关于Dirichlet级数及随机Dirichlet级数的增长性, 国内已作过许多重要研究^[3-8], 对于全平面上的零级Dirichlet级数的增长性和正规增长性, 文[3]作过专门研究并得到了很好的结果, 但对于半平面, 由于其特殊性至今尚未见很好的结果, 本文对半平面零级Dirichlet级数的增长性和正规增长性进行研究, 得到了两个充要条件.并且还证明了零级随机Dirichlet级数, 几乎必然与其在每条水平直线上的增长性相同.

考虑Dirichlet级数

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

(2.1)

其中$s=\sigma+it, \{b_n\}$为复常数列, $0=\lambda_0 < \lambda_1 < \lambda_2\cdots < \lambda_n\uparrow+\infty$.若

$ \overline{\lim\limits_{n\rightarrow\infty}}\frac{\ln |b_n|}{\lambda_n}=0, ~~\overline{\lim\limits_{n\rightarrow\infty}} \frac{\ln n}{\lambda_n}=0. $

(2.2)

则级数$(2.1)$在半平面上是收敛与绝对收敛的, 它表示的函数$f(s)$为一半平面解析函数.令

$ M(\sigma, f)=\sup\{|f(\sigma+it)|;t\in R\}, ~~m(\sigma, f)=\max\{|b_n|e^{-\lambda_n\sigma};n\in N\}, $

(2.3)

由文^[1]知$m(\sigma, f)\leq M(\sigma, f)$.

定义2.1  $f(s)$在半平面上的增长级$\rho$定义为$\rho=\overline{\lim\limits_{\sigma\rightarrow0^{+}}} \frac{\ln^+\ln^+ M(\sigma, f)}{\frac{1}{\sigma}}$.若$\rho=0$则称级数(2.1) 为零级Dirichlet级数.

定义2.2^[2]  对于零级Dirichlet级数(2.1), 由文献[2]引进型函数$U(r)=r^{\rho(r)}(r=\frac{1}{\sigma})$, 单调上升且满足:

1)$\rho(r)$单调趋于$0$, $\lim\limits_{r\rightarrow0^{+}}r\rho^{'}(r)\ln r=0;$

2)$U(kr)=(1+o(1))U(r); $

3)$U(r^k)=U^{k+o(1)}(r)$.

称$U(r)$为零级型函数.

定理2.1  设Dirichlet级数$(2.1)$满足条件$(2.2)$及

$ \overline{\lim\limits_{n\to\infty}}\frac{\ln\ln n}{\ln\lambda_n}=d<1. $

(2.4)

则有

$ \overline{\lim\limits_{\sigma\rightarrow 0^{+}}}\frac{\ln\ln M(\sigma, f)}{\ln U(\frac{1}{\sigma})}= \overline{\lim\limits_{\sigma\rightarrow 0^{+}}}\frac{\ln\ln m(\sigma, f)}{\ln U(\frac{1}{\sigma})}. $

(2.5)

证  设$\overline{\lim\limits_{\sigma\rightarrow 0^{+}}}\frac{\ln\ln m(\sigma, f)}{\ln U(\frac{1}{\sigma})}=A$, 则$\forall\varepsilon>0, $当$\sigma$充分接近$0$时

$ \label{1.5}\ln m(\sigma, f)\leq U^{A+\varepsilon}(\frac{1}{\sigma}). $

(2.6)

由$(2.4)$式, 对任意$\varepsilon_1\in (0, 1-d)$, 存在自然数$N>3$当$n\geq N$时,

$ -\lambda_n\leq -{(\ln n)}^{\frac {1}{1-\varepsilon_1}}=-\ln n{(\ln n)}^{\frac {\varepsilon_1}{1-\varepsilon_1}}. $

所以

$ M(\sigma, f)\leq\sum\limits_{n=0}^{+\infty}|b_n|e^{-\lambda_n\sigma}=\sum\limits_{n=0}^{+\infty}( |b_n|e^{-\lambda_n \Delta}e^{-\lambda_n\delta})\leq m(\Delta, f)(k+\sum\limits_{n=k}^{+\infty}e^{-\lambda_n\delta} )\\ \leq m(\Delta, f)(k+\sum\limits_{n=k}^{+\infty}\exp{[-\delta\ln n{(\ln n)}^{\frac {\varepsilon_1}{1-\varepsilon_1}}]}), $

其中$\Delta=\frac{\sigma\ln U(\frac{1}{\sigma})}{1+\ln U(\frac{1}{\sigma})}, \delta=\frac{\sigma}{1+\ln U(\frac{1}{\sigma})} $.令$T=\exp\{({\frac{2}{\delta}})^{\frac{1-\varepsilon_1}{\varepsilon_1}}\}.$

当$n>T$时, $\delta({\ln n})^{\frac{\varepsilon_1}{1-\varepsilon_1}}>2$, 从而

$ M(\sigma, f)\leq m(\Delta, f)(k+\sum\limits_{n=k}^{T}n^{-\delta}+\sum\limits_{n=T+1}^{\infty}n^{-2})\\ \leq m(\Delta, f)(k+\int_1^{T}t^{-\delta }dt+c_1)\leq m(\Delta, f)(k_1+\frac{1}{1-\delta}T^{1-\delta}). $

故有

$ \ln M(\sigma, f)\leq\ln m(\Delta, f)+\ln c+(1-\delta)\ln T\\ \leq U^{A+\varepsilon}(\frac{1}{\Delta})+\ln c+(1-\delta)(\frac{2}{\delta})^{\frac{\varepsilon_1}{1-\varepsilon_1}} \leq U^{A+2\varepsilon}(\frac{1}{\sigma}), $

从而易得$\overline{\lim\limits_{\sigma\rightarrow 0^{+}}}\frac{\ln\ln M(\sigma, f)}{\ln U(\frac{1}{\sigma})}\leq A.$

另一方面, 由$ m(\sigma, f)\leq M(\sigma, f)$可得

$ A=\overline{\lim\limits_{\sigma\rightarrow 0^{+}}}\frac{\ln\ln m(\sigma, f)}{\ln U(\frac{1}{\sigma})} \leq\overline{\lim\limits_{\sigma\rightarrow 0^{+}}}\frac{\ln\ln M(\sigma, f)}{\ln U(\frac{1}{\sigma})}. $

因此$\overline{\lim\limits_{\sigma\rightarrow 0^{+}}} \frac{\ln\ln m(\sigma, f)}{\ln U(\frac{1}{\sigma})} =\overline{\lim\limits_{n\to\infty}}\frac{\ln\ln M(\sigma, f)}{\ln U(\frac{1}{\sigma})}.$>

定理2.2  设Dirichlet级数$(2.1)$满足条件(2.2), (2.3), (2.4), 则

(a)$\overline{\lim\limits_{\sigma\rightarrow 0^{+}}}\frac{\ln\ln M(\sigma, f)}{\ln U(\frac{1}{\sigma})}=1 \Leftrightarrow \overline{\lim\limits_{n\to\infty}}\frac{\ln\ln |b_n|}{\ln U(\frac{\lambda_n}{\ln |b_n|})}=1;$

(b)$\lim\limits_{\sigma\rightarrow 0^{+}}\frac{\ln\ln M(\sigma, f)}{\ln U(\frac{1}{\sigma})}=1\Leftrightarrow \overline{\lim\limits_{n\to\infty}}\frac{\ln\ln |b_n|}{\ln U(\frac{\lambda_n}{\ln |b_n|})}=1, $并且存在$\{\lambda_n\}$的子列$\{\lambda_{n(p)}\}$使

$ \lim\limits_{p\to\infty}\frac{\lambda_{n(p)}}{\lambda_{n(p+1)}}=1, \lim\limits_{p\to\infty}\frac{\ln\ln |b_{n(p)}|}{\ln U(\frac{\lambda_{n(p)}}{\ln |b_{n(p)}|})}=1. $

证  (a)设$\overline{\lim\limits_{n\to\infty}}\frac{\ln\ln |b_n|}{\ln U(\frac{\lambda_n}{\ln |b_n|})}=1, $则对任意$\varepsilon >0$和充分大的$n$, 有$\lambda_n < \frac{\lambda_n}{\ln |b_n|}U^{1+\varepsilon}(\frac{\lambda_n}{\ln |b_n|}).$令$v=uU^{1+\varepsilon}(u), u=h(v)$互为反函数, 则有$h(\lambda_n) < \frac{\lambda_n}{\ln |b_n|}, \ln |b_n|e^{-\lambda_n\sigma} < \frac{\lambda_n}{h(\lambda_n)}-\lambda_n\sigma, $当$\sigma>0$充分小时, 令$H=\frac{1}{\sigma}U^{1+\varepsilon}(\frac{1}{\sigma})$.

下面我们将分两种情形讨论

情形(ⅰ)当$\lambda_n>H$时, 有$\ln |b_n|e^{-\lambda_n\sigma}\leq\lambda_n(\frac{1}{h(H)}-\sigma)=0.$

情形(ⅱ)当$\lambda_n\leq H$时, 存在$k\in N$使

$ \frac{1}{k\sigma}U^{1+\varepsilon}(\frac{1}{k\sigma})\geq\lambda_n\geq \frac{1}{(k+1)\sigma}U^{1+\varepsilon}(\frac{1}{(k+1)\sigma}), $

从而$\frac{1}{k\sigma}\geq h(\lambda_n)\geq\frac{1}{(k+1)\sigma}, $则

$ \ln |b_n|e^{-\lambda_n\sigma}<\lambda_n(\frac{1}{h(\lambda_n)}-\sigma)\leq U^{1+\varepsilon}(\frac{1}{k\sigma}) \leq U^{1+\varepsilon}({\frac{1}{\sigma}}). $

由情形(ⅰ)和(ⅱ)及定理$1$, 有$\overline{\lim\limits_{\sigma\rightarrow 0^{+}}}\frac{\ln\ln M(\sigma, f)}{\ln U(\frac{1}{\sigma})}\leq1.$

若$\overline{\lim\limits_{\sigma\rightarrow 0^{+}}}\frac{\ln\ln M(\sigma, f)}{\ln U(\frac{1}{\sigma})}=L < 1, $取$0 < \varepsilon < 1$使$L+3\varepsilon < 1$, 存在$\sigma_0$, 当$\sigma>\sigma_0$时

$ \ln M(\sigma, f)<U^{L+\varepsilon(\frac{1}{\sigma})}. $

从而有对任意充分小的$\sigma$, 对任意的$n$有$\ln |b_n|e^{-\lambda_n\sigma}\leq U^{L+\varepsilon}(\frac{1}{\sigma}).$

由假设前提知, 存在子序列$\{\lambda_{n(p)}\}$使$\ln |b_{n(p)}|>U^{L+2\varepsilon}(\frac{\lambda_{n(p)}}{\ln |b_{n(p)}|}), $对任意充分大的$p$取$\{\sigma_p\}$使

$ U^{L+\varepsilon}(\frac{1}{\sigma_p})=\frac{1}{2}\ln |b_{n(p)}|\rightarrow\infty~~ (p\rightarrow\infty), $

$ \ln |b_{n(p)}|e^{-\lambda_{n(p)}\sigma_p}\leq U^{L+\varepsilon}(\frac{1}{\sigma_p})=\frac{1}{2}\ln |b_{n(p)}|, $

从而$\frac{1}{2}\ln |b_{n(p)}|\leq\lambda_{n(p)}\sigma_p, $则

$ \ln |b_{n(p)}|=2U^{L+\varepsilon}(\frac{1}{\sigma_p})\leq 2U^{L+\varepsilon}(\frac{2\lambda_{n(p)}}{\ln |b_{n(p)}|}) \leq U^{L+2\varepsilon}(\frac{\lambda_{n(p)}}{\ln |b_{n(p)}|}), $

这与$\ln |b_{n(p)}|\geq U^{1+\varepsilon}(\frac{\lambda_{n(p)}}{\ln |b_{n(p)}|})$矛盾.

于是充分性得证, 必要性可由充分性的证明过程看出.

(b)$\Leftarrow$设对任意充分小的$\varepsilon\in(0, 1)$, 存在子序列$\{\lambda_{n(p)}\}$满足

$ \ln |b_{n(p)}|\geq U^{1-\varepsilon}(\frac{\lambda_{n(p)}}{\ln |b_{n(p)}|}). $

即$\lambda_{n(p)}>\frac{\lambda_{n(p)}}{\ln |b_{n(p)}|}U^{1-\varepsilon}(\frac{\lambda_n{(p)}}{\ln |b_{n(p)}|}), $并且$\lambda_{n(p+1)}^{1-\varepsilon} < \lambda_{n(p)} < \lambda_{n(p+1)}, $则有

$ h(\lambda_{n(p)})>\frac{\lambda_{n(p)}}{\ln |b_{n(p)}|}, \ln |b_{n(p)}|>\frac{\lambda_{n(p)}}{h(\lambda_{n(p)})}, $

其中$u=h(v), v=uU^{1-\varepsilon}(u)$互为反函数, 取单调上升的序列$\{\lambda_{n(p)}\}$, 使$\lambda_{n(p)}=\frac{1}{2\sigma_p}U^{1-\varepsilon} (\frac{1}{2\sigma_p})$, 则有$h(\lambda_{n(p)})=\frac{1}{2\sigma_p}$, 任取$\sigma>0$, 则存在正整数$p$使$\sigma_{p+1} < \sigma < \sigma_p$, 对所有的$p$, 有

$ \ln M(\sigma, f)\geq\ln m(\sigma, f)\geq\ln |b_{n(p)}|e^{-\lambda_{n(p)}\sigma_p}\geq\frac{\lambda_{n(p)}}{h(\lambda_{n(p)})}-\lambda_{n(p)}\sigma_p\\ >\lambda_{n(p+1)}^{1-\varepsilon}\sigma_p>\sigma_p(\frac{1}{2\sigma_{p+1}})^{1-\varepsilon}U^{1-2\varepsilon}(\frac{1}{2\sigma_{p+1}}) >U^{1-3\varepsilon}(\frac{1}{\sigma_{p+1}})\geq{U^{1-3\varepsilon}}({\frac{1}{\sigma}}). $

结合(a)的结论即得$\lim\limits_{\sigma\rightarrow 0^{+}}\frac{\ln\ln M(\sigma, f)}{\ln U(\frac{1}{\sigma})}= 1.$

$\Rightarrow$第一条件的必要性从(a)部分得出.假设第二条件不成立, 即存在$\varepsilon, \delta>0$及自然数列$\{n_j\}, \{m_j\}$使

$ m_j<n_{j+1}, \lambda_{m_{j}}>\lambda_{n_{j+1}}^{1+\varepsilon}, $

(2.7)

且对任意充分大的$n(n_j\leq n < m_j)$有

$ \lambda_n<\frac{\lambda_n}{\ln |b_n|}U^{1-\delta}(\frac{\lambda_n}{\ln |b_n|}), $

(2.8)

$ h(\lambda_n)<\lambda_n<\frac{\lambda_n}{\ln |b_n|}, $

(2.9)

其中$h$是$v=uU^{1-\delta}(u)$的反函数, 取序列$\{\sigma_p\}$使

$ h(\lambda_{m_j})=\frac{1}{\sigma_p}. $

(2.10)

$(1)$当$n_j\leq n < m_j$对任意的$\sigma_p$有

$ \ln |b_n|e^{-\lambda_n\sigma_p}\leq\frac{\lambda_n}{h(\lambda_n)}-\lambda_n\sigma_p. $

(2.11)

令$H=\frac{1}{\sigma_p}U^{1-\delta}(\frac{1}{\sigma_p}), h(H)=\frac{1}{\sigma_p}.$

i)若$\lambda_n\geq H$, 则$\ln |b_n|e^{-\lambda_n\sigma_p}\leq\lambda_n(\frac{1}{h(H)}-\sigma_p)=0.$

ii)若$\lambda_n < H$, 则存在$m\in N$使

$ \frac{1}{(m+1)\sigma_p}U^{1-\delta}(\frac{1}{(m+1)\sigma_p})\leq\lambda_n\leq\frac{1}{m \sigma_p}U^{1-\delta}(\frac{1}{m\sigma_p}), \frac{1}{(m+1)\sigma_p}\leq h(\lambda_n)\leq \frac{1}{m \sigma_p}, $

从而

$ \ln |b_n|e^{-\lambda_n\sigma_p}\leq\frac{\lambda_n}{h(\lambda_n)}-\lambda_n\sigma_p \leq U^{1-\delta}(\frac{1}{m\sigma_p})\leq U^{1-\delta}(\frac{1}{\sigma_p}). $

$(2)$当$n>m_j$, 结合$(2.10)$, $(2.11)$式, 有

$ \ln |b_n|e^{-\lambda_n\sigma_p}\leq\frac{\lambda_n}{h(\lambda_n)}-\lambda_n\sigma_p<0. $

(3) 当$n < n_j$, 结合$(2.8), (2.10), $ $(2.11)$式对充分大的$p$及$\ln |b_n|\geq1$有

$ \ln |b_n|e^{-\lambda_n\sigma_p}\leq \ln |b_n|U^{1-\delta}(\frac{\lambda_n}{\ln |b_n|})\leq U^{1-\delta}(\lambda_n)U^{1-\delta}(\lambda_{n_j})\\ U^{1-\delta}(\lambda_{m_j}^{\frac{1}{1+\varepsilon}})\leq U^{1-\delta}[(\frac{1}{\sigma_p})^{\frac{ 1+o(1)}{1+\varepsilon}}]\leq U^{(1-\delta)(\frac{1+o(1)}{1+\varepsilon})+o(1)}(\frac{1}{\sigma_p}). $

对$\ln |b_n| < 1$上面不等式显然成立, 从而有$\overline{\lim\limits_{p\rightarrow \infty}}\frac{\ln\ln m(\sigma_p, f)}{\ln U(\frac{1}{\sigma_p})} < 1.$

结合定理$2.1$有$\lim\limits_{\sigma\rightarrow 0^{+}}\frac{\ln\ln M(\sigma, f)}{\ln U(\frac{1}{\sigma})} < 1$矛盾.

引理3.1^[5]  设$X_n(\omega)$是概率空间$(\Omega, A, P)$上的独立随机变量序列, 它们的数学期望为零$(E(X_n)=0)$, 且方差为$E(|X_n|^2)=\sigma_n^2$.如果$E(\frac{|X_n|}{\sigma_n})=d_n$一致有下界$d>0$, 即对任意$n\in N$, 有$d_n\geq d>0$, 则对任意$H\in A$, 存在$B=B(d, H), K=K(H, X_n)\in N, $使得对任何复数列$b_n\in C$及任何$p>q\geq K$, 恒有

$ \int_H|\sum\limits_{n=q}^pb_nX_n(\omega)|^2P(d\omega)\geq B\sum\limits_{n=q}^p|b_n|^2\sigma_n^2. $

引理3.2^[7]  设$\{X_n(\omega)\}$是概率空间$(\Omega, A, P)$上有有限方差$\int_\Omega|X_n(\omega)|^2d\omega=\delta_n^2 < \infty$的随机变量序列, 则对几乎必然a.s.的$\omega\in\Omega$,

1) 存在$N=N(\omega)$, 当$n\geq N$时, 有$|X_n(\omega)| < n\delta_n$;

2)$\overline{\lim\limits_{n\rightarrow \infty}}\frac{ |X_n(w)|}{\delta_n}\geq \overline{\lim\limits_{n\rightarrow \infty}}\frac{ E(|X_n|)}{\delta_n}$.

考虑与级数$(2.1)$相对应的随机Dirichlet级数$f_\omega(s)=\sum\limits_{n=0}^{\infty}b_nX_n(\omega)e^{-\lambda_ns}, $其中$s=\sigma+it, \{X_n(\omega)\}$是随机变量序列, 它们不一定是同分布的, 且上式满足

$ \overline{\lim\limits_{n\rightarrow \infty}}\frac{\ln |b_n\delta_n|}{\lambda_n}=0, $

(3.1)

则$f_\omega(s)$是一个半平面上的随机函数, 令

$ M(\sigma, f_\omega)=\sup\{|f_\omega(\sigma+it)|;t\in R\}, m(\sigma, f_\omega)=\max\{|b_nX_n(\omega)|e^{-\lambda_n\sigma} ;n\in N\}. $

定理$3.1$  设随机变量序列$\{X_n(\omega)\}$满足引理$3.1$的条件, 若随机Dirichlet级数$(3.1)$满足(3.2), (2.4) 式, 则

$ \overline{\lim\limits_{\sigma\rightarrow 0^+}}\frac{\ln\ln M(\sigma, f_\omega)}{\ln U(\frac{1}{\sigma})}=1~~{\hbox{a.s.}} \Leftrightarrow\overline{\lim\limits_{n\rightarrow \infty}}\frac{\ln\ln |b_n\delta_n|}{\ln U(\frac{\lambda_n}{\ln |b_n\delta_n|})}=1. $

(3.2)

证  令$Z_n(\omega)=\frac{X_n(\omega)}{\delta_n}$, 则$E(|Z_n|^2=1), E(|Z_n|)\geq d, X_n=\delta_nZ_n.$不妨记

$ t_n= \overline{\lim\limits_{n\rightarrow \infty}}\frac{\ln\ln |b_n\delta_n|}{\ln U(\frac{\lambda_n}{\ln |b_n\delta_n|})}. $

我们首先证明$\overline{\lim\limits_{n\rightarrow \infty}}t_n\geq 1\Rightarrow\overline{\lim\limits_{\sigma\rightarrow 0^+}}\frac{\ln\ln M(\sigma, f_\omega)}{\ln U(\frac{1}{\sigma})}\geq $1 a.s..设$\overline{\lim\limits_{n\rightarrow \infty}}t_n=\overline{\lim\limits_{n\rightarrow \infty}}t_{n_k}\geq 1$, 其中$t_{n_k}>0$, 则由引理3.2可得

$ \overline{\lim\limits_{k\rightarrow \infty}}|Z_{n_k}(\omega)|\geq\overline{\lim\limits_{k\rightarrow \infty}} E(|Z_{n_k}|)\geq d~~{\hbox{a.s.}}. $

对几乎必然的$\omega\in\Omega, \{Z_{n_k}(\omega)\}$有子序列$\{Z_{n_{k_l}}(\omega)\}, |Z_{n_{k_l}}|\geq\frac{d}{2}, \{n_{k_l}\}$为$\{n_k\}$的子序列.

令$t_n(\omega)=\frac{\ln\ln |b_nX_n(\omega)|}{\ln U(\frac{\lambda_n}{\ln |b_nX_n(\omega)|})}, $对于

$ \frac{2}{d}f_\omega(s)=\sum\limits_{n=0}^{+\infty}\frac{2}{d}b_nX_n(\omega)e^{-\lambda_ns}, t_n^*(\omega) =\frac{\ln\ln |\frac{2}{d}b_nX_n(\omega)|}{\ln U(\frac{\lambda_n}{\ln |\frac{2}{d}b_nX_n(\omega)|})}. $

故有

$ \overline{\lim\limits_{n\rightarrow \infty}}t_n^*(\omega)=\overline{\lim\limits_{l\rightarrow \infty}}t_{n_{k_l}}^* (\omega)=\overline{\lim\limits_{l\rightarrow \infty}}\frac{\ln\ln |\frac{2}{d}b_{n_{k_l}}Z_{n_{k_l}}(\omega)\delta_{n_{k_l}}|}{\ln U(\frac {\lambda_{n_{k_l}}}{\ln |\frac{2}{d}b_{n_{k_l}}Z_{n_{k_l}}(\omega)\delta_{n_{k_l}}|})}\geq\overline{\lim\limits_{l\rightarrow \infty}} \frac{\ln\ln |b_{n_{k_l}}\delta_{n_{k_l}}|}{\ln U(\frac{\lambda_{n_{k_l}}}{\ln |b_{n_{k_l}}\delta_{n_{k_l}}|})}\\ =\overline{\lim\limits_{l\rightarrow \infty}}t_{n_{k_l}}=\lim\limits_{k\rightarrow\infty}t_{n_k}= \overline{\lim\limits_{n\rightarrow \infty}}t_n\geq 1, $

结合定理$2.2$可知

$ \overline{\lim\limits_{\sigma\rightarrow 0^+}}\frac{\ln\ln M(\sigma, f_\omega)}{\ln U(\frac{1}{\sigma})} =\overline{\lim\limits_{\sigma\rightarrow 0^+}}\frac{\ln\ln M(\sigma, \frac{2}{d}f_\omega)}{\ln U(\frac{1}{\sigma})}= \overline{\lim\limits_{n\rightarrow \infty}}t_n^*(\omega)\geq 1~~{\hbox{a.s.}}. $

下面将证

$ \overline{\lim\limits_{n\rightarrow \infty}}t_n\leq 1\Rightarrow\overline{\lim\limits_{\sigma\rightarrow 0^+}} \frac{\ln\ln M(\sigma, f_\omega)}{\ln U(\frac{1}{\sigma})}\leq 1~~{\hbox{a.s.}}. $

令$g(s)=\sum\limits_{n=0}^{+\infty}b_n\delta_ne^{-\lambda_ns}, \Delta=\frac{\sigma\ln U(\frac{1}{\sigma})}{1+\ln U(\frac{1}{\sigma})}, \delta=\frac{\sigma}{1+\ln U(\frac{1}{\sigma})}$, 由引理$3.2$可得

$ M(\sigma, f_\omega)=\sum\limits_{n=0}^{+\infty}|b_nX_n(\omega)|e^{-\lambda_n\sigma}= \sum\limits_{n=0}^{+\infty}|b_n\delta_nZ_n(\omega)|e^{-\lambda_n\sigma}={\sum\limits_{n=0}^{+\infty}|b_n\delta_n Z_n(\omega)|e^{-\lambda_n\Delta}}e^{{-\lambda_n}{\delta}}\\ \leq m(\Delta, g)\sum\limits_{n=0}^{+\infty}(|Z_n(\omega)|e^{-{\lambda_n}{\delta}})\leq m(\Delta, g)(C+ \sum\limits_{n=1}^{+\infty}C(\omega)ne^{-{\lambda_n}{\delta}}), $

其中$C, C(\omega)$是与$\omega$有关的常数.类似定理$1$证明过程, 结合定理$2.2$可得当$\overline{\lim\limits_{n\rightarrow \infty}}t_n\leq 1$时,

$ \overline{\lim\limits_{\sigma\rightarrow 0^+}}\frac{\ln\ln M(\sigma, f_\omega)}{\ln U(\frac{1}{\sigma})}\leq \overline{\lim\limits_{\sigma\rightarrow 0^+}}\frac{\ln\ln m(\Delta, g)}{\ln U(\frac{1}{\sigma})} =\overline{\lim\limits_{\sigma\rightarrow 0^+}}\frac{\ln\ln M(\Delta, g)}{\ln U(\frac{1}{\sigma})} =\overline{\lim\limits_{\sigma\rightarrow 0^+}}\frac{\ln\ln M(\sigma, g)}{\ln U(\frac{1}{\sigma})}\leq 1. $

定理证毕.

仿照文[7]的证明, 有

定理$3.2$  设随机Dirichlet级数$(3.1)$满足$(2.4)$, $(3.2)$以及$(3.3)$式, $\{X_n(\omega)\}$是引理$3.1$中的随机变量序列, 则级数$(3.1)$几乎(a.s.)对任意的$t_0\in R$有

$ \overline{\lim\limits_{\sigma\rightarrow 0^+}}\frac{\ln\ln |f_\omega(\sigma+it_0)|}{\ln U(\frac{1}{\sigma})}=1, $

其中$U(r)$是定义$2.2$中满足$1), 2)$的型函数.