On some complex differential and difference equations concerning sharing function

Abstract

By using the theory of complex differential equations, the purpose of this paper is to investigate a conjecture of Brück concerning an entire function f and its differential polynomial $L(f)=a_k(z)f^{(k)}+\cdots +a_0(z)f$ sharing a function $\alpha (z)$ and a constant β. We also study the problem on entire function and its difference polynomials sharing a function.

MSC:39A50, 30D35.

1 Introduction and main results

Let f be a nonconstant meromorphic function in the whole complex plane ℂ. We shall use the following standard notations of the value distribution theory:

$$T(r,f),m(r,f),N(r,f),\overline{N}(r,f),\dots$$

(see Hayman [1], Yang [2] and Yi and Yang [3]). We denote by $S(r,f)$ any quantity satisfying $S(r,f)=o(T(r,f))$, as $r\to +\mathrm{\infty }$, possibly outside of a set with finite measure. A meromorphic function $a(z)$ is called a small function with respect to f if $T(r,a)=S(r,f)$. In addition, we will use the notation $\sigma (f)$ to denote the order of meromorphic function $f(z)$, and $\tau (f)$ to denote the type of an entire function $f(z)$ with $0<\sigma (f)=\sigma <+\mathrm{\infty }$, which are defined to be (see [1])

$$\sigma (f)=\underset{r\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{logT(r,f)}{logr},\tau (f)=\underset{r\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{logM(r,f)}{r^{\sigma }}.$$

We use ${\sigma }_2(f)$ to denote the hyper-order of $f(z)$, ${\sigma }_2(f)$ is defined to be (see [3])

$${\sigma }_2(f)=\underset{r\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{loglogT(r,f)}{logr}.$$

In 1976, Rubel and Yang [4] proved the following result.

Theorem 1.1 [4]

Let f be a nonconstant entire function. If f and $f^{\prime }$ share two finite distinct values CM, then $f\equiv f^{\prime }$.

In 1996, Brück [5] gave the following conjecture.

Conjecture 1.1 [5]

Let f be a nonconstant entire function. Suppose that $\sigma (f)$ is not a positive integer or infinite, if f and $f^{\prime }$ share one finite value a CM, then

$$\frac{f^{\prime }-a}{f-a}=c$$

for some nonzero constant c.

In 1998, Gundersen and Yang [6] proved that Brück’s conjecture holds for entire functions of finite order and obtained the following result.

Theorem 1.2 [[6], Theorem 1]

Let f be a nonconstant entire function of finite order. If f and $f^{\prime }$ share one finite value a CM, then $\frac{f^{\prime }-a}{f-a}=c$ for some nonzero constant c.

The shared value problems relative to a meromorphic function f and its derivative $f^{(k)}$ have been a more widely studied subtopic of the uniqueness theory of entire and meromorphic functions in the field of complex analysis (see [7–12]).

In 2009, Chang and Zhu [13] further investigated the problem related to Brück’s conjecture and proved that Theorem 1.2 remains valid if the value a is replaced by a function.

Theorem 1.3 [[13], Theorem 1]

Let f be an entire function of finite order and $a(z)$ be a function such that $\sigma (a)<\sigma (f)<\mathrm{\infty }$. If f and $f^{\prime }$ share $a(z)$ CM, then $\frac{f^{\prime }-a}{f-a}=c$ for some nonzero constant c.

Thus, there are natural questions to ask:

• (i) What would happen when $\sigma (a)<\sigma (f)<\mathrm{\infty }$ is replaced by $0<\sigma (a)=\sigma (f)<\mathrm{\infty }$ in Theorem 1.3?

• (ii) For Theorems 1.1-1.3, what would happen when $f^{\prime }$ is replaced by differential polynomial

  $$L(f)=a_k(z)f^{(k)}+a_{k-1}(z)f^{(k-1)}+\cdots +a_1(z)f^{\prime }+a_0(z)f,$$

  (1)

where $a_k(z)(\not\equiv 0),\dots ,a_0(z)$ are polynomials?

The main purpose of this article is to study the above questions and obtain the following theorems.

Theorem 1.4 Let $f(z)$ and $\alpha (z)$ be two nonconstant entire functions and satisfy $0<\sigma (\alpha )=\sigma (f)<\mathrm{\infty }$ and $\tau (f)>\tau (\alpha )$, and $L(f)$ be stated as in (1) such that

$$\sigma (f)>1+max\{\frac{{deg}_z{a}_j-{deg}_z{a}_k}{k-j},0\}.$$

If $f(z)$ and $L(f(z))$ share $\alpha (z)$ CM, then

$$\frac{L(f(z))-\alpha (z)}{f(z)-\alpha (z)}=c$$

for some nonzero constant c.

Theorem 1.5 Let $f(z)$ be a nonconstant transcendental entire function with ${\sigma }_2(f)<\mathrm{\infty }$, let ${\sigma }_2(f)$ be not an integer, and let $L(f)$ be stated as in (1). If f and $L(f)$ share a nonzero constant a CM and $\delta (0,f)>0$, then

$$\frac{L(f(z))-a}{f(z)-a}=c$$

for some nonzero constant c.

Recently, some papers have studied Brück’s conjecture related to difference of entire function (including [14, 15]). In 2009, Heittokangas et al. [14] got the following result.

Theorem 1.6 [[14], Theorem 1]

Let f be a nonconstant meromorphic function of finite order $\sigma (f)<2$, and let η be a nonzero complex number. If $f(z+\eta )$ and $f(z)$ share a finite complex value a CM, then $f(z+\eta )-a=c(f(z)-a)$ for all $z\in \mathbb{C}$, where c is some nonzero complex number.

In this paper, we further investigate Brück’s conjecture related to entire function and its difference polynomial and obtain the following result.

Theorem 1.7 Let $f(z)$ be a nonconstant entire function of finite order $0<\sigma (f)<\mathrm{\infty }$, $L_1(f)$ be difference polynomial of f of the form

$$L_1(f(z))=f(z+{\eta }_k)+f(z+{\eta }_{k-1})+\cdots +f(z+{\eta }_1),$$

where ${\eta }_k,{\eta }_{k-1},\dots ,{\eta }_1$ are nonzero complex numbers. If $L_1(f(z))=cf(z)$ and ξ (≠0) is a Borel exceptional value of $f(z)$, then $L_1(f)=kf(z)$.

2 Some lemmas

To prove our theorems, we will require some lemmas as follows.

Lemma 2.1 [16]

Let $f(z)$ be a transcendental entire function, $\nu (r,f)$ be the central index of $f(z)$. Then there exists a set $E\subset (1,+\mathrm{\infty })$ with finite logarithmic measure, we choose z satisfying $|z|=r\notin [0,1]\cup E$ and $|f(z)|=M(r,f)$, we get

$$\frac{f^{(j)}(z)}{f(z)}={\left\{\frac{\nu (r,f)}{z}\right\}}^j(1+o(1))\mathit{\text{for }}j\in N.$$

Lemma 2.2 [17]

Let $f(z)$ be an entire function of finite order $\sigma (f)=\sigma <\mathrm{\infty }$, and let $\nu (r,f)$ be the central index of f. Then, for any ε (>0), we have

$$\underset{r\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{log\nu (r,f)}{logr}=\sigma .$$

Lemma 2.3 [18]

Let f be a transcendental entire function, and let $E\subset [1,+\mathrm{\infty })$ be a set having finite logarithmic measure. Then there exists $\{z_n=r_n{e}^{i{\theta }_n}\}$ such that $|f(z_n)|=M(r_n,f)$, ${\theta }_n\in [0,2\pi )$, ${lim}_{n\to \mathrm{\infty }}{\theta }_n={\theta }_0\in [0,2\pi )$, $r_n\notin E$ and if $0<\sigma (f)<\mathrm{\infty }$, then, for any given $\epsilon >0$ and sufficiently large $r_n$,

$$r_n^{\sigma (f)-\epsilon }<\nu (r_n,f)<r_n^{\sigma (f)+\epsilon }.$$

Lemma 2.4 [16]

Let $P(z)=b_n{z}^n+b_{n-1}{z}^{n-1}+\cdots +b_0$ with $b_n\ne 0$ be a polynomial. Then, for every $\epsilon >0$, there exists $r_0>0$ such that for all $r=|z|>r_0$ the inequalities

$$(1-\epsilon )|b_n|r^n\le |P(z)|\le (1+\epsilon )|b_n|r^n$$

hold.

Lemma 2.5 Let $f(z)$ and $A(z)$ be two entire functions with $0<\sigma (f)=\sigma (A)=\sigma <\mathrm{\infty }$, $0<\tau (A)<\tau (f)<\mathrm{\infty }$, then there exists a set $E\subset [1,+\mathrm{\infty })$ that has infinite logarithmic measure such that for all $r\in E$ and a positive number $\kappa >0$, we have

$$\frac{M(r,A)}{M(r,f)}<exp\{-\kappa r^{\sigma }\}.$$

Proof By definition, there exists an increasing sequence $\{r_m\}\to \mathrm{\infty }$ satisfying $(1+\frac{1}{m})r_m<r_{m+1}$ and

$$\underset{m\to \mathrm{\infty }}{lim}\frac{logM(r_m,f)}{r_m^{\sigma }}=\tau (f).$$

(2)

For any given β ($\tau (A)<\beta <\tau (f)$), there exists some positive integer $m_0$ such that for all $m\ge m_0$ and for any given ε ($0<\epsilon <\tau (f)-\beta$), we have

$$logM(r_m,f)>(\tau (f)-\epsilon )r_m^{\sigma }.$$

(3)

Thus, there exists some positive integer $m_1$ such that for all $m\ge m_1$, we have

$${\left(\frac{m}{m+1}\right)}^{\sigma }>\frac{\beta }{\tau (f)-\epsilon }.$$

(4)

From (2)-(4), for all $m\ge m_2=max\{m_0,m_1\}$ and for any $r\in [r_m,(1+\frac{1}{m})r_m]$, we have

$$\begin{array}{rl}M(r,f)& \ge M(r_m,f)>exp\left\{(\tau (f)-\epsilon )r_m^{\sigma }\right\}\\ \ge exp\left\{(\tau (f)-\epsilon ){\left(\frac{m}{m+1}r\right)}^{\sigma }\right\}>exp\left\{\beta r^{\sigma }\right\}.\end{array}$$

(5)

Set $E={\bigcup }_{m=m_2}^{\mathrm{\infty }}[r_m,(1+\frac{1}{m})r_m]$, then

$$m_{l}E=\sum _{m=m_2}^{\mathrm{\infty }}{\int }_{r_m}^{(1+\frac{1}{m})r_m}\frac{dt}{t}=\sum _{m=m_2}^{\mathrm{\infty }}log(1+\frac{1}{m})=\mathrm{\infty }.$$

From the definition of type of entire function, for any sufficiently small $\epsilon >0$, we have

$$M(r,A)<exp\left\{(\tau (A)+\epsilon )r^{\sigma }\right\}.$$

(6)

By (5) and (6), set $\kappa =\beta -\tau (A)-\epsilon$, for all $r\in E$, we have

$$\frac{M(r,A)}{M(r,f)}<exp\{-(\beta -\tau (A)-\epsilon )r^{\sigma }\}=e^{-\kappa r^{\sigma }}.$$

Thus, this completes the proof of this lemma. □

Lemma 2.6 [[19], Theorem 2.1]

Let $f(z)$ be a meromorphic function of finite order σ, and let η be a fixed nonzero complex number, then, for each $\epsilon >0$, we have

$$m(r,\frac{f(z+c)}{f(z)})+m(r,\frac{f(z)}{f(z+c)})=O\left(r^{\sigma -1+\epsilon }\right).$$

Lemma 2.7 [[19], Corollary 2.5]

Let $f(z)$ be a meromorphic function with order $\sigma =\sigma (f)$, $\sigma <+\mathrm{\infty }$, and let η be a fixed nonzero complex number, then, for each $\epsilon >0$, we have

$$T(r,f(z+\eta ))=T(r,f)+O\left(r^{\sigma -1+\epsilon }\right)+O(logr).$$

Lemma 2.8 [1, 20]

Let $g:(0,+\mathrm{\infty })\to R$, $h:(0,+\mathrm{\infty })\to R$ be monotone increasing functions such that $g(r)\le h(r)$ outside of an exceptional set E with finite linear measure, or $g(r)\le h(r)$, $r\notin H\cup (0,1]$, where $H\subset (1,\mathrm{\infty })$ is a set of finite logarithmic measure. Then, for any $\alpha >1$, there exists $r_0$ such that $g(r)\le h(\alpha r)$ for all $r\ge r_0$.

3 The proof of Theorem 1.4

Since $f(z)$ is an entire function, and $f(z)$ and $L(f(z))$ share $\alpha (z)$ CM, then there is an entire function $\gamma (z)$ such that

$$\frac{L(f(z))-\alpha (z)}{f(z)-\alpha (z)}=e^{\gamma (z)}.$$

(7)

Next, we will claim that $\gamma (z)$ is a constant.

Suppose that $\gamma (z)$ is transcendental. It follows that $\sigma (e^{\gamma (z)})=\mathrm{\infty }$. However, since $0<\sigma (f)=\sigma (\alpha )<\mathrm{\infty }$, it follows from the left-hand side of (7) that $\sigma (\frac{L(f(z))-\alpha (z)}{f(z)-\alpha (z)})<\mathrm{\infty }$, a contradiction. Thus, $\gamma (z)$ is not transcendental.

Suppose that $\gamma (z)$ is a nonconstant polynomial, let

$$\gamma (z)=b_m{z}^m+b_{m-1}{z}^{m-1}+\cdots +b_0,$$

where $b_m,\dots ,b_0$ are constants and $b_m\ne 0$, $m\ge 1$. Thus, it follows from (7) and Lemma 2.4 that

$$|b_m|r^m(1+o(1))=|\gamma (z)|=|log\frac{\frac{L(f(z))}{f(z)}-\frac{\alpha (z)}{f(z)}}{1-\frac{\alpha (z)}{f(z)}}|.$$

(8)

Since $L(f)=a_k{f}^k+a_{k-1}{f}^{(k-1)}+\cdots +a_{0}f$, from Lemma 2.1, then there exists a subset $E_1\subset (1,+\mathrm{\infty })$ with finite logarithmic measure such that for some point $z=r{e}^{i\theta }$ ($\theta \in [0,2\pi )$), $r\notin E_1$ and $M(r,f)=|f(z)|$, we have

$$\frac{f^{(j)}(z)}{f(z)}={\left\{\frac{\nu (r,f)}{z}\right\}}^j(1+o(1)),1\le j\le k.$$

Thus, it follows that

$$\begin{array}{rl}\frac{L(f(z))}{f(z)}& =a_k{\left\{\frac{\nu (r,f)}{z}\right\}}^k(1+o(1))+\cdots +a_1\left\{\frac{\nu (r,f)}{z}\right\}(1+o(1))+a_0\\ =\frac{a_k}{z^k}(1+o(1))[\nu {(r,f)}^k+\sum _{j=1}^k\frac{a_{k-j}}{a_k}{z}^j\nu {(r,f)}^{k-j}(1+o(1))].\end{array}$$

(9)

From Lemma 2.3, there exists $\{z_n=r_n{e}^{i{\theta }_n}\}$ such that $|f(z_n)|=M(r_n,f)$, ${\theta }_n\in [0,2\pi )$, ${lim}_{n\to \mathrm{\infty }}{\theta }_n={\theta }_0\in [0,2\pi )$, $r_n\notin E_1$, then, for any given ε satisfying

$$0<\epsilon <\underset{1\le j\le k}{min}\frac{j\sigma (f)-j-d_{k-j}}{3k-j},$$

where $d_{k-j}={deg}_z{a}_{k-j}-{deg}_z{a}_k$, and sufficiently large $r_n$, we have

$$r_n^{\sigma (f)-\epsilon }<\nu (r_n,f)<r_n^{\sigma (f)+\epsilon }.$$

(10)

Since $a_j(z)$, $j=0,1,\dots ,k$, are polynomials, let $a_j(z)={\sum }_{t=0}^{s_j}{l}_{jt}{z}^t$, where $s_j={deg}_z{a}_j$, $j=0,1,\dots ,k$. Then, from Lemma 2.4 and (10), we have

$$\begin{array}{rl}\left|\frac{a_{k-j}}{a_k}{z}^j\nu {(r,f)}^{k-j}(1+o(1))\right|& \le M\frac{|l_{k-j,s_{k-j}}|r_n^{s_{k-j}}}{|l_{k,s_k}|r_n^{s_k}}{r}_n^j{r}_n^{(\sigma (f)+\epsilon )(k-j)}\\ =M\frac{|l_{k-j,s_{k-j}}|}{|l_{k,s_k}|}{r}_n^{d_{k-j}+j+(\sigma (f)+\epsilon )(k-j)}\\ \le M\frac{|l_{k-j,s_{k-j}}|}{|l_{k,s_k}|}{r}_n^{k\sigma (f)-j\sigma (f)+d_{k-j}+j+(k-j)\epsilon },\end{array}$$

(11)

where $d_{k-j}=s_{k-j}-s_k$ and M is a positive constant. Since $-j\sigma (f)+d_{k-j}+j+(k-j)\epsilon <-2k\epsilon <0$, it follows from (11) that

$$\left|\frac{a_{k-j}}{a_k}{z}^j\nu {(r,f)}^{k-j}(1+o(1))\right|<M\frac{|l_{k-j,s_{k-j}}|}{|l_{k,s_k}|}{r}_n^{k(\sigma (f)-2\epsilon )},r_n\notin E_1.$$

(12)

Since $0<\sigma (\alpha )=\sigma (f)<\mathrm{\infty }$ and $\tau (\alpha )<\tau (f)<\mathrm{\infty }$, from Lemma 2.5, there exists a set $E\subset [1,+\mathrm{\infty })$ that has infinite logarithmic measure such that for a sequence ${\{r_n\}}_1^{\mathrm{\infty }}\in E_2=E-E_1$, we have

$$\frac{M(r,\alpha )}{M(r,f)}<exp\{-\kappa r_n^{\sigma (f)}\}\to 0\text{as }n\to \mathrm{\infty }.$$

(13)

From (8), (9), (12), (13) and Lemma 2.2, we can get that

$$|b_m|r_n^m(1+o(1))=|\gamma (z)|=O(log{r}_n),$$

(14)

which is impossible. Thus, $\gamma (z)$ is not a polynomial.

Therefore, $\gamma (z)$ is a constant, that is, there exists some nonzero constant c such that

$$\frac{L(f(z))-\alpha (z)}{f(z)-\alpha (z)}=c.$$

Thus, this completes the proof of Theorem 1.4.

4 The proof of Theorem 1.5

Since $L(f)$ and f share the constant a CM, then there exists an entire function $\phi (z)$ such that

$$\frac{L(f)-a}{f-a}=e^{\phi }.$$

(15)

We will consider two cases as follows.

Case 1. If $a=0$, it follows from (15) that

$$\frac{L(f(z))}{f(z)}=e^{\phi (z)}.$$

(16)

Since $L(f(z))=a_k(z)f^{(k)}(z)+\cdots +a_1(z)f^{\prime }(z)+a_0(z)$ and $a_j(z)$, $j=0,1,\dots ,k$, are polynomials, it follows from (16) that

$$T(r,e^{\phi })=m(r,e^{\phi })=m(r,\frac{L(f)}{f})\le \sum _{j=1}^{k}m(r,\frac{f^{(j)}}{f})+\sum _{i=0}^{k}m(r,a_i)=O(logrT(r,f)),$$

outside of an exceptional set $E_3$ with finite linear measure. Thus, there exists a constant K such that

$$T(r,e^{\phi })\le Klog(rT(r,f))\text{for }r\notin E_3.$$

By Lemma 2.8, there exists an $r_0>0$, and for all $r\ge r_0$, we have

$$T(r,e^{\phi })\le Klog(\zeta rT(\zeta r,f))\text{for any }\zeta >1.$$

(17)

Thus, we can deduce from (17) that $\sigma (e^{\phi })\le {\sigma }_2(f)<\mathrm{\infty }$, that is, $\phi (z)$ is a polynomial.

By using the same argument as in [[21], Theorem 1.1], we can get that ${\sigma }_2(f)={deg}_z\phi$, which is a contradiction to ${\sigma }_2(f)$ is not a positive integer. Thus, $\phi (z)$ is only a constant, it follows from (15) that $L(f(z))=cf(z)$, where c is a nonzero constant.

Case 2. If $a\ne 0$, from the derivation of (15) and eliminating $e^{\phi }$, we can get

$${\phi }^{\prime }(z)=\frac{L^{\prime }(f(z))}{L(f(z))-a}-\frac{f^{\prime }(z)}{f(z)-a}.$$

(18)

If ${\phi }^{\prime }(z)\equiv 0$, that is, $\phi (z)\equiv c$, c is a constant. Thus, we can prove the conclusion of Theorem 1.5 easily.

If ${\phi }^{\prime }(z)\not\equiv 0$, then it follows from (18) that

$$m(r,{\phi }^{\prime })=S(r,f).$$

(19)

We can rewrite (18) in the following form:

$$\begin{array}{rl}{\phi }^{\prime }& =f[\frac{L(f)}{f}\frac{1}{L(f)}\frac{L^{\prime }(f)}{L(f)-a}-\frac{1}{f}\frac{f^{\prime }}{f-a}]\\ =\frac{f}{a}[\frac{L(f)}{f}\frac{L^{\prime }(f)}{L(f)-a}-\frac{L^{\prime }(f)}{f}-\frac{f^{\prime }}{f-a}+\frac{f^{\prime }}{f}].\end{array}$$

(20)

Since ${\phi }^{\prime }\not\equiv 0$ and f is transcendental, set

$$\mathrm{\Psi }:=\frac{L(f)}{f}\frac{L^{\prime }(f)}{L(f)-a}-\frac{L^{\prime }(f)}{f}-\frac{f^{\prime }}{f-a}+\frac{f^{\prime }}{f},$$

(21)

then we have $m(r,\mathrm{\Psi })=S(r,f)$. Thus, it follows from (20) and (21) that

$$\frac{a}{f(z)}=\frac{\mathrm{\Psi }(z)}{{\phi }^{\prime }(z)}.$$

(22)

Since $\phi (z)$ is an entire function, from (18)-(22), then we have

$$\begin{array}{rl}m(r,\frac{1}{f})& \le m(r,\mathrm{\Psi })+m(r,\frac{1}{{\phi }^{\prime }})\le S(r,f)+T(r,{\phi }^{\prime })\\ =S(r,f)+m(r,{\phi }^{\prime })=S(r,f).\end{array}$$

It follows that

$$\delta (0,f)=\underset{r\to \mathrm{\infty }}{lim\hspace{0.17em}inf}\frac{m(r,\frac{1}{f})}{T(r,f)}=0,$$

which is a contradiction to the assumption of Theorem 1.5.

Thus, from Case 1 and Case 2, we complete the proof of Theorem 1.5.

5 The proof of Theorem 1.7

Since $f(z)$ is an entire function of finite order $0<\sigma (f)<\mathrm{\infty }$ and ξ (≠0) is a Borel exceptional value of $f(z)$, then $f(z)$ can be written in the form

$$f(z)=\xi +p(z)e^{h(z)},$$

(23)

where $h(z)$ is a polynomial of degree l and $p(z)$ is an entire function satisfying $\sigma (p(z))<\sigma (f(z))={deg}_{z}h(z)=l$. Thus, we have

$$f(z+{\eta }_j)=\xi +p(z+{\eta }_j)e^{h(z+{\eta }_j)},j=1,2,\dots ,k.$$

(24)

From Lemma 2.7, we have $\sigma (p(z+{\eta }_j))<\sigma (f(z+{\eta }_j))=\sigma (f(z))$ and ${deg}_{z}h(z+{\eta }_j)={deg}_{z}h(z)=l$ for $j=1,2,\dots ,k$. Since $L_1(f(z))=cf(z)$, it follows from (23) and (24) that

$$\sum _{j=1}^{k}p(z+{\eta }_j)e^{h(z+{\eta }_j)}=(c-k)d+cp(z)e^{h(z)}.$$

(25)

Set $h(z)={\mu }_l{z}^l+\cdots$ and ${\mu }_l\ne 0$, then we can deduce from (25) that

$$\sum _{j=1}^{k}p(z+{\eta }_j)e^{{\mu }_{m-1}^j{z}^{m-1}+\cdots }=\frac{(c-k)d+cp(z)e^{h(z)}}{e^{{\mu }_l{z}^l}}.$$

(26)

Let $\mathrm{\Phi }:={\sum }_{j=1}^{k}p(z+{\eta }_j)e^{{\mu }_{m-1}^j{z}^{m-1}+\cdots }$, it is easy to see that $\mathrm{\Phi }\not\equiv 0$ and $\sigma (\mathrm{\Phi })<\sigma (f)$, that is, $T(r,\mathrm{\Phi })=o(T(r,f))=o(T(r,e^{h(z)}))$.

Suppose that $c\ne k$. Since $\xi \ne 0$, it follows from (26) that

$$N(r,\frac{1}{e^{h(z)}-\frac{(c-k)\xi }{cp(z)}})=N(r,\frac{1}{\mathrm{\Phi }})\le T(r,\mathrm{\Phi })=S(r,e^{h(z)}).$$

By the second fundamental theorem concerning small functions, for any $\epsilon >0$, we have

$$\begin{array}{rl}T(r,e^{h(z)})& \le N(r,\frac{1}{e^{h(z)}-\frac{(c-k)\xi }{cp(z)}})+\epsilon T(r,e^{h(z)})+S(r,e^{h(z)})\\ =\epsilon T(r,e^{h(z)})+S(r,e^{h(z)}).\end{array}$$

Since ε is arbitrary, we can get a contradiction from the above inequality. Thus, we can get that $c=k$.

Therefore, we prove that $L_1(f(z))=kf(z)$, that is, the conclusion of Theorem 1.7 holds.