f(z+\frac{\pi}{4})g(z+\frac{\pi}{3})^{2}+zg(z+\frac{\pi}{4}) \\ \quad =\frac{(3z+1)g^{4}+[(2\sqrt{3}-6)z+2-2\sqrt{3}]g^{3}+ (4-4\sqrt{3})(z+1)g^{2}+[(2\sqrt{3}-2)z+6-2\sqrt{3}]g+z+3}{3g^{4}+2\sqrt {3}g^{3}-2g^{2}-2\sqrt{3}g-1}, \\ f(z+\frac{\pi}{3})^{2}g(z+\frac{\pi}{4})+f(z+\frac{\pi}{3})g(z+\frac{\pi }{4})^{2}= \frac{-(\sqrt{3}+1)f^{4}-2f^{3}+2f^{2}-2f+3+\sqrt{3}}{3f^{4}+(6-2\sqrt {3})f^{3}+( 4-4\sqrt{3})f^{2}+(2-2\sqrt{3})f+1}. \end{array} \right .$$

In Example 1, we have $$\max\{p,q\}=4$$, $$\max\{s,t\}=4$$, $$\sigma_{11}=1$$, $$\sigma_{12}=3$$, $$\sigma_{21}=2$$, $$\sigma_{22}=2$$, $$\rho(f)=\rho(g)=1<+\infty$$, and $$(\max\{p,q\}-\sigma_{12})\cdot(\max\{ s,t\}-\sigma_{21})=(4-3)(4-2)=2=1\times2= \sigma_{11}\sigma_{22}$$. Therefore, the estimation in Theorem 1 is sharp.

### Remark 1

Obviously, if the condition $$\max\{p,q\}>\sigma_{12}$$, $$\max\{s,t\} >\sigma_{21}$$ in Theorem 1 is replaced by $$(\max\{p,q\}-\sigma _{12})(\max\{s,t\}-\sigma_{21})=0$$ or $$(\max\{p,q\}-\sigma_{12})(\max\{ s,t\}-\sigma_{21})<0$$, the estimation $$(\max\{p,q\}-\sigma_{12})\cdot (\max\{s,t\}-\sigma_{21})\leq \sigma_{11}\sigma_{22}$$ is still correct. If $$\sigma_{11}=0$$ or $$\sigma _{22}=0$$, then the first or second equation in (2) gets the form of (1). For some results as regards (1), the reader may refer to the paper [18].

### Remark 2

If the condition $$\max\{p,q\}>\sigma_{12}$$, $$\max\{s,t\}>\sigma _{21}$$ in Theorem 1 is replaced by $$\max\{p,q\}<\sigma_{12}$$, $$\max\{s,t\}<\sigma_{21}$$, then the estimation $$(\max\{p,q\}-\sigma_{12})\cdot(\max\{s,t\}-\sigma_{21})\leq \sigma_{11}\sigma_{22}$$ is not true generally. For example, $$(f(z), g(z))=(\tan z, \cot z)$$ satisfies the following system of complex difference equations:

$$\left \{ \begin{array}{l} f(z+\frac{\pi}{4})^{2}g(z-\frac{\pi}{4})g(z+\frac{\pi }{4})^{5}+2g(z+\frac{\pi}{4})^{2} =\frac{g^{2}-2g+1}{g^{2}+2g+1}, \\ f(z+\frac{\pi}{4})^{4}f(z-\frac{\pi}{4})g(z+\frac{\pi}{4})+zf(z+\frac {\pi}{4})=\frac{-(z+1)f^{2}-2f+z+1}{f^{2}-2f+1}, \end{array} \right .$$
(3)

where $$\max\{p,q\}=2$$, $$\max\{s,t\}=2$$, $$\sigma_{11}=2$$, $$\sigma_{12}=6$$, $$\sigma_{21}=5$$, $$\sigma_{22}=1$$, $$\max\{p,q\}<\sigma_{12}$$, $$\max\{s,t\}<\sigma_{21}$$. However, $$(\max\{ p,q\}-\sigma_{12})\cdot(\max\{s,t\}-\sigma_{21})=(-4)\times(-3)=12>2= \sigma_{11}\sigma_{22}$$.

If $$\sigma_{12}=\sigma_{21}=0$$, then we have the following simpler result.

### Corollary 1

Let $$c_{1}, c_{2}, \ldots, c_{n}$$ be distinct, nonzero complex numbers, and suppose that $$(f(z), g(z))$$ is a transcendental meromorphic solution of a system of complex difference equations of the form

$$\left \{ \textstyle\begin{array}{l} \sum_{\lambda\in I}\alpha_{\lambda}(z) (\prod_{\nu=1}^{n}f(z+c_{\nu})^{l_{\lambda, \nu}} ) = \frac{\sum_{i=0}^{p}a_{i}(z)g(z)^{i} }{\sum_{j=0}^{q}b_{j}(z)g(z)^{j}} , \\ \sum_{\mu\in J}\beta_{\mu}(z) (\prod_{\nu=1}^{n}g(z+c_{\nu})^{m_{\mu, \nu}} ) = \frac{\sum_{k=0}^{s}d_{k}(z)f(z)^{k} }{\sum_{l=0}^{t}e_{l}(z)f(z)^{l}}, \end{array} \right .$$
(4)

with coefficients $$\alpha_{\lambda}(z)$$ ($$\lambda\in I$$), $$\beta_{\mu}(z)$$ ($$\mu\in J$$), $$a_{i}(z)$$ ($$i=0,1,\ldots, p$$), $$b_{j}(z)$$ ($$j=0,1,\ldots, q$$), $$d_{k}(z)$$ ($$k=0,1,\ldots, s$$), and $$e_{l}(z)$$ ($$l=0,1,\ldots, t$$) are small functions relative to $$f(z)$$ and $$g(z)$$, $$a_{p}(z), b_{q}(z), d_{s}(z), e_{t}(z)\not\equiv0$$, where $$I = \{\lambda=(l_{\lambda, 1}, l_{\lambda, 2}, \ldots, l_{\lambda,n})| l_{\lambda, \nu}\in{N} \cup\{0\}, \nu=1,2, \ldots,n\}$$ and $$J = \{\mu=(m_{\mu, 1}, m_{\mu, 2}, \ldots, m_{\mu,n})|m_{\mu, \nu}\in{N} \cup\{0\},\nu=1,2, \ldots,n\}$$ are two finite index sets, and denote

$$\xi_{\nu}= \max_{\lambda}\{l_{\lambda, \nu}\}\quad ( \nu=1,2,\ldots,n),\qquad \sigma_{1} =\sum_{\nu=1}^{n} \xi_{\nu}$$

and

$$\eta_{\nu}= \max_{\mu}\{m_{\mu, \nu}\}\quad ( \nu=1,2,\ldots,n), \qquad \sigma_{2} =\sum _{\nu=1}^{n}\eta_{\nu}.$$

If $$\rho(f)<+\infty$$ or $$\rho(g)<+\infty$$, then $$\rho(f)=\rho(g)$$ and $$\max\{p,q\}\cdot\max\{s,t\}\leq \sigma_{1} \sigma_{2}$$.

### Example 2

Let $$c_{1}=\arctan2$$, $$c_{2}=\arctan(-2)$$. It is easy to check that $$(f(z), g(z))=(\tan z, \cot z)$$ satisfies the following system of difference equations:

$$\left \{ \begin{array}{l} f(z+c_{1})^{2}f(z+c_{2}) + f(z+c_{1})f(z+c_{2})^{2} =\frac{-40g^{3}+10 g}{g^{4}-8g^{2}+16}, \\ g(z+c_{1})g(z+c_{2})+g(z+c_{1})^{2}=\frac{-20f^{2}+10f}{f^{3}+2f^{2}-4f-8}. \end{array} \right .$$

In Example 2, we have $$\max\{p,q\}=4$$, $$\max\{s,t\}=3$$, $$\sigma_{1}=4$$, $$\sigma_{2}=3$$, $$\rho(f)=\rho(g)=1<+\infty$$, and $$\max\{p,q\}\cdot\max\{s,t\}=\sigma _{1}\cdot\sigma_{2}=12$$. Therefore, the estimation in Corollary 1 is sharp.

In [15], Laine et al. also considered the growth of meromorphic solutions of some classes of complex difference functional equations and obtained the following result.

### Theorem E

Suppose that f is a transcendental meromorphic solution of the equation

$$\sum_{\{J\}}\alpha_{J}(z) \biggl(\prod _{j \in J}f(z+c_{j}) \biggr)=f\bigl(p(z)\bigr),$$

where $$p(z)$$ is a polynomial of degree $$k \geq2$$, $$\{J\}$$ is the collection of all subsets of $$\{1,2,\ldots,n\}$$. Moreover, we assume that the coefficients $$\alpha_{J}(z)$$ are small functions relative to f and that $$n\geq k$$. Then

$$T(r,f)=O\bigl((\log r)^{\alpha+\varepsilon}\bigr),$$

where $$\alpha=\frac{\log n}{\log k}$$.

In 2010, Zhang et al. [18] got a more generalized result than Theorem E. Next we give the growth of meromorphic solutions of a system of complex functional equations as follows.

### Theorem 2

Let $$c_{1}, c_{2}, \ldots, c_{n}$$ be distinct, nonzero complex numbers and suppose that $$(f(z), g(z))$$ is a transcendental meromorphic solution of a system of complex functional equations of the form

$$\left \{ \textstyle\begin{array}{l} \sum_{\lambda_{1} \in I_{1}, \mu_{1}\in J_{1}}\alpha_{\lambda_{1}, \mu_{1}}(z) (\prod_{\nu=1}^{n}f(z+c_{\nu})^{l_{\lambda_{1}, \nu}}\prod_{\nu=1}^{n}g(z+c_{\nu})^{m_{\mu_{1}, \nu}} ) = f(p(z)) , \\ \sum_{\lambda_{2} \in I_{2}, \mu_{2}\in J_{2}}\beta_{\lambda_{2}, \mu_{2}}(z) (\prod_{\nu =1}^{n}f(z+c_{\nu})^{l_{\lambda_{2}, \nu}}\prod_{\nu=1}^{n}g(z+c_{\nu})^{m_{\mu_{2}, \nu}} ) = g(p(z)), \end{array} \right .$$
(5)

where $$p(z)$$ is a polynomial of degree $$k\geq2$$, $$I_{i} = \{\lambda_{i}=(l_{\lambda_{i}, 1}, l_{\lambda_{i}, 2}, \ldots, l_{\lambda_{i},n})| l_{\lambda_{i}, \nu} \in{N} \cup\{0\}, \nu=1,2, \ldots,n\}$$ ($$i=1,2$$) and $$J_{j} = \{\mu_{j}=(m_{\mu_{j}, 1}, m_{\mu_{j}, 2}, \ldots, m_{\mu_{j},n})| m_{\mu_{j}, \nu}\in{N} \cup\{0\},\nu=1,2, \ldots,n\}$$ ($$j=1, 2$$) are finite index sets, and denote

\begin{aligned}& \xi_{1,\nu}= \max_{\lambda_{1}\in I_{1}}\{l_{\lambda_{1}, \nu}\},\qquad \eta_{1,\nu}= \max_{\mu_{1}\in J_{1}}\{m_{\mu_{1}, \nu}\}, \\& \xi_{2,\nu}= \max_{\lambda_{2}\in I_{2}}\{l_{\lambda_{2}, \nu}\},\qquad \eta_{2,\nu}= \max_{\mu_{2}\in J_{2}}\{m_{\mu_{2}, \nu}\} \end{aligned}

($$\nu=1,2,\ldots,n$$),

$$\sigma_{11} =\sum_{\nu=1}^{n} \xi_{1,\nu},\qquad \sigma_{12} =\sum _{\nu =1}^{n}\eta_{1,\nu},\qquad \sigma_{21} =\sum_{\nu=1}^{n} \xi_{2,\nu},\qquad \sigma_{22} =\sum _{\nu =1}^{n}\eta_{2,\nu}$$

and

$$\sigma=\max\{\sigma_{11},\sigma_{12},\sigma_{21}, \sigma_{22}\}.$$

Moreover, we assume that the coefficients $$\alpha_{\lambda_{1}, \mu _{1}}(z)$$ ($$\lambda_{1} \in I_{1}$$, $$\mu_{1}\in J_{1}$$), $$\beta_{\lambda_{2}, \mu_{2}}(z)$$ ($$\lambda_{2} \in I_{2}$$, $$\mu_{2}\in J_{2}$$) are small functions relative to $$f(z)$$ and $$g(z)$$, and that $$2{\sigma}\geq k$$. Then

$$T(r,f)=O\bigl((\log r)^{\alpha+\varepsilon}\bigr),\qquad T(r,g)=O\bigl((\log r)^{\alpha +\varepsilon}\bigr),$$

where $$\alpha=\frac{\log{2\sigma}}{\log k}$$.

If $$\sigma_{12}=\sigma_{21}=0$$, then we can obtain the following result easily.

### Corollary 2

Let $$c_{1}, c_{2}, \ldots, c_{n}$$ be distinct, nonzero complex numbers and suppose that $$(f(z), g(z))$$ is a transcendental meromorphic solution of a system of complex functional equations of the form

$$\left \{ \textstyle\begin{array}{l} \sum_{\lambda\in I}\alpha_{\lambda}(z) (\prod_{\nu=1}^{n}f(z+c_{\nu})^{l_{\lambda, \nu}} ) = g(p(z)) , \\ \sum_{\mu\in J}\beta_{\mu}(z) (\prod_{\nu=1}^{n}g(z+c_{\nu})^{m_{\mu, \nu}} ) = f(p(z)), \end{array} \right .$$
(6)

where $$p(z)$$ is a polynomial of degree $$k\geq2$$, $$I = \{\lambda=(l_{\lambda, 1}, l_{\lambda, 2}, \ldots, l_{\lambda,n})| l_{\lambda, \nu}\in{N} \cup\{0\}, \nu=1,2, \ldots,n\}$$ and $$J = \{\mu=(m_{\mu, 1}, m_{\mu, 2}, \ldots, m_{\mu,n})| m_{\mu, \nu}\in{N} \cup\{0\},\nu=1,2, \ldots,n\}$$ are two finite index sets, and denote

\begin{aligned}& \xi_{\nu}= \max_{\lambda}\{l_{\lambda, \nu}\}\quad ( \nu=1,2,\ldots,n), \qquad \sigma_{1} =\sum _{\nu=1}^{n}\xi_{\nu}, \\& \eta_{\nu}= \max_{\mu}\{m_{\mu, \nu}\}\quad ( \nu=1,2,\ldots,n),\qquad \sigma_{2} =\sum_{\nu=1}^{n} \eta_{\nu} \end{aligned}

and

$$\sigma=\max\{\sigma_{1},\sigma_{2}\}.$$

Moreover, we assume that the coefficients $$\alpha_{\lambda}(z)$$ ($$\lambda \in I$$), $$\beta_{\mu}(z)$$ ($$\mu\in J$$) are small functions relative to $$f(z)$$ and $$g(z)$$, and that $$2{\sigma}\geq k$$. Then

$$T(r,f)=O\bigl((\log r)^{\alpha+\varepsilon}\bigr),\qquad T(r,g)=O\bigl((\log r)^{\alpha +\varepsilon}\bigr),$$

where $$\alpha=\frac{\log{2\sigma}}{\log k}$$.

## 2 Some lemmas

In order to prove our results, we need the following lemmas.

### Lemma 1

(see [3])

Let $$f(z)$$ be a meromorphic function. Then for all irreducible rational functions in f,

$$R(z,f)=\frac{P(z,f)}{Q(z,f)}=\frac{\sum_{i=0}^{p}a_{i}(z)f^{i}}{\sum_{j=0}^{q}b_{j}(z)f^{j}},$$

such that the meromorphic coefficients $$a_{i}(z)$$, $$b_{j}(z)$$ satisfy

$$\left \{ \begin{array}{l@{\quad}l} T(r,a_{i})=S(r,f),& i=0,1,\ldots,p , \\ T(r,b_{j})=S(r,f),& j=0,1,\ldots,q, \end{array} \right .$$

we have

$$T\bigl(r,R(z,f)\bigr)=\max\{p,q\}\cdot T(r,f)+ S(r,f).$$

In [22], AZ Mokhon’ko and VD Mokhon’ko gave an estimation of Nevanlinna’s characteristic function of

$$F(z)=\frac{P(z)}{Q(z)}=\frac{\sum_{\lambda\in I}f_{1}^{l_{\lambda, 1}}f_{2}^{l_{\lambda, 2}}\cdots\, f_{n}^{l_{\lambda, n}}}{\sum_{\mu \in J}f_{1}^{m_{\mu, 1}}f_{2}^{m_{\mu, 2}}\cdots \, f_{n}^{m_{\mu, n}}},$$

where $$f_{1},f_{2},\ldots,f_{n}$$ are distinct meromorphic functions, $$I=\{\lambda=(l_{\lambda, 1}, l_{\lambda , 2}, \ldots, l_{\lambda, n})| l_{\lambda, \nu}\in{N} \cup\{0\}, \nu=1,2,\ldots,n\}$$ and $$J=\{\mu=(m_{\mu, 1}, m_{\mu, 2}, \ldots, m_{\mu, n})| m_{\mu, \nu}\in{N} \cup\{0\}, \nu=1,2, \ldots,n\}$$ are two finite index sets. However, the method of the proof was too complex. For $$F(z)$$ of the form $$\sum_{\lambda\in I}f_{1}^{l_{\lambda, 1}}f_{2}^{l_{\lambda, 2}}\cdots \, f_{n}^{l_{\lambda, n}}$$, Zheng et al. [20] gave a simpler proof, but the estimation of $$T(r,F)$$ was not sharp. For completeness, we give the proof of the following lemma.

### Lemma 2

Let $$f_{1},f_{2},\ldots,f_{n}$$ be distinct meromorphic functions. Then

$$T \biggl(r,\sum_{\lambda\in I}f_{1}^{l_{\lambda, 1}}f_{2}^{l_{\lambda, 2}} \cdots\, f_{n}^{l_{\lambda, n}} \biggr) \leq \sum _{j=1}^{n}\sigma_{j}T(r, f_{j}) + \log t,$$

where $$I=\{(l_{\lambda, 1}, l_{\lambda, 2}, \ldots, l_{\lambda, n})| l_{\lambda, j}\in{N} \cup\{0\}, j=1,2, \ldots,n\}$$ is an finite index set consisting of t elements and $$\sigma_{j} = \max_{\lambda\in I}\{l_{\lambda, j}\}$$ ($$j=1,2,\ldots,n$$).

### Proof

All the poles of the function $$\sum_{\lambda\in I} f_{1}^{l_{\lambda, 1}}f_{2}^{l_{\lambda, 2}}\cdots\, f_{n}^{l_{\lambda, n}}$$ are generated by the poles of the functions $$f_{j}$$ ($$j=1,2,\ldots,n$$), and every pole of multiplicity k of $$f_{j}$$ ($$j=1,2,\ldots,n$$) has order at most $$k\sigma_{j}$$. This implies that

$$n \biggl(r, \sum_{\lambda\in I}f_{1}^{l_{\lambda, 1}}f_{2}^{l_{\lambda, 2}} \cdots\, f_{n}^{l_{\lambda, n}} \biggr)\leq\sum _{j=1}^{n} \sigma_{j}n(r, f_{j}).$$

Thus we obtain

$$N \biggl(r,\sum_{\lambda\in I}f_{1}^{l_{\lambda, 1}}f_{2}^{l_{\lambda, 2}} \cdots\, f_{n}^{l_{\lambda, n}} \biggr) \leq \sum _{j=1}^{n}\sigma_{j}N(r, f_{j}).$$
(7)

We next prove that

$$m \biggl(r,\sum_{\lambda\in I}f_{1}^{l_{\lambda, 1}}f_{2}^{l_{\lambda, 2}} \cdots\, f_{n}^{l_{\lambda, n}} \biggr) \leq \sum _{j=1}^{n}\sigma_{j}m(r, f_{j})+ \log t,$$
(8)

and we define

$$\left \{ \begin{array}{l@{\quad}l} f^{\ast}_{j}(z)=f_{j}(z),& |f_{j}(z)|>1 , \\ f^{\ast}_{j}(z)=1,& |f_{j}(z)|\leq1, \end{array} \right .$$

for $$j=1,2,\ldots,n$$. Thus we have

\begin{aligned} \biggl\vert \sum_{\lambda\in I}f_{1}^{l_{\lambda, 1}}f_{2}^{l_{\lambda, 2}} \cdots\, f_{n}^{l_{\lambda, n}}\biggr\vert \leq& \sum _{\lambda\in I}\bigl\vert f_{1}^{l_{\lambda, 1}}f_{2}^{l_{\lambda, 2}} \cdots\, f_{n}^{l_{\lambda, n}}\bigr\vert \\ \leq& \sum_{\lambda\in I}\bigl\vert f_{1}^{\ast l_{\lambda, 1}}f_{2}^{\ast l_{\lambda, 2}} \cdots\, f_{n}^{\ast l_{\lambda, n}}\bigr\vert \\ =& \bigl\vert f_{1}^{\ast\sigma_{1}}f_{2}^{\ast\sigma_{ 2}} \cdots\, f_{n}^{\ast\sigma_{n}}\bigr\vert \biggl(\sum _{\lambda\in I} \frac{\vert f_{1}^{\ast l_{\lambda, 1}}f_{2}^{\ast l_{\lambda, 2}}\cdots \, f_{n}^{\ast l_{\lambda, n}}\vert }{\vert f_{1}^{\ast\sigma_{1}} f_{2}^{\ast \sigma_{ 2}}\cdots\, f_{n}^{\ast\sigma_{n}}\vert } \biggr) \\ \leq& t\bigl\vert f_{1}^{\ast\sigma_{1}}f_{2}^{\ast\sigma_{ 2}} \cdots\, f_{n}^{\ast\sigma_{n}}\bigr\vert . \end{aligned}

By the definition of $$m(r,f)$$, we immediately conclude that

\begin{aligned} m \biggl(r,\sum_{\lambda\in I}f_{1}^{l_{\lambda, 1}}f_{2}^{l_{\lambda, 2}} \cdots\, f_{n}^{l_{\lambda, n}} \biggr) =& \frac{1}{2\pi}\int _{0}^{2 \pi}\log^{+}\biggl\vert \sum _{\lambda\in I}f_{1}^{l_{\lambda, 1}}f_{2}^{l_{\lambda, 2}} \cdots\, f_{n}^{l_{\lambda, n}}\biggr\vert \, d\theta \\ \leq& \frac{1}{2\pi}\int_{0}^{2 \pi} \log^{+}\bigl\vert f_{1}^{\ast \sigma_{1}}f_{2}^{\ast\sigma_{ 2}} \cdots\, f_{n}^{\ast \sigma_{n}}\bigr\vert \, d\theta+ \log t \\ =& \sum_{j=1}^{n}\sigma_{j}m(r, f_{j})+\log t. \end{aligned}

By (7) and (8), the assertion follows. □

### Remark 3

If we suppose that $$\alpha_{\lambda}(z)=o(T(r,f_{j}))$$ ($$\lambda\in I$$) hold for all $$j\in\{1,2,\ldots,n\}$$, and denote $$T(r,a_{\lambda})=S(r,f)$$ ($$\lambda \in I$$), then we have the following estimation:

$$T \biggl(r,\sum_{\lambda\in I}\alpha_{\lambda}(z)f_{1}^{l_{\lambda, 1}}f_{2}^{l_{\lambda, 2}} \cdots\, f_{n}^{l_{\lambda, n}} \biggr) \leq \sum _{j=1}^{n}\sigma_{j}T(r, f_{j}) + S(r,f).$$

### Lemma 3

(see [6])

Let $$f(z)$$ be a meromorphic function with order $$\rho= \rho(f)$$, $$\rho< + \infty$$, and c be a fixed non zero complex number, then for each $$\varepsilon> 0$$, we have

$$T\bigl(r, f(z+c)\bigr) = T(r, f) + O\bigl(r^{\rho-1 + \varepsilon}\bigr) + O(\log r).$$

### Lemma 4

(see [3])

Let $$g:(0,+\infty)\rightarrow{R}$$, $$h:(0,+\infty)\rightarrow{R}$$ be monotone increasing functions such that $$g(r)\leq h(r)$$ outside of an exceptional set E of finite linear measure. Then, for any $$\alpha>1$$, there exists $$r_{0}>0$$ such that $$g(r)\leq h(\alpha r)$$ for all $$r_{0}$$.

### Lemma 5

(see [23])

Let f be a transcendental meromorphic function, and $$p(z)=a_{k}z^{k}+a_{k-1}z^{k-1}+\cdots+a_{1}z+a_{0}$$, $$a_{k}\neq0$$, be a nonconstant polynomial of degree k. Given $$0<\delta<|a_{k}|$$, denote $$\lambda= |a_{k}|+\delta$$ and $$\mu=|a_{k}|-\delta$$. Then given $$\varepsilon>0$$ and $$a\in{C}\cup\{\infty\}$$, we have

\begin{aligned}& kn\bigl(\mu r^{k},a,f\bigr)\leq n\bigl(r,a,f\bigl(p(z)\bigr)\bigr) \leq kn\bigl(\lambda r^{k}, a, f\bigr), \\& N\bigl(\mu r^{k},a,f\bigr)+O(\log r)\leq N\bigl(r,a,f\bigl(p(z)\bigr) \bigr)\leq N\bigl(\lambda r^{k}, a, f\bigr)+O(\log r), \\& (1-\varepsilon)T\bigl(\mu r^{k},f\bigr)\leq T\bigl(r,f\bigl(p(z) \bigr)\bigr)\leq (1+\varepsilon)T\bigl(\lambda r^{k},f\bigr), \end{aligned}

for all r large enough.

## 3 Proofs of theorems

### Proof of Theorem 1

We assume that $$(f(z), g(z))$$ is a transcendental meromorphic solution of the system of complex difference equations (2). By the first equation in (2), Lemma 1, Lemma 2, and Lemma 3, we have, for each $$\varepsilon>0$$,

\begin{aligned}& \max\{p,q\}T(r,g) \\& \quad = T \Biggl(r,\sum_{\lambda_{1} \in I_{1}, \mu_{1}\in J_{1}}\alpha_{\lambda_{1}, \mu_{1}}(z) \Biggl(\prod_{\nu=1}^{n}f(z+c_{\nu})^{l_{\lambda_{1}, \nu}} \prod_{\nu=1}^{n}g(z+c_{\nu})^{m_{\mu_{1}, \nu}} \Biggr) \Biggr) + S(r,g) \\& \quad \leq \sum_{\nu=1}^{n} \xi_{1,\nu}T\bigl(r, f(z+c_{\nu})\bigr) + \sum _{\nu =1}^{n} \eta_{1,\nu}T\bigl(r, g(z+c_{\nu})\bigr)+S(r,f)+ S(r,g) \\& \quad = \sum_{\nu=1}^{n} \xi_{1,\nu}T \bigl(r, f(z)\bigr) + O\bigl(r^{\rho(f) -1 + \varepsilon}\bigr) + \sum _{\nu=1}^{n} \eta_{1,\nu}T\bigl(r, g(z)\bigr)+O \bigl(r^{\rho(g) -1 + \varepsilon}\bigr) \\& \qquad {} +O(\log r) + S(r,f)+ S(r,g) \\& \quad = \Biggl(\sum_{\nu=1}^{n} \xi_{1,\nu} \Biggr)T\bigl(r, f(z)\bigr) + \Biggl(\sum _{\nu =1}^{n} \eta_{1,\nu} \Biggr)T\bigl(r, g(z) \bigr) \\& \qquad {} +O\bigl(r^{\rho(g) -1 + \varepsilon}\bigr) + O(\log r) + S(r,f)+ S(r,g) \\& \quad = \sigma_{11}T\bigl(r, f(z)\bigr) + \sigma_{12}T \bigl(r, g(z)\bigr)+ O\bigl(r^{\rho(f) -1 + \varepsilon}\bigr)+O\bigl(r^{\rho(g) -1 + \varepsilon}\bigr) \\& \qquad {}+ O(\log r) + S(r,f)+ S(r,g). \end{aligned}
(9)

By the above inequality, we get, for each $$\varepsilon>0$$,

\begin{aligned}& \bigl(\max\{p,q\}-\sigma_{12} \bigr)T(r,g) \\& \quad \leq \sigma_{11}T\bigl(r, f(z)\bigr)+ O\bigl(r^{\rho(f) -1 + \varepsilon} \bigr)+O\bigl(r^{\rho(g) -1 + \varepsilon}\bigr) \\& \qquad {} + O(\log r) + S(r,f)+ S(r,g). \end{aligned}
(10)

Since $$\max\{p,q\}>\sigma_{12}$$ by the assumption, we have, for each $$\varepsilon>0$$,

\begin{aligned} T(r,g) \leq&\frac{\sigma_{11}}{\max\{p,q\}-\sigma_{12}}T\bigl(r, f(z)\bigr) + O \bigl(r^{\rho(f) -1 + \varepsilon}\bigr)+O\bigl(r^{\rho(g) -1 + \varepsilon}\bigr) \\ &{} + O(\log r) + S(r,f)+ S(r,g). \end{aligned}
(11)

Similarly, by the second equation in (2), we obtain, for each $$\varepsilon>0$$,

\begin{aligned}& \max\{s,t\}T(r,f) \\& \quad = T \Biggl(r,\sum_{\lambda_{2} \in I_{2}, \mu_{2}\in J_{2}}\beta_{\lambda_{2}, \mu_{2}}(z) \Biggl(\prod_{\nu =1}^{n}f(z+c_{\nu})^{l_{\lambda_{2}, \nu}} \prod_{\nu=1}^{n}g(z+c_{\nu})^{m_{\mu_{2}, \nu}} \Biggr) \Biggr) + S(r,f) \\& \quad \leq \sum_{\nu=1}^{n} \xi_{2,\nu}T\bigl(r, f(z+c_{\nu})\bigr) + \sum _{\nu =1}^{n} \eta_{2,\nu}T\bigl(r, g(z+c_{\nu})\bigr)+S(r,f)+ S(r,g) \\& \quad = \sum_{\nu=1}^{n} \xi_{2,\nu}T \bigl(r, f(z)\bigr) + O\bigl(r^{\rho(f) -1 + \varepsilon}\bigr) + \sum _{\nu=1}^{n} \eta_{2,\nu}T\bigl(r, g(z)\bigr) \\& \qquad {} + O\bigl(r^{\rho(g) -1 + \varepsilon}\bigr) +O(\log r) + S(r,f)+ S(r,g) \\& \quad = \Biggl(\sum_{\nu=1}^{n} \xi_{2,\nu} \Biggr)T\bigl(r, f(z)\bigr) + \Biggl(\sum _{\nu =1}^{n} \eta_{2,\nu} \Biggr)T\bigl(r, g(z) \bigr) \\& \qquad {} + O(\log r) + S(r,f)+ S(r,g) \\& \quad = \sigma_{21}T\bigl(r, f(z)\bigr) + \sigma_{22}T \bigl(r, g(z)\bigr)+ O\bigl(r^{\rho(f) -1 + \varepsilon}\bigr)+O\bigl(r^{\rho(g) -1 + \varepsilon}\bigr) \\& \qquad {}+ O(\log r) + S(r,f)+ S(r,g). \end{aligned}
(12)

By (12) and $$\max\{s,t\}>\sigma_{21}$$, we have, for each $$\varepsilon>0$$,

\begin{aligned}& \bigl(\max\{s,t\}-\sigma_{21} \bigr)T(r,f) \\& \quad \leq \sigma_{22}T\bigl(r, g(z)\bigr)+ O\bigl(r^{\rho(f) -1 + \varepsilon} \bigr)+O\bigl(r^{\rho(g) -1 + \varepsilon}\bigr) \\& \qquad {}+ O(\log r) + S(r,f)+ S(r,g) \end{aligned}
(13)

and

\begin{aligned} T(r,f) \leq& \frac{\sigma_{22}}{\max\{s,t\}-\sigma_{21}}T\bigl(r, g(z)\bigr) + O \bigl(r^{\rho(f) -1 + \varepsilon}\bigr)+O\bigl(r^{\rho(g) -1 + \varepsilon}\bigr) \\ &{} + O(\log r) + S(r,f)+ S(r,g). \end{aligned}
(14)

Using (11), we can obtain $$\rho(g)\leq\rho(f)$$. Similarly, we can get $$\rho(f)\leq\rho (g)$$ from (14). Therefore, we have $$\rho(f)=\rho(g)$$.

It follows from (10) and (13) that

\begin{aligned}& \bigl(\max\{p,q\}-\sigma_{12} \bigr) \bigl(\max\{s,t \}-\sigma_{21} \bigr)T(r,f)T(r,g) \\& \quad \leq \sigma_{11}\sigma_{22}T\bigl(r, f(z)\bigr)T \bigl(r, g(z)\bigr)+ o\bigl(T(r,f)T(r,g)\bigr). \end{aligned}
(15)

From (15), we conclude that

$$\bigl(\max\{p,q\}-\sigma_{12}\bigr)\cdot\bigl(\max\{s,t\}- \sigma_{21}\bigr)\leq \sigma_{11}\sigma_{22}.$$

This yields the asserted result. □

### Proof of Theorem 2

We assume that $$(f(z), g(z))$$ is a transcendental meromorphic solution of a system of complex functional equations (5). Let $$C=\max\{|c_{1}|,|c_{2}|,\ldots,|c_{n}|\}$$. According to the first equation in (5), Lemma 2, Lemma 3, and the last assertion of Lemma 5, we get

\begin{aligned}& (1-\varepsilon)T\bigl(\mu r^{k},f\bigr) \\& \quad \leq T\bigl(r,f\bigl(p(z)\bigr)\bigr) \\& \quad = T \Biggl(r,\sum_{\lambda_{1} \in I_{1}, \mu_{1}\in J_{1}}\alpha_{\lambda_{1}, \mu_{1}}(z) \Biggl(\prod_{\nu=1}^{n}f(z+c_{\nu})^{l_{\lambda_{1}, \nu}} \prod_{\nu=1}^{n}g(z+c_{\nu})^{m_{\mu_{1}, \nu}} \Biggr) \Biggr) \\& \quad \leq \sum_{\nu=1}^{n} \xi_{1,\nu}T\bigl(r, f(z+c_{\nu})\bigr) + \sum _{\nu =1}^{n} \eta_{1,\nu}T\bigl(r, g(z+c_{\nu})\bigr)+S(r,f)+ S(r,g) \\& \quad \leq \sum_{\nu=1}^{n} \xi_{1,\nu}T\bigl(r+C, f(z)\bigr) + \sum_{\nu=1}^{n} \eta _{1,\nu}T\bigl(r+C, g(z)\bigr) + S(r,f)+ S(r,g) \\& \quad = \Biggl(\sum_{\nu=1}^{n} \xi_{1,\nu} \Biggr)T\bigl(r+C, f(z)\bigr) + \Biggl(\sum _{\nu=1}^{n}\eta_{1,\nu} \Biggr)T\bigl(r+C, g(z) \bigr)+ S(r,f)+ S(r,g) \\& \quad = \sigma_{11} T\bigl(r+C, f(z)\bigr) + \sigma_{12} T \bigl(r+C, g(z)\bigr)+S(r,f)+S(r,g). \end{aligned}

Since $$T(r+C,f) \leq T(\beta r,f)$$ and $$T(r+C,g) \leq T(\beta r,g)$$ hold for r large enough for $$\beta>1$$, we may assume r to be large enough to satisfy

$$(1-\varepsilon)T\bigl(\mu r^{k},f\bigr)\leq\sigma_{11} (1+ \varepsilon)T(\beta r,f)+\sigma_{12} (1+\varepsilon)T(\beta r,g)$$

outside a possible exceptional set of finite linear measure. By Lemma 4, we know that, whenever $$\gamma>1$$,

$$(1-\varepsilon)T\bigl(\mu r^{k},f\bigr)\leq \sigma_{11} (1+\varepsilon)T(\gamma \beta r,f)+\sigma_{12} (1+\varepsilon)T(\gamma\beta r,g)$$
(16)

holds for all r large enough. Let $$t=\gamma\beta r$$, then the inequality (16) may be written in the form

$$T \biggl(\frac{\mu}{(\gamma\beta)^{k}}t^{k},f \biggr)\leq \frac{\sigma_{11} (1+\varepsilon)}{1-\varepsilon}T(t,f)+\frac{\sigma_{12} (1+\varepsilon)}{1-\varepsilon}T(t,g).$$
(17)

Similarly, by the second equation in (5), for all r large enough and $$\beta>1$$, $$\gamma>1$$, we have

\begin{aligned}& (1-\varepsilon)T\bigl(\mu r^{k},g\bigr) \\& \quad \leq T\bigl(r,g\bigl(p(z)\bigr)\bigr) \\& \quad = T \Biggl(r,\sum_{\lambda_{2} \in I_{2}, \mu_{2}\in J_{2}}\beta_{\lambda_{2}, \mu_{2}}(z) \Biggl(\prod_{\nu =1}^{n}f(z+c_{\nu})^{l_{\lambda_{2}, \nu}} \prod_{\nu=1}^{n}g(z+c_{\nu})^{m_{\mu_{2}, \nu}} \Biggr) \Biggr) \\& \quad \leq \sum_{\nu=1}^{n} \xi_{2,\nu}T\bigl(r, f(z+c_{\nu})\bigr) + \sum _{\nu =1}^{n} \eta_{2,\nu}T\bigl(r, g(z+c_{\nu})\bigr)+S(r,f)+ S(r,g) \\& \quad \leq \sum_{\nu=1}^{n} \xi_{2,\nu}T\bigl(r+C, f(z)\bigr) + \sum_{\nu=1}^{n} \eta _{2,\nu}T\bigl(r+C, g(z)\bigr) + S(r,f)+ S(r,g) \\& \quad = \Biggl(\sum_{\nu=1}^{n} \xi_{2,\nu} \Biggr)T\bigl(r+C, f(z)\bigr) + \Biggl(\sum _{\nu=1}^{n}\eta_{2,\nu} \Biggr)T\bigl(r+C, g(z) \bigr)+ S(r,f)+ S(r,g) \\& \quad = \sigma_{21} T\bigl(r+C, f(z)\bigr) + \sigma_{22} T \bigl(r+C, g(z)\bigr)+S(r,f)+S(r,g) \\& \quad \leq \sigma_{21} (1+\varepsilon)T(\gamma\beta r,f)+ \sigma_{22} (1+\varepsilon)T(\gamma\beta r,g). \end{aligned}

Let $$t=\gamma\beta r$$, we have

$$T \biggl(\frac{\mu}{(\gamma\beta)^{k}}t^{k},g \biggr)\leq \frac{\sigma_{21} (1+\varepsilon)}{1-\varepsilon}T(t,f)+\frac{\sigma_{22} (1+\varepsilon)}{1-\varepsilon}T(t,g).$$
(18)

Letting $$s=\log t + \frac{\log\frac{\mu}{(\gamma\beta)^{k}}}{k-1}$$, then $$t=e^{s} (\frac{\mu}{(\gamma\beta)^{k}} )^{\frac{1}{1-k}}$$ and

\begin{aligned} ks =& k\log t+ \frac{k}{k-1}\log\frac{\mu}{(\gamma\beta)^{k}} \\ =& k\log t+\log\frac{\mu}{(\gamma\beta)^{k}}+\frac{\log\frac{\mu }{(\gamma\beta)^{k}}}{k-1} \\ =& \log\frac{\mu}{(\gamma\beta)^{k}}t^{k}+\log \biggl(\frac{\mu}{(\gamma \beta)^{k}} \biggr)^{\frac{1}{k-1}}. \end{aligned}

So

$$\frac{\mu}{(\gamma\beta)^{k}}t^{k}=e^{ks} \biggl( \frac{\mu}{(\gamma\beta )^{k}} \biggr)^{\frac{1}{1-k}}.$$
(19)

Let $$T(t,f)=T (e^{s} (\frac{\mu}{(\gamma\beta)^{k}} )^{\frac {1}{1-k}},f )=\Phi(s,f)$$, $$T(t,g)=T (e^{s} (\frac{\mu}{(\gamma\beta)^{k}} )^{\frac {1}{1-k}},g )=\Phi(s,g)$$, $$M= \max \{\frac{\sigma_{11} (1+\varepsilon)}{1-\varepsilon},\frac{\sigma_{12} (1+\varepsilon)}{1-\varepsilon},\frac{\sigma_{21} (1+\varepsilon)}{1-\varepsilon},\frac{\sigma_{22} (1+\varepsilon)}{1-\varepsilon} \}$$, then from (17) and (19), we have

\begin{aligned} \Phi(ks,f) =& T \biggl(e^{ks} \biggl(\frac{\mu}{(\gamma\beta)^{k}} \biggr)^{\frac{1}{1-k}},f \biggr) \\ =& T \biggl(\frac{\mu}{(\gamma\beta)^{k}}t^{k},f \biggr) \\ \leq& \frac{\sigma_{11} (1+\varepsilon)}{1-\varepsilon}T(t,f)+\frac{\sigma_{12} (1+\varepsilon)}{1-\varepsilon}T(t,g) \\ \leq& M\Phi(s,f) + M\Phi(s,g). \end{aligned}
(20)

Similarly, from (18), we can get

$$\Phi(ks,g)\leq M\Phi(s,f) + M\Phi(s,g).$$
(21)

The inequalities (20) and (21) hold for all s large enough.

Letting now $$\alpha=\frac{\log2M}{\log k}$$, namely, $$2M=k^{\alpha}$$. Write $$\Psi(s,f)=\frac{\Phi(s,f)}{s^{\alpha}}$$ and $$\Psi(s,g)=\frac{\Phi(s,g)}{s^{\alpha}}$$, thus we have

$$\Psi(ks,f)\leq\frac{1}{2}\Psi(s,f) + \frac{1}{2} \Psi(s,g)$$
(22)

and

$$\Psi(ks,g)\leq\frac{1}{2}\Psi(s,f) + \frac{1}{2} \Psi(s,g).$$
(23)

The inequalities (22) and (23) hold for all s large enough, we may assume that (22) and (23) hold for all $$s\geq s_{0}$$.

Let $$M_{1}=\sup_{s_{0}\leq s \leq k s_{0}}\Psi(s,f)$$ and $$M_{2}=\sup_{s_{0}\leq s \leq k s_{0}}\Psi(s,g)$$, then by (22) and (23) we can obtain

\begin{aligned} \begin{aligned} &\sup_{s\in[ks_{0},k^{2}s_{0}]}\Psi(s,f)=\sup_{s\in[s_{0},ks_{0}]}\Psi (ks,f) \leq\frac{1}{2}\sup_{s\in[s_{0},ks_{0}]}\Psi(s,f) + \frac {1}{2} \sup_{s\in[s_{0},ks_{0}]}\Psi(s,g)\leq\frac{M_{1}}{2}+\frac{M_{2}}{2}, \\ &\sup_{s\in[ks_{0},k^{2}s_{0}]}\Psi(s,g)=\sup_{s\in[s_{0},ks_{0}]}\Psi (ks,g) \leq\frac{1}{2}\sup_{s\in[s_{0},ks_{0}]}\Psi(s,f) + \frac {1}{2} \sup_{s\in[s_{0},ks_{0}]}\Psi(s,g)\leq\frac{M_{1}}{2}+\frac{M_{2}}{2}. \end{aligned} \end{aligned}

Similarly, we have

\begin{aligned}& \sup_{s\in[k^{2}s_{0},k^{3}s_{0}]}\Psi(s,f)=\sup_{s\in [ks_{0},k^{2}s_{0}]}\Psi(ks,f) \leq\frac{1}{2}M_{1}+\frac{1}{2}M_{2}, \\& \sup_{s\in[k^{2}s_{0},k^{3}s_{0}]}\Psi(s,g)=\sup_{s\in [ks_{0},k^{2}s_{0}]}\Psi(ks,g) \leq\frac{1}{2}M_{1}+\frac{1}{2}M_{2}, \\& \ldots. \end{aligned}

Thus, we deduce that

\begin{aligned}& \sup_{s\geq ks_{0}}\Psi(s,f)\leq\frac{1}{2}M_{1}+ \frac {1}{2}M_{2}< +\infty, \\& \sup_{s\geq ks_{0}}\Psi(s,g)\leq\frac{1}{2}M_{1}+ \frac {1}{2}M_{2}<+\infty. \end{aligned}

Therefore, $$\Psi(s,f)$$ and $$\Psi(s,g)$$ are bounded for all $$s\geq s_{0}$$. There exist some constants $$K_{1}$$, $$K_{2}$$, $$K_{3}$$, $$K_{4}$$, such that, for any $$\varepsilon>0$$,

$$T(t,f)=\Phi(s,f)=\Psi(s,f)s^{\alpha}\leq K_{1}s^{\alpha}=K_{1} \biggl(\log t + \frac{\log\frac{\mu}{(\gamma\beta)^{k}}}{k-1} \biggr)^{\alpha }\leq K_{2}( \log t)^{\alpha+\varepsilon}$$

and

$$T(t,g)=\Phi(s,g)=\Psi(s,g)s^{\alpha}\leq K_{3}s^{\alpha}=K_{3} \biggl(\log t + \frac{\log\frac{\mu}{(\gamma\beta)^{k}}}{k-1} \biggr)^{\alpha }\leq K_{4}( \log t)^{\alpha+\varepsilon}.$$

Therefore, we have

$$T(r,f)=O\bigl((\log r)^{\alpha+\varepsilon}\bigr)$$

and

$$T(r,g)=O\bigl((\log r)^{\alpha+\varepsilon}\bigr),$$

where

$$\alpha=\frac{\log 2M}{\log k} = \frac{\log 2\sigma}{\log k} + o(1).$$

Letting now $$\alpha= \frac{\log2\sigma}{\log k}$$, we obtain the required form. Theorem 2 is proved. □