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On a conjecture of R. Brück and some linear differential equations

Abstract

In this paper, we mainly investigate the Brück conjecture concerning entire function f and its differential polynomial \(L_1(f)=a_k(z)f^{(k)}+\cdots +a_0(z)f\) sharing an entire function \(\alpha (z)\) with \(\sigma (\alpha )\le \sigma (f)\), by using the theory of complex differential equation.

Introduction and some results

It is assumed that the reader is familiar with the standard symbols and fundamental results of Nevanlinna value distribution theory of meromorphic functions. For a meromorphic function f in the whole complex plane \(\mathbb {C}\), we shall use the following standard notations of the value distribution theory:

$$\begin{aligned} T(r,f),m(r,f), N(r,f),\overline{N}(r,f),\ldots \end{aligned}$$

(see Hayman 1964; Yang 1993; Yi and Yang 2003, 1995). We use S(rf) to denote any quantity satisfying \(S(r,f)=o(T(r,f)),\) as \(r\rightarrow +\infty\), 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 \(\sigma (f), \mu (f)\) to denote the order and the lower order of meromorphic function f(z), which are defined by

$$\begin{aligned} \sigma (f)=\limsup _{r\rightarrow +\infty }\frac{\log T(r,f)}{\log r} =\limsup _{r\rightarrow +\infty }\frac{\log \log M(r,f)}{\log r}, \end{aligned}$$

and

$$\begin{aligned} \mu (f)=\liminf _{r\rightarrow +\infty }\frac{\log T(r,f)}{\log r} =\liminf _{r\rightarrow +\infty }\frac{\log \log M(r,f)}{\log r}, \end{aligned}$$

where \(M(r,f)=\max _{|z|=r}|f(z)|\). We also use \(\tau (f)\) to denote the type of an entire function f(z) with \(0<\sigma (f)=\sigma <+\infty\), which is defined to be (see Hayman 1964)

$$\begin{aligned} \tau (f)=\limsup _{r\rightarrow +\infty }\frac{\log M(r,f)}{r^\sigma }. \end{aligned}$$

We use \(\sigma _2(f)\) to denote the hyper-order of f(z), which is defined to be (see Yi and Yang 2003, 1995)

$$\begin{aligned} \sigma _2(f)=\limsup _{r\rightarrow +\infty }\frac{\log \log T(r,f)}{\log r}. \end{aligned}$$

Let f(z) and g(z) be two nonconstant meromorphic functions, for some \(a\in \mathbb {C}\cup \{\infty \}\), if the zeros of \(f(z)-a\) and \(g(z)-a\) (if \(a=\infty\), zeros of \(f(z)-a\) and \(g(z)-a\) are the poles of f(z) and g(z) respectively) coincide in locations and multiplicities we say that f(z) and g(z) share the value a CM (counting multiplicities) and if coincide in locations only we say that f(z) and g(z) share a IM (ignoring multiplicities).

Rubel and Yang (1977) proved the following result.

Theorem 1.1

Rubel and Yang (1977). Let f be a nonconstant entire function. If f and \(f'\) share two finite distinct values CM, then \(f\equiv f'\).

In 1996, Brück proposed the following conjecture Brück (1996):

Conjecture 1.1

Brück (1996). Let f be a non-constant entire function. Suppose that \(\sigma _2(f)\) is not a positive integer or infinite, if f and \(f'\) share one finite value a CM, then

$$\begin{aligned} \frac{f'-a}{f-a}=c\end{aligned}$$

for some non-zero constant c.

Gundersen and Yang (1998) proved that Brück conjecture holds for entire functions of finite order and obtained the following result.

Theorem 1.2

[Gundersen and Yang (1998), Theorem 1]. Let f be a nonconstant entire function of finite order. If f and \(f'\) share one finite value a CM, then \(\frac{f'-a}{f-a}=c\) for some non-zero constant c.

The shared value problems related 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 Chen et al. 2014; Li and Yi 2007; Liao 2015; Mues and Steinmetz 1986; Zhang and Yang 2009; Zhang 2005; Zhao 2012).

Li and Cao (2008) improved the Brück conjecture for entire function and its derivation sharing polynomials and obtained the following result:

Theorem 1.3

Li and Cao (2008). Let \(Q_1\) and \(Q_2\) be two nonzero polynomials, and let P be a polynomial. If f is a nonconstant entire solution of the equation

$$\begin{aligned} f^{(k)}-Q_1=(f-Q_2)e^P, \end{aligned}$$

then \(\sigma _2(f)=\deg P\) , where and in the following, \(\deg P\) is the degree of P.

Mao (2009) studied the problem on Brück conjecture when \(f^{(k)}\) is replaced by differential polynomial \(L(f)=A_kf^{(k)}+\cdots +A_1f'+A_0f\) in Theorem 1.3.

Theorem 1.4

Mao (2009). Let P(z) be a polynomial, \(A_k(z)(\not \equiv 0),\ldots ,A_0(z)\) be polynomials, and f be an entire function of order

$$\begin{aligned} \sigma (f)>1+\max _{0\le j\le k-1}\left\{ \frac{\deg A_j-\deg A_k}{k-j},0\right\} \end{aligned}$$

and hyper-order \(\sigma _2(f)<\frac{1}{2}\). If f and L(f) share P CM, then

$$\begin{aligned} \frac{L(f)-P(z)}{f(z)-P(z)}=c, \end{aligned}$$

for some constant \(c\ne 0\) , where, and in the sequel, \(\deg A_j\) denotes the degree of \(A_j(z)\), k is a positive integer.

Chang and Zhu (2009) further investigated the problem related to Brück conjecture and proved that Theorem 1.2 remains valid if the value a is replaced by a function a(z).

Theorem 1.5

[ Chang and Zhu (2009), Theorem 1] . Let f be an entire function of finite order and a(z) be a function such that \(\sigma (a)<\sigma (f)<+\infty\). If f and \(f'\) share a(z) CM, then \(\frac{f'-a}{f-a}=c\) for some non-zero constant c.

Thus, an interesting subject arises naturally about this problem: what would happen when \(\sigma (a)<\sigma (f)<+\infty\) is replaced by \(0<\sigma (a)=\sigma (f)<+\infty\) in Theorems 1.2–1.5?

Conclusions

Motivated by the above question, the main purpose of this article is to study the growth of solution of differential equation on entire function f and its linear differential polynomial

$$\begin{aligned} L_1(f)=a_k(z)f^{(k)}+a_{k-1}(z)f^{(k-1)}+\cdots +a_1(z)f'+a_0(z)f, \end{aligned}$$
(1)

where k is a positive integer, \(a_k(z)(\not \equiv 0),a_{k-1}(z),\ldots ,a_1(z)\) and \(a_0(z)\) are polynomials, and obtain the following theorems.

Theorem 2.1

Let f(z) and \(\alpha (z)\) be two nonconstant entire functions and satisfy \(0<\sigma (\alpha )=\sigma (f)<+\infty\) and \(\tau (f)>\tau (\alpha )\) , and let P(z) be a polynomial such that

$$\begin{aligned} \sigma (f)>\deg P+\max \left\{ \frac{deg a_j-deg a_k}{k-j},0\right\} . \end{aligned}$$
(2)

If f is a nonconstant entire solution of the following differential equation

$$\begin{aligned} L_1(f)-\alpha (z)=(f(z)-\alpha (z))e^{P(z)}, \end{aligned}$$
(3)

where \(L_1(f)\) is stated as in (1). Then P(z) is a constant.

If \(L_1(f)\) is replaced by the following linear differential polynomial \(L_2(f)\)

$$\begin{aligned} L_2(f)=a_k(z)f^{(k)}+a_{k-1}(z)f^{(k-1)}+\cdots +a_1(z)f'+a_0(z)f+\beta (z), \end{aligned}$$
(4)

where k is a positive integer, \(a_k(z)(\not \equiv 0),a_{k-1}(z),\ldots ,a_1(z)\) and \(a_0(z)\) are polynomials, and \(\beta\) is an entire function satisfying either \(\sigma (\beta )<\mu (f)\) or \(0<\sigma (\beta )=\sigma (f)<+\infty\) and \(\tau (\beta )<\tau (f)\), then we obtain the following results.

Theorem 2.2

Let f(z) and \(\alpha (z)\) be two nonconstant entire functions and satisfy \(0<\sigma (\alpha )=\sigma (f)<+\infty\) and \(\tau (f)>\tau (\alpha )\) , and let P(z ) be a polynomial satisfying (2). If f is a nonconstant entire solution of the following differential equation

$$\begin{aligned} L_2(f)-\alpha (z)=(f(z)-\alpha (z))e^{P(z)}, \end{aligned}$$
(5)

where \(L_2(f)\) is stated as in (4) and \(\beta\) is an entire function satisfying \(0<\sigma (\beta )=\sigma (f)<+\infty\) and \(\tau (\beta )<\tau (f)\). Then P(z) is a constant.

Theorem 2.3

Let f(z) and \(\alpha (z)\) be two nonconstant entire functions and satisfy \(\sigma (\alpha )<\mu (f)\) , and let P(z) be a polynomial satisfying (2). If f is a nonconstant entire solution of Eq. (5), where \(L_2(f)\) is stated as in (4) and \(\beta\) is an entire function satisfying \(\sigma (\beta )<\mu (f)\) . Then \(\sigma _2(f)=\deg P\).

Corollary 2.1

Let f(z) and \(\alpha (z)\) be two nonconstant entire functions and satisfy \(\sigma (\alpha )<\mu (f)\) , and let P(z be a polynomial satisfying) (2). If f is a nonconstant entire solution of Eq. (3),where \(L_1(f)\) is stated as in (1). Then \(\sigma _2(f)=\deg P\).

Some Lemmas

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

Lemma 3.1

Laine (1993). Let f(z) be a transcendental entire function, \(\nu (r,f)\) be the central index of f(z). Then there exists a set \(E\subset (1,+\infty )\) with finite logarithmic measure, we choose z satisfying \(|z|=r\not \in [0,1]\cup E\) and \(|f(z)|=M(r,f)\) , we get

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

Lemma 3.2

He and Xiao (1988). Let f(z) be an entire function of finite order \(\sigma (f)=\sigma <+\infty\) , and let \(\nu (r,f)\) be the central index of f . Then

$$\begin{aligned} \limsup _{r\rightarrow +\infty }\frac{\log \nu (r,f)}{\log r}=\sigma (f). \end{aligned}$$

And if f is a transcendental entire function of hyper order \(\sigma _2(f)\) , then

$$\begin{aligned} \limsup _{r\rightarrow +\infty }\frac{\log \log \nu (r,f)}{\log r}=\sigma _2(f). \end{aligned}$$

Lemma 3.3

Mao (2009). Let f be a transcendental entire function and let \(E\subset [1,+\infty )\) be a set having finite logarithmic measure. Then there exists \(\{z_n=r_ne^{i\theta _n}\}\) such that \(|f(z_n)|=M(r_n,f),\theta _n\in [0,2\pi )\), \(\lim _{n\rightarrow +\infty }\theta _n=\theta _0\in [0,2\pi ), r_n\not \in E\) and if \(0<\sigma (f)<+\infty\) , then for any given \(\varepsilon >o\) and sufficiently large \(r_n\),

$$\begin{aligned} r_n^{\sigma (f)-\varepsilon }<\nu (r_n,f)<r_n^{\sigma (f)+\varepsilon }. \end{aligned}$$

If \(\sigma (f)=+\infty\) , then for any given large \(M>0\) and sufficiently large \(r_n\), \(\nu (r_n,f)>r_n^M\).

Lemma 3.4

Laine (1993). Let \(P(z)=b_nz^n+b_{n-1}z^{n-1}+\cdots +b_0\) with \(b_n\ne 0\) be a polynomial. Then, for every \(\varepsilon >0\) , there exists \(r_0>0\) such that for all \(r=|z|>r_0\) the inequalities

$$\begin{aligned} (1-\varepsilon )|b_n|r^n\le |P(z)|\le (1+\varepsilon )|b_n|r^n \end{aligned}$$

hold.

Lemma 3.5

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

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

Proof

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

$$\begin{aligned} \lim \limits _{m\rightarrow +\infty }\frac{\log M(r_m,f)}{r_m^{\sigma }}=\tau (f). \end{aligned}$$
(6)

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

$$\begin{aligned} \log M(r_m,f)>(\tau (f)-\varepsilon )r_m^{\sigma }. \end{aligned}$$
(7)

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

$$\begin{aligned} \left( \frac{m}{m+1}\right) ^{\sigma }>\frac{\beta }{\tau (f)-\varepsilon }. \end{aligned}$$
(8)

From (68), 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

$$\begin{aligned} M(r,f)&\ge M(r_m,f)>\exp \{(\tau (f)-\varepsilon )r_m^{\sigma }\} \\&\ge \exp \left\{ (\tau (f)-\varepsilon )\left( \frac{m}{m+1}r\right) ^{\sigma }\right\} >\exp \{\beta r^{\sigma }\}.\nonumber \end{aligned}$$
(9)

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

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

From the definition of type of entire function, for any sufficiently small \(\varepsilon >0\), we have

$$\begin{aligned} M(r,A)<\exp \{\tau (A)+\varepsilon )r^\sigma \}. \end{aligned}$$
(10)

By (9) and (10), set \(\kappa =\beta -\tau (A)-\varepsilon\), for all \(r\in E\), we have

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

Thus, this completes the proof of this lemma. \(\square\)

The proof of Theorem 2.1

Proof

Since P(z) is a polynomial, assume that \(\deg P=m\ge 1\). Let

$$\begin{aligned} P(z)=b_mz^m+b_{m-1}z^{m-1}+\cdots +b_0, \end{aligned}$$

where \(b_m,\ldots ,b_0\) are constants and \(b_m\ne 0, m\ge 1\). Thus, it follows from (3) and Lemma 3.4 that

$$\begin{aligned} |b_m|r^m(1+o(1))=|P(z)|=\left| \log \frac{\frac{L_1(f(z))}{f(z)}-\frac{\alpha (z)}{f(z)}}{1-\frac{\alpha (z)}{f(z)}}\right| . \end{aligned}$$
(11)

Since \(L_1(f)=a_kf^k+a_{k-1}f^{(k-1)}+\cdots +a_0f\), from Lemma 3.1, then there exists a subset \(E_1\subset (1,+\infty )\) with finite logarithmic measure, such that for some point \(|z|=re^{i\theta } (\theta \in [0,2\pi ))\), \(r\not \in E_1\) and \(M(r,f)=|f(z)|\), we have

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

Thus, it follows that

$$\begin{aligned} \frac{L_1(f(z))}{f(z)}&=a_k\left\{ \frac{\nu (r,f)}{z}\right\} ^k(1+o(1))+\cdots +a_1\left\{ \frac{\nu (r,f)}{z}\right\} (1+o(1))+a_0\nonumber \\&=\frac{a_k}{z^k}(1+o(1))\left[ \nu (r,f)^k+\sum _{j=1}^k\frac{a_{k-j}}{a_k}z^j\nu (r,f)^{k-j}(1+o(1))\right] . \end{aligned}$$
(12)

From Lemma 3.3, there exists \(\{z_n=r_ne^{i\theta _n}\}\) such that \(|f(z_n)|=M(r_n,f),\theta _n\in [0,2\pi )\), \(\lim _{n\rightarrow \infty }\theta _n=\theta _0\in [0,2\pi ), r_n\not \in E_1\), then for any given \(\varepsilon\) satisfying

$$\begin{aligned} 0<\varepsilon <\min _{1\le j\le k}\frac{j[\sigma (f)-\deg P-\frac{d_{k-j}}{j}]}{3k-j}, \end{aligned}$$

where \(d_{k-j}=deg a_{k-j}-deg a_k\), and sufficiently large \(r_n\), we have

$$\begin{aligned} r_n^{\sigma (f)-\varepsilon }<\nu (r_n,f)<r_n^{\sigma (f)+\varepsilon }. \end{aligned}$$
(13)

Since \(a_0(z), \ldots , a_k(z)\) are polynomials, let \(a_j(z)=\sum _{t=0}^{s_j}l_{jt}z^t\), where \(s_j=\deg a_j, j=0,1,\ldots ,k\). Then, from Lemma 3.4 and (13), we have

$$\begin{aligned} \left| \frac{a_{k-j}}{a_k}z^j\nu (r,f)^{k-j}(1+o(1))\right|&\le M\frac{|l_{k-j,s_{k-j}}|r_n^{s_{k-j}}}{|l_{k,s_{k}}|r_n^{s_{k}}}r_n^jr_n^{(\sigma (f)+\varepsilon )(k-j)}\nonumber \\&=M\frac{|l_{k-j,s_{k-j}}|}{|l_{k,s_{k}}|}r_n^{d_{k-j}+j+(\sigma (f)+\varepsilon )(k-j)}\nonumber \\&\le M\frac{|l_{k-j,s_{k-j}}|}{|l_{k,s_{k}}|}r_n^{k\sigma (f)-j\sigma (f)+d_{k-j}+\deg P+(k-j)\varepsilon }, \end{aligned}$$
(14)

where \(d_{k-j}=s_{k-j}-s_k\) and M is a positive constant. Since \(-j\sigma (f)+d_{k-j}+\deg P+(k-j)\varepsilon <-2k\varepsilon <0\), it follows from (14) that

$$\begin{aligned} \left| \frac{a_{k-j}}{a_k}z^j\nu (r_n,f)^{k-j}(1+o(1))\right|&< M\frac{|l_{k-j,s_{k-j}}|}{|l_{k,s_{k}}|}r_n^{k(\sigma (f)-2\varepsilon )}\nonumber \\&= o( \nu (r_n,f)^{k}), ~~as~~r_n\rightarrow +\infty , ~~r_n\not \in E_1. \end{aligned}$$
(15)

Since \(0<\sigma (\alpha )=\sigma (f)<+\infty\) and \(\tau (\alpha )<\tau (f)<+\infty\), from Lemma 3.5, there exists a set \(E\subset [1,+\infty )\) that has infinite logarithmic measure such that for a sequence \(\{r_n\}_1^\infty \in E_2=E-E_1\), we have

$$\begin{aligned} \frac{M(r,\alpha )}{ M(r,f)}<\exp \{-\kappa r_n^{\sigma (f)}\}\rightarrow 0, \quad as~~n\rightarrow +\infty . \end{aligned}$$
(16)

From (11), (12), (15), (16) and Lemma 3.2, we can get that

$$\begin{aligned} |b_m|r_n^m(1+o(1))=|P(z)|=O(\log r_n), \end{aligned}$$
(17)

which is impossible. Thus, P(z) is not a polynomial, that is, P(z) is a constant.

Thus, this completes the proof of Theorem 2.1. \(\square\)

The proof of Theorem 2.2

Proof

First of all, we rewrite (5) as

$$\begin{aligned} \frac{L_2(f)-\alpha (z)}{f(z)-\alpha (z)} =\frac{\frac{L_1(f)}{f}+\frac{\beta (z)}{f(z)}-\frac{\alpha (z)}{f(z)}}{1-\frac{\alpha (z)}{f(z)}}=e^{P(z)}, \end{aligned}$$
(18)

where \(L_1(f)\) is stated as in Theorem 2.1. Since \(0<\sigma (f)=\sigma (\alpha )=\sigma (\beta )<+\infty\), \(\tau (\alpha )<\tau (f)\) and \(\tau (\beta )<\tau (f)\), from Lemma 3.5, there exists a set \(E\subset [1,+\infty )\) that has infinite logarithmic measure such that for a sequence \(\{r_n\}_1^\infty \in E_3=E-E_1\), we have

$$\begin{aligned} \frac{M(r,\alpha )}{M(r,f)}<\exp \{-\kappa r_n^{\sigma (f)}\}\rightarrow 0, ~~and ~~\frac{M(r,\beta )}{M(r,f)}<\exp \{-\kappa r_n^{\sigma (f)}\}\rightarrow 0,~~~as ~~n\rightarrow +\infty . \end{aligned}$$

Then by using the proceeding as in proof of Theorem 2.1, we prove that P(z) is a constant.

This completes the proof of Theorem 2.2. \(\square\)

The proof of Theorem 2.3.

Proof

From P(z) is a polynomial, we will consider two cases (i) \(\sigma (f)<+\infty\) and (ii) \(\sigma (f)=+\infty\).

Case 1. Suppose that \(\sigma (f)<+\infty\). Then \(\sigma _2(f)=0\). Since \(\sigma (\alpha )<\mu (f), \sigma (\beta )<\mu (f)\), from Definitions of the order and the lower order, there exists infinite sequence \(\{z_n\}_1^\infty\), we have

$$\begin{aligned} \frac{|\alpha (z_n)|}{|f(z_n)|}\rightarrow 0, ~~and ~~\frac{|\beta (z_n)|}{|f(z_n)|}\rightarrow 0,~~~as ~~n\rightarrow \infty . \end{aligned}$$

Thus, by using the same argument as in Theorem 2.1, we can get that P(z) is a constant, that is, \(\deg P=0\). Therefore, \(\sigma _2(f)=\deg P\).

Case 2. Suppose that \(\sigma (f)=+\infty\). Set \(F(z)=f(z)-\alpha (z)\). Since \(\sigma (\alpha )<\mu (f)\), it follows from (2) that

$$\begin{aligned} \sigma (F)=+\infty , ~~\sigma _2(F)=\sigma _2(f), \end{aligned}$$
(19)

and

$$\begin{aligned} \sigma (F)>\deg P+\max \left\{ \frac{deg a_j-deg a_k}{k-j},0\right\} . \end{aligned}$$
(20)

Furthermore, we can rewrite (4) as

$$\begin{aligned} a_k(z)\frac{F^{(k)}(z)}{F(z)}+\cdots +a_1(z)\frac{F'(z)}{F(z)}+a_0(z)+\frac{\gamma (z)}{F(z)}=e^{P(z)}, \end{aligned}$$
(21)

where \(\gamma (z)=a_k\alpha ^{(k)}+\cdots +a_1\alpha +\beta -\alpha\). Since \(\sigma (\beta )<\mu (f)\), \(\sigma (\alpha )<\mu (f)\) and \(a_i(z), (i=0,\ldots ,k)\) are polynomials, we have

$$\begin{aligned} \sigma (\gamma )\le \max \{\sigma (\alpha ),\sigma (\beta )\}<\mu (f)\le \sigma (f). \end{aligned}$$
(22)

From Lemma 3.1, there exists a set \(E_4\subset (1,+\infty )\) with finite logarithmic measure, we choose z satisfying \(|z|=r\not \in [0,1]\cup E_4\) and \(|F(z)|=M(r,F)\), we get

$$\begin{aligned} \frac{F^{(j)}(z)}{F(z)}=\left\{ \frac{\nu (r,F)}{z}\right\} ^j(1+o(1)), ~~for ~~j\in 1,2,\ldots ,k. \end{aligned}$$
(23)

Since \(\sigma (F)=+\infty\), then it follows from Lemma 3.3 that there exists \(\{z_n=r_ne^{i\theta _n}\}\) with \(|F(z_n)|=M(r_n,F),\theta _n\in [0,2\pi )\), \(\lim _{n\rightarrow \infty }\theta _n=\theta _0\in [0,2\pi ), r_n\not \in E_5\), such that for any large constant K and for sufficiently large \(r_n\) we have

$$\begin{aligned} \nu (r_n,F)\ge r_n^K. \end{aligned}$$
(24)

From \(M(r_n,F)=|F(z_n)|\), \(F(z), \gamma (z)\) are entire functions and (18), by using definitions of the order and the lower order, we have

$$\begin{aligned} \left| \frac{\gamma (z_n)}{F(z_n)}\right| \rightarrow 0, ~~~as ~~r\rightarrow +\infty . \end{aligned}$$
(25)

Thus, it follows from (21), (23)–(25) that

$$\begin{aligned} a_k\left( \frac{\nu (r_n,F)}{z_n}\right) ^k(1+o(1))=e^{P(z_n)}. \end{aligned}$$
(26)

Let

$$\begin{aligned} P(z)=b_mz^m+b_{m-1}z^{m-1}+\cdots +b_0, \end{aligned}$$

where \(b_m,\ldots ,b_0\) are constants and \(b_m\ne 0, m\ge 1\). From Lemma 3.4, there exists sufficiently large positive number \(r_0\) and \(n_0\in \mathbb {N}_+\), such that for sufficiently large positive integer \(n>n_0\) satisfying \(|z_n|=r_n>r_0\), we have for every \(\varepsilon '>0\)

$$\begin{aligned} \log |b_m|+m\log |z_n|+\log |1-\varepsilon '|\le \log |P(z_n)|\le |\log \log e^{P(z_n)}|. \end{aligned}$$
(27)

It follows from (26) that

$$\begin{aligned} |\log \log e^{P(z_n)}|&\le \log |\log |a_k||+\log \log \nu (r_n,F)+\log \log r_n+O(1)\nonumber \\&\le \log \log \nu (r_n,F)+O(\log \log r_n). \end{aligned}$$
(28)

Thus, we have from (27), (28) and Lemma 3.2 that

$$\begin{aligned} m=\deg P(z)\le \sigma _2(F)=\sigma _2(f). \end{aligned}$$
(29)

On the other hand, since \(a_k\) is a polynomial, it follows from (27) and Lemma 3.4 that

$$\begin{aligned} M(r_n,e^{P(z_n)})\ge K_1r_n^{d_k}\left( \frac{\nu (r_n,F)}{r_n}\right) ^k, \end{aligned}$$

where \(K_1>0\) is a constant. Then we have

$$\begin{aligned} \nu (r_n,F)^k\le K_1^{-1}r_n^{k-d_k}M(r_n,e^{P(z_n)}). \end{aligned}$$
(30)

Thus, it follows from (30) and Lemma 3.2 that

$$\begin{aligned} \sigma _2(f)&=\sigma _2(F)=\limsup _{r_n\rightarrow +\infty }\frac{\log \log \nu (r_n,F)}{\log r_n} =\limsup _{r_n\rightarrow +\infty }\frac{\log \log \nu (r_n,F)^k}{\log r_n}\\&\le \limsup _{r_n\rightarrow +\infty }\frac{\log \log K_1^{-1}r_n^{k-d_k}M(r_n,e^{P(z_n)})}{\log r_n}=\sigma (e^{P}). \end{aligned}$$

Since P(z) is a polynomial, then \(\sigma (e^{P})=\deg P=m\). By combining (29), we have \(\sigma _2(f)=\deg P\).

Therefore, this completes the proof of Theorem 2.3. \(\square\)

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Authors’ contributions

HYX and LZY completed the main part of this article. Both authors read and approved the final manuscript.

Acknowledgements

This first author was supported by the NSF of China (11561033, 11301233), the Natural Science Foundation of Jiangxi Province in China (20132BAB211001,20151BAB201008), and the Foundation of Education Department of Jiangxi (GJJ14644) of China. The second author was supported by the NSF of China (11371225, 11171013 and 11041005).

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Xu, H.Y., Yang, L.Z. On a conjecture of R. Brück and some linear differential equations. SpringerPlus 4, 748 (2015). https://doi.org/10.1186/s40064-015-1530-5

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