*If σ(**f*) ≤ 1, then the result holds obviously. Now, we suppose that σ(*f*) > 1 without loss of generality and suppose that max
$\{\lambda (\frac{1}{f}),\lambda (f)\}<\sigma (f)-1$
on the contrary. Thus, *f*(*z*) can be written by Hadamard decomposition theorem as
$$f(z)={\displaystyle \frac{H(z)}{Q(z)}{e}^{g(z),}}$$(15)

*where **H*(*z*)(≢ 0) is the canonical product (or polynomial) formed by zeros of *f*(*z*), *Q*(*z*)(≢ 0) is the canonical product (or polynomial) formed by poles of *f*(*z*), *g*(*z*) is a polynomial with degree *k* defined by (9), and they satisfy
$$\begin{array}{}\lambda (H)=\sigma (H)=\lambda (f)<\sigma (f)-1=k-1,\\ \lambda (Q)=\sigma (Q)=\lambda (\frac{1}{f})<\sigma (f)-1=k-1.\end{array}$$(16)

*By max
$\{\lambda (\frac{1}{f}),\lambda (f)\}<\sigma (f)-1=k-1,$
, we see that **k* ≥ 2.

*Obviously, **f*
^{(j)}(*z*)(*j* = 0, 1,…, max{*n, m*}) are of the following forms
$${f}^{(j)}(z)=(\frac{H(z)}{Q(z)}{e}^{g(z)}{)}^{(j)}={\psi}_{j}(z){e}^{g(z)},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}j=0,1,\cdots ,\text{max}\{n,m\}.$$(17)

*where
${\psi}_{0}(z)=\frac{H(z)}{Q(z)},{\psi}_{j}(z)(j=1,2,\cdots ,max\{n,m\})$ are rational expressions formed by **H*(*z*), *Q*(*z*), *g*(*z*) and their derivatives, and we have by (16) that

*
$$\sigma ({\psi}_{j})<\sigma (f)-1=k-1,\phantom{\rule{2em}{0ex}}j=0,1,\cdots ,max\{n,m\}.$$(18)
*

*By substituting (15) and (17) into (7), we have
$$\sum _{j=0}^{n}{A}_{j}(z){\psi}_{j}(z){e}^{g(z)}+\sum _{j=1}^{m}{B}_{j}(z){\psi}_{j}(z+{b}_{j}){e}^{g(z+{b}_{j})}+\sum _{j=1}^{l}{C}_{j}(z){\psi}_{0}(z+{c}_{j}){e}^{g(z+{c}_{j})}=0.$$(19)
*

*Since **k* ≥ 2 and *b*_{j}
(≠ 0)(*j* = 1, 2,, *m*), *c*_{j}
(≠ 0)(*j* = 1, 2,…, l) are unequal to each other, we have by (9) that
$$\left\{\begin{array}{l}g(z+{b}_{i})-g(z+{b}_{s})=k{d}_{k}({b}_{i}-{b}_{s}){z}^{k-1}+\cdots ,1\underset{\_}{<}i<s\underset{\_}{<}m,\\ g(z+{c}_{i})-g(z+{c}_{s})=k{d}_{k}({c}_{i}-{c}_{s}){z}^{k-1}+\cdots ,1\underset{\_}{<}i<s\underset{\_}{<}l,\\ g(z+{b}_{i})-g(z+{c}_{s})=k{d}_{k}({b}_{i}-{c}_{s}){z}^{k-1}+\cdots ,1\underset{\_}{<}i\underset{\_}{<}m,1\underset{\_}{<}s\underset{\_}{<}l,\\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}g(z+{b}_{i})-g(z)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=k{d}_{k}{b}_{i}{z}^{k-1}+\cdots ,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}1\underset{\_}{<}i\underset{\_}{<}m,\\ \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}g(z+{c}_{s})-g(z)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=k{d}_{k}{c}_{s}{z}^{k-1}+\cdots ,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}1\underset{\_}{<}s\underset{\_}{<}l\end{array}\right.$$(20)

*are all polynomials with degree **k* – 1(≥ 1) . By (18) and Remark 2.2, we have *σ*(*ψ*_{j}(*z* + *b*_{j}
)) < *k* - 1(*j* = 1, 2,…, *m*) and *σ*(*ψ*_{0}(*z* + *c*_{j}
)) < *k* - 1(*j* = 1, 2,…, l) . Thus, by (18) and (20), we have
$$\left\{\begin{array}{l}T(r,\underset{j=0}{\overset{n}{\sum !}}{A}_{j}(z){\psi}_{j}(z))=0\{T(r,\mathrm{exp}\{g(z+{b}_{i})-g(z+{b}_{s})\})\},1\underset{\_}{<}i<s\underset{\_}{<}m,\\ T(r,\sum _{j=0}^{n}{A}_{j}(z){\psi}_{j}(z))=0\{T(r,\mathrm{exp}\{g(z+{c}_{i})-g(z+{c}_{s})\})\},1\underset{\_}{<}i<s\underset{\_}{<}l,\\ T(r,\sum _{j=0}^{n}{A}_{j}(z){\psi}_{j}(z))=0\{T(r,\mathrm{exp}\{g(z+{b}_{i})-g(z+{c}_{s})\})\},1\underset{\_}{<}i\underset{\_}{<}m,1\underset{\_}{<}s\underset{\_}{<}l,\\ T(r,\sum _{j=0}^{n}{A}_{j}(z){\psi}_{j}(z))=0\{T(r,\mathrm{exp}\{g(z+{b}_{i})-g(z)\}){\}}_{,}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}1\underset{\_}{<}i\underset{\_}{<}{m}_{,}\\ T(r,\sum _{j=0}^{n}{A}_{j}(z){\psi}_{j}(z))=0\{T(r,\mathrm{exp}\{g(z+{c}_{s})-g(z)\}){\}}_{,}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}1\underset{\_}{<}s\underset{\_}{<}{l}_{,}\end{array}\right.$$(21)

*and for **j* = 1, 2,…, *m*, we have
$$\left\{\begin{array}{l}T(r,{B}_{j}(z){\psi}_{j}(z+{b}_{j}))=0\{T(r,\mathrm{exp}\{g(z+{b}_{i})-g(z+{b}_{s})\})\},1\underset{\_}{<}i<s\underset{\_}{<}m,\\ T(r,{B}_{j}(z){\psi}_{j}(z+{b}_{j}))=0\{T(r,\mathrm{exp}\{g(z+{c}_{i})-g(z+{c}_{s})\})\},1\underset{\_}{<}i<s\underset{\_}{<}l,\\ T(r,{B}_{j}(z){\psi}_{j}(z+{b}_{j}))=0\{T(r,\mathrm{exp}\{g(z+{b}_{i})-g(z+{c}_{s})\})\},1\underset{\_}{<}i\underset{\_}{<}m,1\underset{\_}{<}s\underset{\_}{<}l,\\ T(r,{B}_{j}(z){\psi}_{j}(z+{b}_{j}))=0\{T(r,\mathrm{exp}\{g(z+{b}_{i})-g(z)\})\},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}1\underset{\_}{<}i\underset{\_}{<}m,\\ T(r,{B}_{j}(z){\psi}_{j}(z+{b}_{j}))=0\{T(r,\mathrm{exp}\{g(z+{c}_{s})-g(z)\})\},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}1\underset{\_}{<}s\underset{\_}{<}l,\end{array}\right.$$(22)

*for ** j* = 1, 2, …, *l*, we have
$$\left\{\begin{array}{l}T({r}_{,}{C}_{j}(z){\psi}_{0}(z+{c}_{j}))=0\{T({r}_{,}\mathrm{exp}\{g(z+{b}_{i})-g(z+{b}_{s})\}){\}}_{,}1\underset{\_}{<}i<s\underset{\_}{<}{m}_{,}\\ T({r}_{,}{C}_{j}(z){\psi}_{0}(z+{c}_{j}))=0\{T({r}_{,}\mathrm{exp}\{g(z+{c}_{i})-g(z+{c}_{s})\}){\}}_{,}1\underset{\_}{<}i<s\underset{\_}{<}{l}_{,}\\ T({r}_{,}{C}_{j}(z){\psi}_{0}(z+{c}_{j}))=0\{T({r}_{,}\mathrm{exp}\{g(z+{b}_{i})-g(z+{c}_{s})\}){\}}_{,}1\underset{\_}{<}i\underset{\_}{<}{m}_{,}1\underset{\_}{<}s\underset{\_}{<}{l}_{,}\\ T({r}_{,}{C}_{j}(z){\psi}_{0}(z+{c}_{j}))=0\{T({r}_{,}\mathrm{exp}\{g(z+{b}_{i})-g(z))\}){\}}_{,}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}1\underset{\_}{<}i\underset{\_}{<}{m}_{,}\\ T({r}_{,}{C}_{j}(z){\psi}_{0}(z+{c}_{j}))=0\{T({r}_{,}\mathrm{exp}\{g(z+{c}_{s})-g(z))\}){\}}_{,}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}1\underset{\_}{<}s\underset{\_}{<}l.\end{array}\right.$$(23)

*By combining ((20) – (23), we see that Lemma 2.1 holds for (19), that is,
$$\left\{\begin{array}{l}\sum _{j=0}^{n}{A}_{j}(z){\phi}_{j}(z)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\equiv 0,\\ {B}_{j}(z){\phi}_{j}(z+{b}_{j})\equiv 0,\phantom{\rule{thinmathspace}{0ex}}j=1,2,\cdots ,m,\\ {C}_{j}(z){\psi}_{0}(z+{c}_{j})\equiv 0,\phantom{\rule{thinmathspace}{0ex}}j=1,2,\cdots ,l.\end{array}\right.$$(24)
*

*Since **B*_{m}
(*z*) *C*_{l}
(*z*) ≢ 0, (24) can not hold. Hence,
$max\{\lambda (\frac{1}{f}{)}_{,}\lambda (f)\}\underset{\_}{>}\sigma (f)-1.$

*Here, we use the similar reasoning method as the one of (i). Suppose that **σ*(*f*) >1 without loss of generality and suppose that
$max\{\lambda (\frac{1}{f-a}{)}_{,}\lambda (f-a)\}<\sigma (f)-1$
on the contrary. Thus, *f*(*z*) can be written by Hadamard decomposition theorem as
$$f(z)={\displaystyle \frac{H(z)}{Q(z)}{e}^{g(z)}+{a}_{,}}$$(25)

*where **H*(*z*)(*≢ O*) is the canonical product (or polynomial) formed by zeros of *f*(*z*)-*a*, *Q*(*z*)(*≢ O*) is the canonical product (or polynomial) formed by poles of *f*(*z*)-*a*, *g*(*z*) is a polynomial with degree *k*(≥ 2) defined by (9), and they satisfy
$$\begin{array}{}\lambda \mathit{(}\mathit{H}\mathit{)}\mathit{=}\sigma \mathit{(}\mathit{H}\mathit{)}\mathit{=}\lambda \mathit{(}\mathit{f}\mathit{-}\mathit{a}\mathit{)}\mathit{<}\sigma \mathit{(}\mathit{f}\mathit{)}\mathit{-}1\mathit{=}\mathit{k}\mathit{-}{1}_{,}\\ \lambda \mathit{(}\mathit{Q}\mathit{)}\mathit{=}\sigma \mathit{(}\mathit{Q}\mathit{)}\mathit{=}\lambda \mathit{(}\frac{1}{\mathit{f}\mathit{-}\mathit{a}}\mathit{)}\mathit{<}\sigma \mathit{(}\mathit{f}\mathit{)}\mathit{-}1\mathit{=}\mathit{k}\mathit{-}\mathit{1.}\end{array}$$(26)

*Obviously,
${f}^{(j)}(z)(j={0}_{,}{1}_{,}\cdots ,max\{n,m\})$
are of the following forms
$${f}^{(j)}(z)=(\frac{H(z)}{Q(z)}{e}^{g(z)}+a{)}^{(j)}={\psi}_{j}(z){e}^{g(z)},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}j=1,2,\cdots ,\text{max}\{n,m\},$$(27)
*

*where **ψ*_{j}
(*z*)(*j* = 1, 2, …, max{*n, m*}) are rational expressions formed by *H*(*z*), *Q*(*z*),*g*(*z*) and their derivatives, and we have by (26) that
$$\sigma ({\psi}_{j})<\sigma (f)-1=k-{1}_{,}\phantom{\rule{2em}{0ex}}j={0}_{,}{1}_{,}\cdots ,max\{n,m\},$$(28)

*where
${\psi}_{0}(z)=\frac{H(z)}{Q(z)}.$
By substituting (25) and (27) into (7), we have
$$\begin{array}{}\sum _{j=0}^{n}{A}_{j}(z){\psi}_{j}(z){e}^{g(z)}+\sum _{j=1}^{m}{B}_{j}(z){\psi}_{j}(z+{b}_{j}){e}^{g(z+{b}_{j})}\\ +{\displaystyle \sum _{j=1}^{l}{C}_{j}(z){\psi}_{0}(z+{c}_{j}){e}^{g(z+{c}_{j})}+a({A}_{0}(z)+\sum _{j=1}^{l}{C}_{j}(z)){e}^{0}=0.}\end{array}$$(29)
*

*By using the similar reasoning method as the one dealing with (19) in (i), we see that Lemma 2.1 holds for (29), that is, (24)and
${A}_{0}(z)+{\displaystyle \sum _{j=1}^{l}{C}_{j}(z)\equiv 0}$
hold, which is a contradiction. Hence,
$max\{\lambda \mathit{(}\frac{\mathit{1}}{\mathit{f}\mathit{-}\mathit{a}}{\mathit{)}}_{\mathit{,}}\lambda \mathit{(}\mathit{f}\mathit{-}\mathit{a}\mathit{)}\}\underset{\_}{\mathit{>}}\sigma \mathit{(}\mathit{f}\mathit{)}\mathit{-}\mathrm{1.}$
*

*Further, if
${A}_{0}(z)+{\displaystyle \sum _{j=1}^{l}{C}_{j}(z)\not\equiv 0,}$
then (7) can be rewritten as
$$\sum _{j=0}^{n}{A}_{j}(z){h}^{(j)}(z)+\sum _{j=1}^{m}{B}_{j}(z){h}^{(j)}(z+{b}_{j})+\sum _{j=1}^{l}{C}_{j}(z)h(z+{c}_{j})=-a({A}_{0}(z)+\sum _{j=1}^{l}{C}_{j}(z){)}_{,}$$
*

*where **h*(*z*) = *f*(*z*) − *a*. By Corollary 1.2 we have
$max\{\lambda (\frac{1}{h}{)}_{,}\lambda (h)\}=\sigma (h),$that is,
$max\{\lambda (\frac{1}{f-a}{)}_{;}\lambda (f-a)\}=\sigma (f).$

*(iii) By using the similar reasoning method as the ones of (i) and (ii), we have
$max\{\lambda (\frac{1}{f-\phi}{)}_{,}\lambda (f-\phi ,)\}\underset{\_}{>}\sigma (f)-1\phantom{\rule{thinmathspace}{0ex}}\text{when}\phantom{\rule{thinmathspace}{0ex}}\sigma (\phi )<\sigma (f)-1.$In the meantime, we have by Theorem 1.1 that
$\text{max}\{\lambda (\frac{1}{f-\phi}{)}_{,}\lambda (f-\phi )\}=\sigma (f),$ when **σ*(*φ*) < *σ*(*f*) and *φ*(*z*) is not a solution of (7).

*Therefore, the proof of Theorem 1.4 is complete.*

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