Theorem A Suppose that $f$ is a non-constant meromorphic function in $\mathbb {C}$ and $g$ is an entire function satisfying the equation

\[ f'(z)=a(f\circ g)(z)+bf(z)+c, \]

where $a(\neq 0),$ $b,$ $c$ are constants. Then,

1. (i) $g$ must be linear, if $f$ is transcendental;

2. (ii) $g$ must be a polynomial of degree less than or equal to 2, if $f$ is rational; furthermore, the degree of $g$ is 2 if and only if $f=\frac {\alpha }{z-w_0}+\beta,$ $g=w_0-a(z-w_0)^{2}$ and $b = a\beta + c = 0,$ where $\alpha (\neq 0),$ $\beta,$ $w_0$ are complex numbers.

Theorem 1 Suppose that $f$ is a non-constant rational function in $\mathbb {C}$ and $g$ is a non-constant entire function satisfying the equation

(1.5)\begin{equation} f^{(k)}(z)=a(f^{n}\circ g)(z)+bf(z)+c, \end{equation}

where $a(\neq 0),$ $b,$ $c$ are constant. Then, $g$ is a polynomial. If $g$ is nonlinear, then

\[ g(z)=A(z-a_1)^{\mu}+a_1,\quad f(z)=A_0+\sum_{j=1}^{q}\frac{A_{j}}{(z-a_1)^{j}}, \]

where $a_1,$ $A,$ $A_0$ and $A_j$ are constants, $\mu \in \{2,\, ~3,\,\ldots,\,~[\tfrac {k+1}{n}]\},$ $q=\frac {k}{n\mu -1}$ is an integer.

Theorem 2 Suppose that $f$ is a transcendental meromorphic function in $\mathbb {C}$ and $g$ is a non-constant entire function satisfying the equation

(1.6)\begin{equation} f^{(k)}(z)=a(f^{n}\circ g)(z)+bf(z)+c, \end{equation}

where $a(\neq 0),$ $b,$ $c$ are polynomials in $\mathbb {C}$. Then $g$ must be linear. Further, Equation (1.6) does not admit any transcendental meromorphic solution if $n\geq 2$ and $\frac {k}{n-1}$ is not an integer.

Proof of Theorem 1. Clearly, $g$ is a polynomial, since $f$ is a rational function. Firstly, suppose that $f$ is a polynomial. Comparing the degrees of both sides of Equation (1.5), one has $n=1$ and $\deg g=1$, which means that $g$ is linear. Next, we assume $f$ is a non-polynomial rational function, which means that $f$ has at least one pole. We below characterize the forms of $f$ and $g$. Set $f(z)=\frac {P_1(z)}{Q(z)}$, where $P_1,\,~Q$ are two co-prime polynomials. By decomposition of rational function, we then can set

(1.8)\begin{equation} f(z)=P(z)+R(z)=P(z)+\displaystyle\sum_{i=1}^{s} \sum_{j=1}^{m_i}\displaystyle\frac{A_{ij}}{(z-a_i)^{j}} \end{equation}

where $P(z)$ is a polynomial, $s$ and $m_i$ are positive integers and $A_{ij}$ and $a_i~ (i=1,\,\ldots ~s)$ are constant with $A_{i m_i}\neq 0$ for all $i=1,\,\ldots,\,~s$. In addition, without loss of generality, we suppose that $a_i$ and $a_j$ are distinct for any $1\leq i\neq j\leq s$, and $m_1 \leq m_2 \leq m_3 \leq \ldots \leq m_s$. The above form of $f$ implies that $f$ has distinct poles $a_1,\,~a_2,\,\ldots,\, ~a_s$, with orders (or multiplicities) $m_1,\,~ m_2,\,\ldots,\, ~ m_s$ respectively. By differentiating the function $f(z)$ $k$-times, we obtain

(1.9)\begin{equation} f^{(k)}(z)=P^{(k)}(z)+\displaystyle\sum_{i=1}^{s} \sum_{j=k+1}^{m_i+k}\displaystyle\frac{B_{ij}}{(z-a_i)^{j}}, \end{equation}

where $B_{ij}$ are constants and $B_{i (m_i+k)}\neq 0$ for $i=1,\,\ldots,\,s$. Combining (1.8) and (1.9) yields

\begin{align*} & \deg [f^{(k)}-bf-c]\\ & \quad = \deg\bigg[P^{(k)}(z)-bP(z)-c+\sum_{i=1}^{s} \sum_{j=k+1}^{m_i+k}\frac{B_{ij}}{(z-a_i)^{j}}-b\sum_{i=1}^{s} \sum_{j=1}^{m_i}\frac{A_{ij}}{(z-a_i)^{j}}\bigg]\\ & \quad \leq \deg f +ks. \end{align*}

(In fact, if $b\neq 0$, then $\deg [f^{(k)}-bf-c]=\deg f +ks$.) On the other hand, $\deg [(f^{n}\circ g)]=n m\deg f$, where $m=\deg g$. Plus $s\leq \deg f-\deg P\leq \deg f$, one has that

(1.10)\begin{equation} \begin{aligned} n m\deg f & =\deg [(f^{n}\circ g)]=\deg [f^{(k)}-bf-c] \\ & \leq \deg f +ks\leq \deg f +k\deg f=(1+k)\deg f, \end{aligned} \end{equation}

which implies that $m\leq \frac {1+k}{n}$. Clearly, if $n>k+1$, the above inequality is invalid and Equation (1.5) does not admit any rational function $f$. Below, we will prove the theorem under the condition $n\leq k+1$.

From Equation (1.5), we see that all the zeros of $g(z)-a_1$ must be the poles of $f(z)$, which implies that each zero of $g(z)-a_1$ belongs to set $\{a_1,\,~a_2,\,\ldots,\, ~a_s\}$. Without loss of generality, assume that $g(a_t)=a_1$, where $t$ is a fixed integer such that $t\in \{1,\,\ldots,\,s\}$. Suppose that $a_t$ is a zero of $g(z)-a_1$ with multiplicity $\mu$. We consider the following cases.

Case 1. $\mu =1$. Then, we can assume that $g(z)=a_1+h(z)(z-a_t)$, where $h(z)$ is a polynomial and $h(a_t)\neq 0$. Further, substituting the forms of $f$ and $g$ into Equation (1.5), one has

(1.11)\begin{align} & f^{(k)}(z)-bf(z)-c\notag\\ & \quad =P^{(k)}(z)-bP(z)-c+\displaystyle\sum_{i=1}^{s} \sum_{j=k+1}^{m_i+k}\displaystyle\frac{B_{ij}}{(z-a_i)^{j}}-b\sum_{i=1}^{s} \sum_{j=1}^{m_i}\frac{A_{ij}}{(z-a_i)^{j}} \notag\\ & \quad=a(f^{n}\circ g)(z)=a\bigg[P(g(z))+\displaystyle\sum_{i=1}^{s} \sum_{j=1}^{m_i}\displaystyle\frac{A_{ij}}{(g(z)-a_i)^{j}}\bigg]^{n}\notag\\ & \quad=a\bigg[P(g(z))+\sum_{j=1}^{m_1}\frac{A_{1j}}{(g(z)-a_1)^{j}}+\sum_{i=2}^{s} \sum_{j=1}^{m_i}\frac{A_{ij}}{(g(z)-a_i)^{j}}\bigg]^{n}\notag\\ & \quad =a\bigg[P(g(z))+\sum_{j=1}^{m_1}\frac{A_{1j}}{h^{j}(z)(z-a_t)^{j}}+\sum_{i=2}^{s} \sum_{j=1}^{m_i}\frac{A_{ij}}{[a_1-a_i+h(z)(z-a_t)]^{j}}\bigg]^{n}. \end{align}

The first line of (1.11) implies that $a_t$ is a pole of $f^{(k)}(z)-bf(z)-c$ with multiplicity $m_t+k$. And the last line of (1.11) yields that $a_t$ is a pole of $f^{(k)}(z)-bf(z)-c=a(f^{n}\circ g)(z)$ with multiplicity $n m_1$. Therefore, Equation (1.5) leads to

\[ k+m_t=n m_1\geq k+ m_1. \]

The above equality does not hold for $n=1$. So $n\geq 2$ and $m_1\geq \frac {k}{n-1}$, which implies that $f$ has poles with multiplicity at least $q$, where

\[ q=\left\{\begin{array}{@{}ll} \dfrac{k}{n-1}, & \text{if }\dfrac{k}{n-1} \text{ is an integer}\\ \left[\dfrac{k}{n-1}\right]+1, & \text{if }\dfrac{k}{n-1}\text{ is not an integer}, \\ \end{array} \right. \]

where $[x]$ denotes the greatest integer less than or equal to $x$. Thus, $s\leq \frac {\deg f-\deg P}{q}\leq \frac {\deg f}{q}$. By the first part of (1.10), we then have

(1.12)\begin{equation} n m\deg f=\deg f +ks\leq \deg f +k\displaystyle\frac{\deg f}{q}\leq \left(1+\frac{k}{q}\right)\deg f, \end{equation}

Clearly, if $\frac {k}{n-1}$ is an integer, then $(1+\frac {k}{q})=n$. Together with the above inequality (1.12), one gets that $\deg g=m=1$. Further, the above inequalities in (1.12) become equations. So, $s=\frac {\deg f}{q}$, which implies that $\deg P=0$ and $P$ reduces a constant. If $\frac {k}{n-1}$ is not an integer, then $(1+\tfrac {k}{q})< n$. Plus the above inequality (1.12), one gets that $\deg g=m<1$, a contradiction.

Case 2. $\mu =2$. We then set

(1.13)\begin{equation} g(z)=a_1+h(z)(z-a_t)^{2}, \end{equation}

where $h(z)$ is a polynomial and $h(a_t)\neq 0$ (For the sake of simplicity, we still use this notation $h$). Similarly as in Case 1, comparing the multiplicities of both sides of equation $f^{(k)}(z)-bf(z)-c=a(f^{n}\circ g)(z)$ at the pole-point $a_t$ we can get

\[ k+m_t=2 n m_1\geq k+m_1, \]

which implies that $m_1\geq \frac {k}{2n-1}$. The same argument as in Case 1 yields that $f$ has poles with multiplicity at least $p$, where

\[ p=\left\{ \begin{array}{@{}ll} \dfrac{k}{2n-1}, & \text{if }\dfrac{k}{2n-1} \text{ is an integer;} \\ \left[\dfrac{k}{2n-1}\right]+1, & \text{if }\dfrac{k}{2n-1} \text{ is not an integer}. \\ \end{array} \right. \]

Thus, $s\leq \frac {\deg f-\deg P}{p}\leq \frac {\deg f}{p}$. Again by the first part of (1.10), we have

(1.14)\begin{equation} n m\deg f=\deg f +ks\leq \deg f +k\displaystyle\frac{\deg f}{p}\leq \left(1+\frac{k}{p}\right)\deg f. \end{equation}

Obviously, if $\frac {k}{2n-1}$ is an integer, then $(1+\frac {k}{p})=2n$. Then, the above inequality (1.14) yields that $\deg g=m\leq 2$. Together with the form of $g$ in (1.13), we have that $\deg g=2$ and

\[ g(z)=h(z-a_t)^{2}+a_1, \]

where $h$ reduces to a constant. Further, $\deg P=0$ and $P$ reduces to a constant. We claim that $a_t=a_1$. Otherwise, suppose that $a_t\neq a_1$. Below we derive a contradiction. It is pointed out that all the inequalities in (1.14) become equations. Thus, $s=\frac {\deg f}{p}$. Plus the fact $f$ has poles with multiplicity at least $p=\frac {k}{2n-1}$, we derive that any pole of $f$ has multiplicity $p=\frac {k}{2n-1}$. This means $m_1=m_2=\cdots =m_s=p$. Observe that all the zeros of $g(z)-a_t$ must be the poles of $f$. Then, there exists an index $\nu \in \{2,\,\ldots,\, t\}$ such that $g(a_{\mu })=a_t$. Clearly, $a_{\mu }$ is a simple zero of $g(z)-a_t$. Then, Similarly as above discussion, again comparing the multiplicities of both sides of the equation $f^{(k)}(z)-bf(z)-c=a(f^{n}\circ g)(z)$ at the pole-point $a_{\mu }$, we have

\[ k+p=k+m_{\mu}=n m_t=np, \]

which implies that $p=\frac {k}{n-1}$, a contradiction. Therefore, $a_t=a_1$ and $g(z)=h(z-a_1)^{2}+a_1$. Below, we will derive that $f$ just has one pole. Without loss of generality, we assume $f$ has another pole $a_2$ $(\neq a_1)$. Then, Equation (1.5) leads to that $g(a_2)$ is also a pole of $f$. Assume that $g(a_2)=a_{\omega }$. Obviously, $a_2$ is a simple zero of $g(z)-a_\omega$ and $\omega \in \{2,\,\ldots,\,~t\}$. Again comparing the multiplicities of both sides of equation $f^{(k)}(z)-bf(z)-c=a(f^{n}\circ g)(z)$ at the pole-point $a_{2}$, one has

\[ k+p=k+m_2=nm_{\omega}=np, \]

which implies that $p=\frac {k}{n-1}$, a contradiction. Thus, $f$ has just a pole $a_1$ with multiplicity $p=\frac {k}{2n-1}$.

If $\frac {k}{2n-1}$ is not an integer, then $(1+\tfrac {k}{p})<2n$. Plus the above inequality (1.14), one has that $\deg g< 2$, which contradicts the fact $\mu =2$.

Case 3. $\mu \in \{3,\,\ldots,\,~[\tfrac {k+1}{n}]\}$. Then, the same argument as in Case 2 yields that $g(z)=h(z-a_1)^{\mu }+a_1$, $\tfrac {k}{n\mu -1}$ is an integer and $Q$ is a constant. At the same time, $f$ has just one pole $a_1$ with multiplicity $\tfrac {k}{n\mu -1}$. When $\tfrac {k}{n\mu -1}$ is not an integer, we can get a contradiction.

Therefore, all the above discussions yield the desired result and the proof is finished.

Proof of Theorem 2. Based on the idea in [Reference Li and Saleeby8], we will prove the theorem. Assume the functions $f$ and $g$ satisfy Equation (1.6). From Equation (1.6), the facts (2), (3) and (6), one has that

(1.15)\begin{equation} \begin{aligned} nT(r, f\circ g) & =T(r, f^{n}\circ g)\leq T(r, a(f^{n}\circ g))+T\left(r, \displaystyle\frac{1}{a}\right)= T(r, a(f^{n}\circ g))+O(\log r)\\ & =T(r, f^{(k)}-bf-c)+O(\log r)\leq T(r, f^{(k)}-bf)+O(\log r)\\ & \leq m(r, f^{(k)}-bf)+N(r,f^{(k)})+O(\log r)\\ & =m\left(r, \frac{f^{(k)}-bf}{f}\right)+m(r,f)+N(r,f^{(k)})+O(\log r)\\ & \leq m\left(r, \frac{f^{(k)}}{f}\right)+m(r,b)+m(r,f)+N(r,f^{(k)})+O(\log r)\\ & \leq m(r,f)+N(r,f)+k\overline{N}(r,f)+o(T(r,f))\\ & \leq T(r,f )+k\overline{N}(r,f)+o(T(r,f))\\ & \leq T(r,f )+k N(r,f)+o(T(r,f))\\ & \leq (1+k) T(r,f )+o(T(r,f))=(1+k)(1+o(1)) T(r,f), \end{aligned} \end{equation}

outside possibly a set of finite Lebesgue measure. Notice that $f$ is transcendental. We then see that $g$ must be a polynomial, since if $g$ were transcendental, then it is easy to get a contradiction from the fact (1) and (1.15). Suppose that $g=a_m z^{m}+ a_{m_1}z^{m-1}+\cdots + a_1 z + a_0$ with $a_m\neq 0$. Combining the fact (5) and (1.15) yields

\[ nT (r, f)\leq \frac{1}{m}(1+\varepsilon) nT(r, f\circ g)\leq \frac{1+k}{m}(1+\varepsilon)(1+o(1))T(r,f), \]

outside possibly a set of finite Lebesgue measure. This implies that

\[ m\leq \frac{1+k}{n}(1+\varepsilon)(1+o(1)). \]

If $\frac {1+k}{n}<2$, then we get that $m=1$ and $g$ is linear, since $\varepsilon$ can be chosen small enough. For the proof of Theorem 2, it suffices to consider the case $\frac {1+k}{n}\geq 2$. Below we consider two cases.

Case 1. For any positive integer $q$, $f$ just has finitely many poles with multiplicity $q$.

Then, for any fixed integer $q$, we can say that $f$ has poles with multiplicity at least $q$ except finitely many points. Further,

\[ \overline{N}(r,f)\leq \frac{1}{q}N(r,f)+O(\log r)\leq\frac{1}{q}T(r,f)+o(T(r,f)). \]

Substituting the above inequality into (1.15) yields that

(1.16)\begin{equation} \begin{aligned} nT(r, f\circ g) & \leq T(r,f)+k\overline{N}(r,f)+o(T(r,f))\\ & \leq T(r,f )+\displaystyle\frac{k}{q}N(r,f)+o(T(r,f))\\ & \leq \left(1+\frac{k}{q}\right) T(r,f )+o(T(r,f))=\left(1+\frac{k}{q}\right)(1+o(1)) T(r,f ), \end{aligned} \end{equation}

outside possibly a set of finite Lebesgue measure. Furthermore, by the fact (5) again, one has

\[ nT (r, f)\leq \frac{1}{m}(1+\varepsilon) nT(r, f\circ g)\leq \frac{1+\frac{k}{q}}{m}((1+\varepsilon))(1+o(1))T(r,f), \]

outside possibly a set of finite Lebesgue measure. Let $q\rightarrow \infty$, we can derive that $m=n=1$ and $g$ is linear.

Case 2. There exists a positive integer $q$ such that $f$ has infinitely many poles with multiplicity $q$.

Suppose that $P$ is an integer such that $f$ has finitely many poles with multiplicity $< P$ and infinitely many poles with multiplicity $P$. In fact, the number $P$ is the minimum integer which satisfies the assumption of Case 2. We define a set $S$ as follows:

\[ S=\{z: z~ \text{ is a zero of } ~a,~ b,~ c, ~g'~ or~z~ \text{ is a pole of } f~ \text{ with multiplicity } < P\}. \]

Clearly, $S$ is a finite set, since $a$, $b$, $c$ and $g'$ are polynomials. Set $L=\max \{|z|:z\in S\}.$

Assume that $\{z_n\}_{n=1}^{\infty }$ are the poles of $f$ with multiplicity $P$ such that $|z_1|\leq |z_2|\leq \ldots \leq |z_n|\leq \ldots$ and $|z_n|\rightarrow \infty$ as $n\rightarrow \infty$. Note that $g$ is a polynomial. Without loss of generality, we can assume that the modulus of any zero of $g(z)-z_n$ is bigger than $L$ for any $n>N$, where $N$ is an fixed integer. Suppose that $g(a_0)=z_n$ with $n>N$. Then, it follows $a_0\not \in S$, which implies that $a_0$ is neither the pole of $f$ with multiplicity $< P$ nor the zeros of the coefficients $a,\,~b,\,~c$ and $g'$. Moreover, $a_0$ is a simple zero of $g(z)-z_n$. Assume that $g(z)-z_n=h(z)(z-a_0)$, where $h(z)$ is a polynomial with $h(a_0)\neq 0$. Equation (1.6) implies that $a_0$ is also a pole of $f$. Suppose that the multiplicity of $f$ at $a_0$ is $s$. The fact $a_0\not \in S$ yields that $s\geq P$. Then, the Laurent series expansions of $f(z)$ and $f^{(k)}(z)$ at $a_{0}$ are as follows

(1.17)\begin{align} f(z)& =\displaystyle\frac{\alpha_{s}}{(z-a_{0})^{s}}+\frac{\alpha_{s-1}}{(z-a_{0})^{s-1}}+\cdots, \end{align}

(1.18)\begin{align} f^{(k)}(z)& =\displaystyle\frac{\beta_{s}}{(z-a_{0})^{s+k}}+\frac{\beta_{s-1}}{(z-a_{0})^{s+k-1}}+\cdots, \end{align}

where $\alpha _{i}~(i=s,\,~s-1,\,\cdots )$ and $\beta _{i}~(i=s,\,~s-1,\,\cdots )$ are finite complex numbers with $\alpha _{s}\neq 0$ and $\beta _{s}\neq 0$. In fact, $\beta _{s}=(-1)^{k} \alpha _{s} s(s+1)\ldots (s+k-1)$. Suppose that the Laurent series expansion of $f(z)$ at $z_{n}$ is

(1.19)\begin{equation} f(z)=\displaystyle\frac{\theta_{P}}{(z-z_{n})^{P}}+\frac{\theta_{P-1}}{(z-z_{n})^{P-1}}+\cdots, \end{equation}

where $\theta _{i}~(i=P,\,~P-1,\,\cdots )$ are finite complex numbers with $\theta _{P}\neq 0$. Furthermore, by the form of $g(z)$ and (1.19), one has

(1.20)\begin{equation} \begin{aligned} (f^{n}\circ g)(z) & =\left[\displaystyle\frac{\theta_{P}}{(g(z)-z_{n})^{P}}+\frac{\theta_{P-1}}{(g(z)-z_{n})^{P-1}}+\cdots\right]^{n}\\ & =\left[\displaystyle\frac{\theta_{P}}{h(z)^{P}(z-a_{0})^{P}}+\frac{\theta_{P-1}}{h(z)^{P-1}(z-a_{0})^{P-1}}+\cdots\right]^{n}\\ & =\frac{A_1}{(z-a_{0})^{nP}}+\frac{A_2}{(z-a_{0})^{nP-1}}+\ldots, \end{aligned} \end{equation}

where $A_1,\,~A_2,\,\ldots$ are constants and $A_1\neq 0$. In fact, $A_1=\left [\frac {\theta _P}{h(a_0)}\right ]^{n}$. We rewrite Equation (1.6) as

(1.21)\begin{equation} a(f^{n}\circ g)(z)=f^{(k)}(z)-bf(z)-c. \end{equation}

Substitute (1.17), (1.18) and (1.20) into (1.21). Then, by comparing the multiplicities of both sides of Equation (1.21) at the pole-point $a_0$, one has

\[ nP=s+k\geq P+k, \]

which is impossible if $n=1$. When $n\geq 2$, it implies that

\[ P\geq t, \]

where

\[ t=\left\{ \begin{array}{@{}ll} \dfrac{k}{n-1}, & \text{if }\dfrac{k}{n-1}\text{ is an integer}; \\ \left[\dfrac{k}{n-1}\right]+1, & \text{if }\dfrac{k}{n-1}~\text{ is not an integer}. \\ \end{array} \right. \]

The above discussion yields that $f$ has poles with multiplicity at least $t$ except for finitely many points. Then, it follows that

\[ \overline{N}(r,f)\leq \frac{1}{t}N(r,f)+O(\log r)\leq \frac{1}{t}N(r,f)+o(T(r,f)). \]

The similar argument as in (1.16) yields

(1.22)\begin{equation} \begin{aligned} nT(r, f\circ g) & \leq T(r,f)+k\overline{N}(r,f)+o(T(r,f))\\ & \leq T(r,f )+\displaystyle\frac{k}{t}N(r,f)+o(T(r,f))\\ & \leq \left(1+\frac{k}{t}\right) T(r,f )+o(T(r,f))=\left(1+\frac{k}{t}\right)(1+o(1))T(r,f ), \end{aligned} \end{equation}

outside possibly a set of finite Lebesgue measure. Furthermore, plus the fact (5) again, one has

\[ nT (r, f)\leq \frac{1+\frac{k}{t}}{m}(1+\varepsilon)(1+o(1))T(r,f), \]

outside possibly a set of finite Lebesgue measure. If $\frac {k}{n-1}$ is an integer, then $1+\frac {k}{t}=n$. Together with the above inequality, we get that

\[ m n\leq n(1+\varepsilon)(1+o(1)), \]

which implies that $m=1$ and $g$ is linear, since $\varepsilon$ can be taken small enough. Moreover, if $\frac {k}{n-1}$ is not an integer, then $t=[\tfrac {k}{n-1}]+1$ and $1+\frac {k}{t}= \lambda n< n$, where $0<\lambda <1$ is a fixed constant. Again plus the above inequality, one has

\[ m n\leq n\lambda (1+\varepsilon)(1+o(1)), \]

By choosing $\varepsilon$ small enough such that $\lambda (1+\varepsilon )(1+o(1))<1$, then it follows $m <1$, which is impossible. Thus, when $\frac {k}{n-1}$ is not integer, the equation does not admit any transcendental meromorphic function.

Therefore, the proof is finished.

Theorem B Let $g$ be a given non-constant entire function and $P(z,\,y'(z), y''(z),\,\ldots,\,y^{(k)}(z))$ be a given polynomial in variables $z,\,~y(z),\,~y'(z)\ldots,\,y^{(k)}(z)$. If $f$ is a transcendental meromorphic solutions of the equation

(2.1)\begin{equation} P(z,y'(z),y''(z),\ldots,y^{(k)}(z))=(y\circ g)(z), \end{equation}

then $g$ must be polynomial. Furthermore, if $g$ is nonlinear, then $T(r,\,f)=O((\log r)^{\beta })$ as $r\rightarrow \infty$ for some constant $\beta >1$.

Theorem 3 Suppose that $f$ is a transcendental meromorphic function in $\mathbb {C}$ and $g$ is a non-constant entire function satisfying the equation

(2.2)\begin{equation} Q[f](z)=(f^{n}\circ g)(z), \end{equation}

where $n$ is a positive integer and $Q[f]$ is a differential polynomial of total degree $\Lambda$ and total weight $\Gamma,$ respective. Let $\omega =\max \{\Lambda,\, \Gamma \}$. Then $g$ must be a polynomial and the following assertions hold.

1. (i) If $\omega <2n$, then $g$ must be linear;

2. (ii) If $g$ is a polynomial with degree $m\geq 2,$ then

   \[ T(r,f)=O((\log r)^{\beta}), \]
   where $\beta =\frac {\log \frac {\omega }{n}(1+\varepsilon )}{\log m}$ and $\varepsilon$ is any arbitrary positive number.

Proof of Theorem 3. Note that $f$ is transcendental. It is well known that

\[ m(r, Q[f])\leq \Lambda m(r,f)+S(r,f), ~N(r, Q[f])\leq \Gamma N(r,f)+S(r,f), \]

outside possibly a set of finite Lebesgue measure. The fact can be found in [Reference Doeringer3, Lemma 1] and given by Doeringer. Further, from Equation (2.2), one has that

(2.3)\begin{equation} \begin{aligned} nT(r, f\circ g) & =T(r,f^{n}\circ g)= T(r, Q[f])\\ & \leq m(r, Q[f])+N(r,Q[f])\\ & \leq\Lambda m(r,f)+\Gamma N(r,f)+S(r,f)\\ & \leq \omega T(r,f)+S(r,f), \end{aligned} \end{equation}

outside possibly a set of finite Lebesgue measure. Then, together with the fact (1), we then see that $g$ must be a polynomial. Still set $g = a_m z^{m}+ a_{m_1}z^{m-1}+\cdots + a_1 z + a_0$ with $a_m\neq 0$. Then, combining the fact (5) and (2.3) yields

\[ nT (r, f)\leq \frac{1}{m}(1+\varepsilon) nT(r, f\circ g)\leq \frac{\omega}{m}(1+\varepsilon)(1+o(1))T(r,f), \]

outside possibly a set of finite Lebesgue measure. It implies that

\[ m\leq \frac{\omega}{n}(1+\varepsilon). \]

If $\frac {\omega }{n}<2$, then we see that $m=1$ and $g$ is linear by choosing $\varepsilon$ small enough. Thus, the proof of (i) is finished.

Now, we prove the conclusion (ii) under the condition $\omega \geq 2n$. With the above discussion and the fact (4), one has

(2.4)\begin{equation} \begin{aligned} T\left(\displaystyle\frac{|a_m|}{2}r^{m}, f\right) & \leq \frac{1}{1-\epsilon_2}T(r, f\circ g)=\frac{1}{1-\epsilon_2}\frac{1}{n}T(r, (f^{n}\circ g)(z))\\ & \leq \frac{1}{1-\epsilon_2}\frac{\omega}{n}(1+o(1))T(r,f)\leq \frac{1}{1-\epsilon_2}\frac{\omega}{n}(1+\varepsilon_1)T(r,f)\\ & =\frac{\omega}{n}(1+\varepsilon)T(r,f), \end{aligned} \end{equation}

outside possibly a set of finite Lebesgue measure, where $\varepsilon _1$ is an arbitrary positive constant and $\varepsilon =\frac {\varepsilon _1+\epsilon _2}{1-\epsilon _2}$. The form of $\varepsilon$ implies that $\varepsilon$ is also a arbitrary positive number. Next, we need the following result, which is Lemma 1.1.1 in [Reference Laine7].

Set $g(r)=T(\tfrac {|a_m|}{2}r^{m},\, f)$ and $\tfrac {\omega }{n}(1+\varepsilon )T(r,\,f)=h(r)$. Obviously, $g(r)\leq h(r)$ outside possibly a set of finite Lebesgue measure. By Lemma 2.1, for any $\alpha >1$, there exists a positive number $r_0$ such that $g(r)\leq h(\alpha r)$ for all $r>r_0$. That is

(2.5)\begin{equation} T(\displaystyle\frac{|a_m|}{2}r^{m}, f)\leq \frac{\omega}{n}(1+\varepsilon)T(\alpha r,f). \end{equation}

Set $\alpha r=R$. Then, we rewrite (2.5) as

(2.6)\begin{equation} T\left(\displaystyle\frac{|a_m|}{2\alpha ^{m}}R^{m}, f\right)\leq \frac{\omega}{n}(1+\varepsilon)T(R,f),\text{ for }R\geq \alpha r_0. \end{equation}

To end the proof, we will employ a result of Goldstein, which can be found in [Reference Goldstein4, Lemma 3].

Lemma 2.2 Let $\psi (r)$ be a function of $r$ ($r\geq r_0$), positive and bounded in every finite interval. Suppose that

\[ \psi(\mu r^{m})\leq A\psi(r)+B,~~r\geq r_0, \]

where $\mu (>0),$ $m(>1),$ $A(>1),$ $B$ are constants. Then

\[ \psi(r)=O((\log r)^{\alpha}),\text{with}\alpha=\frac{\log A}{\log m}. \]

Applying Lemma 2.2 to the function $T(R,\,f)$, plus the inequality (2.6), we have

\[ T(R,f)=O((\log R)^{\alpha}), \]

where $\alpha =\frac {\log \frac {\omega }{n}(1+\varepsilon )}{\log m}$.

Thus, the proof is finished.