考虑Dirichlet级数

$f(s)=\sum\limits_{n=0}^{\infty}a_ne^{\lambda_ns}, $

(1.1)

其中$\{a_n\}$是复数列, $0<\lambda_n\uparrow\infty$, $s=\sigma+it$ ($\sigma, t$是实变量).当级数(1.1)满足

$\limsup\limits_{n\rightarrow\infty}\frac{\log n}{\lambda_n}=0, \limsup\limits_{n\rightarrow\infty}\frac{\log|a_n|}{\lambda_n}=-\infty. $

(1.2)

这时, 根据文[1-2]的Valiron公式可得级数(1.1)的收敛横坐标及绝对收敛横坐标都是$+\infty$, 那么其和函数$f(s)$在全平面内解析, 即为整函数.

记$f(s)$的最大模, 最大项为

$ \begin{eqnarray*}&&M(\sigma)=M(\sigma, f)=\sup\{|f(\sigma+it)|:t\in R\}, \\ &&m(\sigma)=m(\sigma, f)=\max\{|a_n|e^{\lambda_n\sigma}:n\in N^*\}.\end{eqnarray*} $

定义1.1 ^[3]若$f(s)$是满足(1.2)式的整函数, 那么$f(s)$的级$\rho$定义为

$ \rho=\limsup\limits_{\sigma\rightarrow+\infty}\frac{\log\log M(\sigma, f)}{\sigma}. $

(1.3)

若$\rho=0$, 级数(1.1)是全平面上的零级Dirichlet级数.此时定义该级数(1.1)的对数级$\rho^*$为

$ \rho^*=\limsup\limits_{\sigma\rightarrow+\infty}\frac{\log\log M(\sigma, f)}{\log\sigma}, $

(1.4)

当$\rho^*\in (1, +\infty)$时, Dirichlet级数的对数型$T^*$如下

$T^*=\limsup\limits_{\sigma\rightarrow+\infty}\frac{\log M(\sigma)}{\sigma^{\rho^*}}. $

(1.5)

关于整函数的增长性的问题, Hardy、余家荣、孙道椿、高宗升等已经得到了许多经典的结论^{[1-2, 4-6]}. Sayyed, Metwally ^[7]讨论了泰勒级数的对数级, 而对复平面上的零级Dirichlet级数增长性的研究较少. 2006年, 田宏根、孙道椿、郑承民在相对较宽的条件下, 对该问题进行深入的研究并得到了由系数表示的零级Dirichlet级数的对数级的结果.

定理 A ^[3]若$f(s)$是满足(1.2)式的整函数, 则

$ \rho^*=1+ \limsup\limits_{n\rightarrow\infty}\frac{\log\lambda_n}{\log\log|a_n|^{-1}-\log\lambda_n}. $

(1.6)

本文将继续研究了零级Dirichlet级数的对数型, 得到如下结果.

定理 1.1 若$f(s)$是满足(1.2)式的整函数, 则

$T^*=\frac{(\rho^*-1)^{\rho^*-1}}{(\rho^*)^{\rho^*}}T, $

(1.7)

这里

$ T=\limsup\limits_{n\rightarrow\infty}\frac{\lambda_n}{(\frac{1}{\lambda_n}\log|a_n|^{-1})^{\rho^*-1}}.$

2009年, 孔荫莹在文[9-10]构造了Dirichlet-Hadamard乘积并得到了有限级及无穷级Dirichlet级数在该乘积下的增长性的相关结果. 2015年, 崔永琴等在文[11]构造了新型的Dirichlet-Hadamard乘积进一步推广了文[9, 10]的结果.

然而, 对于零级Dirichlet级数的Hadamard乘积的增长性并未有人涉及.本文将主要考察零级Dirichlet级数的Dirichlet-Hadamard乘积的对数级与对数型, 在介绍主要结果前, 我们先给出如下Dirichlet-Hadamard乘积定义.

定义 1.2 ^[11]若$f_1(s)=\sum\limits_{n=1}^\infty a_ne^{\gamma_ns}$, $f_2(s)=\sum\limits_{n=1}^\infty b_ne^{\xi_ns}$且$f_1(s), f_2(s)$是满足(1.2)式的整函数.若$\alpha, \beta$为两个实常数满足$0<\alpha, \beta<\infty$, 构造它们的Dirichlet-Hadamard乘积如下

$ F(s)=(f_1\Delta f_2)(\mu, v;\alpha, \beta;s)=\sum\limits_{n=1}^\infty c_ne^{\lambda_ns}, c_n=a_n^\mu b_n^v, \lambda_n=\alpha\gamma_n+\beta\xi_n, $

(1.8)

其中$\mu$和$v$是正实数; $\{a_n\}, \{b_n\}\subset \mathbb{C}, 0<\gamma_n, \xi_n\uparrow\infty$.

注 当$\alpha=\beta=\frac{1}{2}$, 则定义1.2中的Dirichlet-Hadamard乘积$F(s)$即为孔荫莹的Dirichlet-Hadamard乘积$G(s)$, 即

$ G(s)=(f_1\Delta f_2)(\mu, v;s)=\sum\limits_{n=1}^\infty c_ne^{\lambda_ns}, c_n=a_n^\mu b_n^v, \lambda_n=\frac{\gamma_n+\xi_n}{2}. $

定理 1.2 若$f_1(s), f_2(s)$是满足(1.2)式的整函数, 它们的对数级分别为$\rho^*_1$和$\rho^*_2$, 且

$\gamma_n\sim\xi_n (n\rightarrow\infty), $

(1.9)

则Dirichlet-Hadamard乘积$F(s)$的对数级$\rho^*$满足$\rho^*\leq\min\{\rho^*_1, \rho^*_2\}.$特别地, 当$\rho^*=\rho^*_1$时, $F(s)$的对数型$T^*$满足

$ T^*\leq \left\{\begin{aligned} &\frac{1}{(\rho_1^*)^{\rho_1^*}}(\frac{\rho_1^*-1}{\mu})^{\rho_1^*-1}, \rho_1^*<\rho_2^*, \\ &\frac{1}{(\rho_1^*)^{\rho_1^*}}(\frac{\rho_1^*-1}{\mu+v})^{\rho_1^*-1}, \rho_1^*=\rho_2^*. \end{aligned}\right. $

推论 1.1 若$f_1(s), f_2(s)$是满足(1.2)式的整函数, 它们的对数级分别为$\rho_1^*$和$\rho_2^*$, 且满足(1.9)式, 则其Dirichlet-Hadamard乘积$G(s)$的对数级$\rho^*$满足$\rho^*\leq\min\{\rho_1^*, \rho_2^*\}.$特别地, 当$\rho^*=\rho_1^*$, $G(s)$对数型$T^*$满足

$ T^*\leq \left\{\begin{aligned} &\frac{1}{(\rho_1^*)^{\rho_1^*}}(\frac{\rho_1^*-1}{\mu})^{\rho_1^*-1}, \rho_1^*<\rho_2^*, \\ &\frac{1}{(\rho_1^*)^{\rho_1^*}}(\frac{\rho_1^*-1}{\mu+v})^{\rho_1^*-1}, \rho_1^*=\rho_2^*. \end{aligned}\right. $

接下来, 在放宽条件的前提下进一步讨论Dirichlet-Hadamard乘积形式的增长性, 得到如下结果.

定理 1.3 若$f_1(s), f_2(s)$是满足(1.2)式的整函数, 它们的对数级分别为$\rho_1^*$和$\rho_2^*$, 且

$\gamma_n=\eta\xi_n, $

(1.10)

则其Dirichlet-Hadamard乘积$F(s)$的对数级$\rho^*$满足$\rho^*\leq\min\{\rho_1^*, \rho_2^*\}.$特别地, 当$\rho^*=\rho_1^*$, $F(s)$对数型$T^*$满足

$ T^*\leq \left\{\begin{aligned} &(\frac{\alpha\eta+\beta}{\eta\rho_1})^{\rho_1^*}(\frac{\rho_1^*-1}{\mu})^{\rho_1^*-1}, \rho_1^*<\rho_2^*, \\ &(\frac{\alpha\eta+\beta}{\rho_1^*})^{\rho_1^*}(\frac{\rho_1^*-1}{\mu\eta^{\rho_1^*} +v})^{\rho_1^*-1}, \rho_1^*=\rho_2^*. \end{aligned}\right. $

推论 1.2 若$f_1(s), f_2(s)$是满足(1.2)式的整函数, 它们的对数级分别为$\rho_1^*$和$\rho_2^*$, 且满足(1.10)式, 则其Dirichlet-Hadamard乘积$G(s)$的对数级$\rho^*$满足$\rho^*\leq\min\{\rho_1^*, \rho_2^*\};$当$\rho^*=\rho_1^*$, $G(s)$对数型$T^*$满足

$ T^*\leq \left\{\begin{aligned} &(\frac{\eta+1}{2\eta\rho_1^*})^{\rho_1^*}(\frac{\rho_1^*-1}{\mu})^{\rho_1^*-1}, \rho_1^*<\rho_2^*, \\ &(\frac{\eta+1}{2\rho_1^*})^{\rho_1^*}(\frac{\rho_1^*-1}{\mu\eta^{\rho_1^*}+v})^{\rho_1^*-1}, \rho_1^*=\rho_2^*. \end{aligned}\right. $

引理2.1 ^[11]若$f_1(s), f_2(s)$是满足(1.2)式的整函数, 且满足(1.9)式, 那么其Dirichlet-Hadamard乘积$F(s)$是整函数.

引理 2.2 若$f_1(s), f_2(s)$是满足(1.2)式的整函数, 且满足(1.10)式, 那么其Dirichlet-Hadamard乘积$F(s)$是整函数.

证

$ \limsup\limits_{n\rightarrow\infty}\frac{\log n}{\lambda_n} =\limsup\limits_{n\rightarrow\infty}\frac{\log n}{\alpha\gamma_n+\beta\xi_n} \leq \frac{1}{\alpha}\limsup\limits_{n\rightarrow\infty}\frac{\log n}{\gamma_n}=0, $

又

$ \limsup\limits_{n\rightarrow\infty}\frac{\log |c_n|}{\lambda_n} =\limsup\limits_{n\rightarrow\infty}\frac{\mu\log |a_n|+\nu\log |b_n|}{\alpha\gamma_n+\beta\xi_n} \leq \limsup\limits_{n\rightarrow\infty}\frac{\mu\log |a_n|}{\alpha\gamma_n+\beta\xi_n}=-\infty. $

所以其Dirichlet-Hadamard乘积$F(s)$是整函数.

引理 2.3 若$a, b (b>1)$是一正的常数, $x$是任一正实数, 那么函数$\psi(\sigma)=a\sigma^b-x\sigma (-\infty<\sigma<+\infty)$在$\sigma=(\frac{x}{ab})^{\frac{1}{b-1}}$时达到最小值$a(\frac{x}{ab})^{\frac{b}{b-1}}-x(\frac{x}{ab})^{\frac{1}{b-1}}.$

证 由$\psi'(\sigma)=ab\sigma^{b-1}-x$, 令$\psi'(\sigma)=0$解得$\sigma=(\frac{x}{ab})^{\frac{1}{b-1}}$.

可验证当$\sigma=(\frac{x}{ab})^{\frac{1}{b-1}}$时$\psi(\sigma)$取得最小值$a(\frac{x}{ab})^{\frac{b}{b-1}}-x(\frac{x}{ab})^{\frac{1}{b-1}}.$

引理 2.4 若$a, b (b>1)$是一正的常数, $\sigma$是任一实数, 那么函数$\varphi(x)=-\frac{1}{a}x^b+\sigma x$在$x=(\frac{a\sigma}{b})^{\frac{1}{b-1}}$时达到最大值$-\frac{1}{a}(\frac{a\sigma}{b})^{\frac{b}{b-1}}+\sigma(\frac{a\sigma}{b})^{\frac{1}{b-1}}.$

证 由$\varphi'(x)=-\frac{b}{a}x^{b-1}+\sigma$, 令$\varphi'(x)=0$解得$x=(\frac{a\sigma}{b})^{\frac{1}{b-1}}$.

可验证当$x=(\frac{a\sigma}{b})^{\frac{1}{b-1}}$时$\varphi(x)$达到最大值$-\frac{1}{a}(\frac{a\sigma}{b})^{\frac{b}{b-1}}+\sigma(\frac{a\sigma}{b})^{\frac{1}{b-1}}.$

定理1.1的证明 先证$T^*\geq T\frac{(\rho^*-1)^{\rho^*-1}}{(\rho^*)^{\rho^*}}$.

由$T^*$的定义知, $\forall \varepsilon>0$, 有充分大的$\sigma$使

$\frac{\log M(\sigma)}{\sigma^{\rho^*}}<T^*+\varepsilon, $

从而

$\log |a_n|<(T^*+\varepsilon)\sigma^{\rho^*}-\lambda_n\sigma.$

由引理2.3知, 取$\sigma=(\frac{\lambda_n}{(T^*+\varepsilon)\rho^*})^{\frac{1}{\rho^*-1}}$, 则

$ \begin{aligned} \log|a_n| &<(T^*+\varepsilon)[(\frac{\lambda_n}{(T^*+\varepsilon)\rho^*})^{\frac{1}{\rho^*-1}}]^{\rho^*}- \lambda_n(\frac{\lambda_n}{(T^*+\varepsilon)\rho^*})^{\frac{1}{\rho^*-1}}\\ &=\lambda_n^{\frac{\rho^*}{\rho^*-1}}\frac{1}{[(T^*+\varepsilon)\rho^*]^{\frac{1}{\rho^*-1}}}\frac{1-\rho^*}{\rho^*}. \end{aligned} $

所以

$ \begin{aligned} \frac{\lambda_n}{(\frac{1}{\lambda_n}\log|a_n|^{-1})^{\rho^*-1}} &<\frac{\lambda_n}{(\frac{1}{\lambda_n}\lambda_n^{\frac{\rho^*}{\rho^*-1}})^{\rho^*-1}}(\frac{\rho^*}{\rho^*-1})^{\rho^*-1}(T^*+\varepsilon)\rho^*\\ &=\frac{(\rho^*)^{\rho^*}}{(\rho^*-1)^{\rho^*-1}}(T^*+\varepsilon). \end{aligned} $

由$\varepsilon$的任意性知

$T\leq\frac{(\rho^*)^{\rho^*}}{(\rho^*-1)^{\rho^*-1}}T^*.$

假设等号不成立, 即存在$T_1$使得$T<T_1<\frac{(\rho^*)^{\rho^*}}{(\rho^*-1)^{\rho^*-1}}T^*$, 于是存在$N_1>0$, 当$n>N_1$时,

$\frac{\lambda_n}{(\frac{1}{\lambda_n}\log|a_n|^{-1})^{\rho^*-1}}<T_1, $

即

$|a_n|<e^{-\lambda_n^{\frac{\rho^*}{\rho^*-1}}T_1^{-\frac{1}{\rho^*-1}}}.$

由(1.2)式知存在一常数$M$, $N_2>N_1$, 使得$n>N_2$时有$\lambda_n>M\log n$, 于是

$M(\sigma, f)\leq \sum\limits_{n=1}^{N_1}|a_n|e^{\lambda_n\sigma}+\sum\limits_{N_1+1}^{N_2}|a_n|e^{\lambda_n\sigma}+\sum\limits_{N_2+1}^{\infty}|a_n|e^{\lambda_{n+1}\sigma}, $

(3.1)

其中$\sum\limits_{n=1}^{N_1}|a_n|e^{\lambda_n\sigma}$为有界量,

$\sum\limits_{N_1+1}^{N_2}|a_n|e^{\lambda_n\sigma}\leq\sum\limits_{N_1+1}^{N_2}e^{-\lambda_n^{\frac{\rho^*}{\rho^*-1}}T_1^{-\frac{1}{\rho^*-1}}}e^{\lambda_n\sigma} =\sum\limits_{N_1+1}^{N_2}e^{-\lambda_n^{\frac{\rho^*}{\rho^*-1}}T_1^{-\frac{1}{\rho^*-1}}+\lambda_n\sigma}, $

由引理2.4知, 取$\lambda_n=(\frac{\sigma(\rho^*-1)}{\rho^*})^{\rho^*-1}T_1$, 有

$\sum\limits_{N_1+1}^{N_2}|a_n|e^{\lambda_n\sigma}\leq N_2e^{T_1\frac{(\rho^*-1)^{\rho^*-1}}{(\rho^*)^{\rho^*}}\sigma^{\rho^*}}.$

(3.2)

再由(1.2)式知$\lambda_{n+1}\leq (1+\varepsilon)\lambda_n, $对所有的$n\in N_+$成立, 记$\lambda_n>T_1((1+\varepsilon)\sigma+\frac{2}{M})^{\rho^*-1}$.所以

$\begin{aligned} \sum\limits_{N_2+1}^{\infty}|a_n|e^{\lambda_{n+1}\sigma}\leq \sum\limits_{N_2+1}^{\infty}e^{-\lambda_n^{\frac{\rho^*}{\rho^*-1}}T_1^{-\frac{1}{\rho^*-1}}}e^{\lambda_{n+1}\sigma} \leq \sum\limits_{N_2+1}^{\infty} e^{(\lambda_{n+1}-(1+\varepsilon))\sigma-\lambda_n\frac{2}{M}} \leq\sum\limits_{N_2+1}^{\infty} \frac{1}{n^2}. \end{aligned} $

(3.3)

由(3.1)-(3.3)式知, 对充分大的$\sigma$有

$\frac{\log M(\sigma, f)}{\sigma^{\rho^*}}\leq T_1\frac{(\rho^*-1)^{\rho^*-1}}{(\rho^*)^{\rho^*}}(1+o(1)).$

从上式得到$T^*\leq T_1 \frac{(\rho^*-1)^{\rho^*-1}}{(\rho^*)^{\rho^*}}$与假设矛盾, 故$T= \frac{(\rho^*)^{\rho^*}}{(\rho^*-1)^{\rho^*-1}}T^*, $定理1.1得证.

定理1.2的证明 由定理A可知$\forall \varepsilon>0$, 存在两个正整数$N_1, N_2$, 当$n>N =\max\{N_1, N_2\}$时, 有

$\frac{\log\gamma_n}{\log(\frac{1}{\gamma_n}\log|a_n|^{-1})} <\rho^*_1-1+\varepsilon, \frac{\log\xi_n}{\log(\frac{1}{\xi_n}\log|b_n|^{-1})} <\rho^*_2-1+\varepsilon, $

即

$\log|a_n|^{-1}>\gamma_n\gamma_n^{\frac{1}{\rho^*_1-1+\varepsilon}}, \log|b_n|^{-1}>\xi_n\xi_n^{\frac{1}{\rho^*_2-1+\varepsilon}}. $

由$c_n$的定义有

$ \begin{aligned} \log|c_n|^{-1}=\mu\log|a_n|^{-1}+v\log|b_n|^{-1} >\mu\gamma_n\gamma_n^{\frac{1}{\rho^*_1-1+\varepsilon}}+v\xi_n\xi_n^{\frac{1}{\rho^*_2-1+\varepsilon}}, \end{aligned} $

(3.4)

则

$ \begin{aligned} \frac{\log\lambda_n}{\log(\frac{1}{\lambda_n}\log|c_n|^{-1})} &<\frac{\log\lambda_n}{\log(\mu\frac{\gamma_n}{\lambda_n}\gamma_n^{\frac{1}{\rho^*_1-1+\varepsilon}}+v \frac{\xi_n}{\lambda_n}\xi_n^{\frac{1}{\rho^*_2-1+\varepsilon}})}\\ &=\frac{\log\lambda_n}{\frac{\rho^*_1+\varepsilon}{\rho^*_1-1+\varepsilon}\log\gamma_n+\log(\mu+v \frac{\xi_n}{\gamma_n}\frac{\xi_n^{\frac{1}{\rho^*_2-1+\varepsilon}}}{\gamma_n^{\frac{1}{\rho^*_1-1+\varepsilon}}})-\log\lambda_n}. \end{aligned} $

由于$\lambda_n=\alpha\gamma_n+\beta\xi_n, \gamma_n\sim\xi_n (n\rightarrow\infty)$, 可得

$\log\gamma_n\sim\log\xi_n\sim\log\lambda_n (n\rightarrow\infty).$

由引理2.1知$F(s)$是整函数, 不妨设$\rho^*_1<\rho^*_2$, 又由$\varepsilon$的任意性, 得

$\rho^*=1+\limsup\limits_{n\rightarrow\infty} \frac{\log\lambda_n}{\log(\frac{1}{\lambda_n}\log|c_n|^{-1})}\leq 1+\frac{1}{\frac{\rho^*_1}{\rho^*_1-1}-1}=\rho^*_1, $

所以$\rho^*\leq\min\{\rho^*_1, \rho^*_2\}.$

特别地, 当$\rho^*=\rho^*_1$, 由定理1.1可知

$ \begin{aligned} &\frac{(\rho^*_1-1)^{\rho^*_1-1}}{(\rho^*_1)^{\rho^*_1}}\frac{\lambda_n}{(\frac{1}{\lambda_n}\log|c_n|^{-1})^{\rho^*_1-1}}\\ <&\frac{(\rho^*_1-1)^{\rho^*_1-1}}{(\rho^*_1)^{\rho^*_1}}\frac{\lambda_n}{(\mu\frac{\gamma_n}{\lambda_n}\gamma_n^{\frac{1}{\rho^*_1-1+\varepsilon}} +v\frac{\xi_n}{\lambda_n}\xi_n^{\frac{1}{\rho^*_2-1+\varepsilon}})^{\rho^*_1-1}}\\ =&\frac{(\rho^*_1-1)^{\rho^*_1-1}}{(\rho^*_1)^{\rho^*_1}}\frac{\lambda_n}{\frac{1}{\lambda_n^{\rho^*_1-1}}\gamma_n^{\frac{\rho^*_1+\varepsilon} {\rho^*_1-1+\varepsilon}(\rho^*_1-1)}(\mu+v\frac{\xi_n}{\gamma_n}\frac{\xi_n^{\frac{1}{\rho^*_2-1+\varepsilon}}}{\gamma_n^{\frac{1}{\rho^*_1-1+\varepsilon}}}) ^{\rho^*_1-1}}. \end{aligned} $

若$\rho^*_1=\rho^*_2$, 由$\varepsilon$的任意性可得

$T^*\leq\frac{(\rho^*_1-1)^{\rho^*_1-1}}{(\rho^*_1)^{\rho^*_1}}\frac{1}{(\mu+v)^{\rho^*_1-1}}.$

若$\rho^*_1<\rho^*_2$, 则有

$T^*\leq\frac{(\rho^*_1-1)^{\rho^*_1-1}}{(\rho^*_1)^{\rho^*_1}}\frac{1}{\mu^{\rho^*_1-1}}, $

故定理1.2得证.

定理1.3的证明 类似于定理1.2的证明: $\forall \varepsilon>0$, 存在两个正整数$N_1, N_2$, 当$n>N=\max\{N_1, N_2\}$时有

$ \begin{aligned} \log|c_n|^{-1}=\mu\log|a_n|^{-1}+v\log|b_n|^{-1} >\mu\gamma_n\gamma_n^{\frac{1}{\rho^*_1-1+\varepsilon}}+v\xi_n\xi_n^{\frac{1}{\rho^*_2-1+\varepsilon}}. \end{aligned} $

由$\gamma_n=\eta\xi_n$, 有$\gamma_n=\frac{\eta}{\alpha\eta+\beta}\lambda_n$, $\xi_n=\frac{1}{\alpha\eta+\beta}\lambda_n$.于是

$\frac{1}{\lambda_n}\log|c_n|^{-1}>\lambda_n^{\frac{1}{\rho_1^*+\varepsilon-1}}[\mu(\frac{\eta}{\alpha\eta+\beta})^{\frac{\rho_1^*+\varepsilon} {\rho_1^*+\varepsilon-1}}+v(\frac{1}{\alpha\eta+\beta})^{\frac{\rho_2^*+\varepsilon} {\rho_2^*+\varepsilon-1}}\lambda_n^{\frac{1}{\rho_2^*+\varepsilon-1}-\frac{1}{\rho_1^*+\varepsilon-1}}].$

由引理2.2知$F(s)$是整函数, 不妨设$\rho^*_1<\rho^*_2$, 又由$\varepsilon$的任意性, 得

$ \begin{aligned} \rho^*&=1+\limsup\limits_{n\rightarrow\infty} \frac{\log\lambda_n}{\log(\frac{1}{\lambda_n}\log|c_n|^{-1})}\\ &\leq 1+\limsup\limits_{n\rightarrow\infty}\frac{\log\lambda_n}{\frac{1}{\rho_1^*+\varepsilon-1} \log\lambda_n+\log[\mu(\frac{\eta}{\alpha\eta+\beta})^{\frac{\rho_1^*+\varepsilon} {\rho_1^*+\varepsilon-1}}+v(\frac{1}{\alpha\eta+\beta})^{\frac{\rho_2^*+\varepsilon} {\rho_2^*+\varepsilon-1}}\lambda_n^{\frac{1}{\rho_2^*+\varepsilon-1}-\frac{1}{\rho_1^*+\varepsilon-1}}]}\\ &\leq 1+\frac{1}{\frac{1}{\rho^*_1-1}}=\rho^*_1. \end{aligned} $

所以$\rho^*\leq\min\{\rho^*_1, \rho^*_2\}.$

特别地, 当$\rho^*=\rho^*_1$, 由定理1.1可知

$ \begin{aligned} &\frac{(\rho^*_1-1)^{\rho^*_1-1}}{(\rho^*_1)^{\rho^*_1}}\frac{\lambda_n}{(\frac{1}{\lambda_n}\log|c_n|^{-1})^{\rho^*_1-1}}\\ <&\frac{(\rho^*_1-1)^{\rho^*_1-1}}{(\rho^*_1)^{\rho^*_1}}\frac{\lambda_n}{\{\lambda_n^{\frac{1}{\rho_1^*+\varepsilon-1}} [\mu(\frac{\eta}{\alpha\eta+\beta})^{\frac{\rho_1^*+\varepsilon} {\rho_1^*+\varepsilon-1}}+v(\frac{1}{\alpha\eta+\beta})^{\frac{\rho_2^*+\varepsilon} {\rho_2^*+\varepsilon-1}}\lambda_n^{\frac{1}{\rho_2^*+\varepsilon-1}-\frac{1}{\rho_1^*+\varepsilon-1}}]\}^{\rho_1^*-1}}. \end{aligned} $

若$\rho^*_1=\rho^*_2$, 由$\varepsilon$的任意性可得

$T^*\leq\frac{1}{(\rho_1^*)^{\rho_1^*}}(\frac{\rho_1^*-1}{\mu(\frac{\eta}{\alpha\eta+\beta})^{\frac{\rho_1^*}{\rho_1^*-1}} +v(\frac{1}{\alpha\eta+\beta})^{\frac{\rho_1^*}{\rho_1^*-1}}} )^{\rho_1^*-1}=(\frac{\alpha\eta+\beta}{\rho_1^*})^{\rho_1^*}(\frac{\rho_1^*-1}{\mu\eta^{\rho_1^*} +v})^{\rho_1^*-1}.$

若$\rho^*_1<\rho^*_2$, 则有

$T^*\leq(\frac{\alpha\eta+\beta}{\eta\rho_1^*})^{\rho_1^*}(\frac{\rho_1^*-1}{\mu})^{\rho_1^*-1}. $

故定理1.3得证.