Laplace-Stieltjes变换(以下简称L-S变换)

(1.1) $ \begin{equation} G(s) = \int_0^{+\infty}e^{-sx}{\rm d}\alpha(x), s = \sigma+{\rm i}t, \end{equation} $

其中$ \alpha(x) $为区间$ [0, Y](0<Y<+\infty) $上的有界变差函数, $ \sigma, t $为实数.若选取$ \{\lambda_n\}_0^{\infty} $满足

(1.2) $ \begin{equation} 0\leq \lambda_1<\lambda_2<\cdots<\lambda_n<\cdots, \lambda_n\rightarrow\infty, \; \; n\rightarrow\infty \end{equation} $

与

$ \alpha(x) = \left\{\begin{array}{ll} a_1+a_2+\cdots+a_n, &\lambda_n\leq x<\lambda_{n+1}, \\ 0, &x = 0, \\ \frac{\alpha(x+)+\alpha(x-)}{2}, \; \; &x>0, \end{array}\right. $

那么

$ G(s) = \int_0^{+\infty}e^{-sx}{\rm d}\alpha(x) = \sum\limits_{n = 0}^{+\infty}a_ne^{-s\lambda_n}: = f(s), $

上述级数被称为Dirichlet级数.若$ a_n, \lambda_n, n $满足某些特定条件, $ f(s) $于全平面或半平面内解析.过去的几十年间, Dirichlet级数所表示的整函数的增长性与值分布研究经典而有趣,出现了许多重要的成果(参见文献[3, 13-14, 17, 20-21, 27-28]).

1963年,余家荣^[26]首次介绍了全平面内收敛的L-S变换所表示的整函数的``最大模" $ M_u(\sigma, G) $、``最大项" $ \mu(\sigma, G) $、Borel线以及级等概念,并分析了它们的性质,同时证明了L-S变换的收敛横坐标,绝对收敛横坐标以及一致收敛横坐标的Valiron-Knopp-Bohr公式.

定理A^[26]  若L-S变换(0.1)满足条件(0.2),且

(1.3) $ \begin{equation} \limsup\limits_{n\rightarrow+\infty}(\lambda_{n+1}-\lambda_n) = h<+\infty, \; \; \limsup\limits_{n\rightarrow+\infty}\frac{\log n}{\lambda_n} = D<+\infty, \end{equation} $

则

$ \limsup\limits_{n\rightarrow+\infty}\frac{\log B_n^\ast}{\lambda_n}\leq \sigma_u^G\leq \limsup\limits_{n\rightarrow+\infty}\frac{\log B_n^\ast}{\lambda_n}+\limsup\limits_{n\rightarrow+\infty}\frac{\log n}{\lambda_n}, $

这里$ \sigma_u^G $为$ G(s) $一致收敛横坐标,

$ B_n^* = \sup\limits_{\lambda_n<x\leq \lambda_{n+1}, -\infty<t<+\infty}\left|\int_{\lambda_n}^xe^{-{\rm i}ty}{\rm d}\alpha(y)\right|. $

随后,国内外许多学者对L-S变换所表示的解析函数的增长性与值分布等问题投入了大量的关注,得到了很多有意义的结果(参见文献[1, 4, 9, 12]).文献[5-6, 8]讨论了零级,有限级以及无限级L-S变换所表示解析函数的增长性,获得了L-S变换具有各种增长级的等价关系;文献[15-16, 22-23, 29]研究了L-S变换的值分布性质,得到了L-S变换具有Borel点与Borel线等各种奇异方向的条件.

2012年与2014年,罗茜,刘兴臻以及孔荫莹^[10-11]讨论了形式与变换(1.1)不同的L-S变换

(1.4) $ \begin{equation} F(s) = \int_0^{+\infty}e^{sx}{\rm d}\alpha(x), s = \sigma+{\rm i}t, \end{equation} $

其中$ \alpha(x) $如(0.1)式所述, $ \{\lambda_n\} $满足条件(1.2)和(1.3).若

(1.5) $ \begin{equation} \limsup\limits_{n\rightarrow+\infty}\frac{\log A_n^\ast}{\lambda_n} = -\infty, \end{equation} $

这里

$ A_n^* = \sup\limits_{\lambda_n<x\leq \lambda_{n+1}, -\infty<t<+\infty}\left|\int_{\lambda_n}^xe^{{\rm i}ty}{\rm d}\alpha(y)\right|, $

利用文献[26]中的讨论,易得$ \sigma_u^F = +\infty $,即, $ F(s) $为整函数.为简便,记$ \overline{L}_\beta $为形如(1.4)式并于半平面$ \Re s<\beta(-\infty <\beta<\infty) $内收敛,以及序列$ \{\lambda_n\} $满足条件(0.2)与(0.3); $ L_\infty $为形如(1.4)式并于全平面$ \Re s<+\infty $内收敛,以及序列$ \{\lambda_n\} $满足条件(0.2), (0.3)和(0.5).这样,若$ F(s)\in L_\infty $,则$ F(s)\in \overline{L}_\beta $,对$ -\infty<\beta<+\infty $.

令

$ M_u(\sigma, F) = \sup\limits_{0<x<+\infty, -\infty<t<+\infty}\left|\int_0^xe^{(\sigma+{\rm i}t)y}{\rm d}\alpha(y)\right|, $

$ \mu(\sigma, F) = \max\limits_{n\in N}\{A_n^*e^{\lambda_n\sigma}\}, N(\sigma, F) = \max\{\lambda_n: \; A_n^*e^{\lambda_n\sigma} = \mu(\sigma, F)\}. $

在叙述文献[10-11]结果前,先介绍L-S变换的级,下级,型与下型的概念如下(参见文献[10-11, 24]).

定义1.1  若L-S变换$ F(s)\in L_\infty $满足

$ \limsup\limits_{\sigma\rightarrow+\infty}\frac{\log^+\log^+M_u(\sigma, F)}{\sigma} = \rho, $

则称$ \rho $为$ F(s) $的级,这里$ \log^+x = \max\{\log x, 0\} $.如果$ \rho\in (0, +\infty) $,则称$ F(s) $为有限级整函数. $ F(s) $的下级定义为

$ \liminf\limits_{\sigma\rightarrow+\infty}\frac{\log^+\log^+M_u(\sigma, F)}{\sigma} = \mu. $

注1.1  若$ \rho = \mu $,则称$ F(s) $具有正规增长;若$ \rho\neq\mu $,则称$ F(s) $具有非正规增长.

定义1.2  若L-S变换$ F(s)\in L_\infty $具有级$ \rho (0<\rho<\infty) $,那么L-S变换$ F(s) $的型与下型分别定义为

$ \limsup\limits_{\sigma\rightarrow+\infty}\frac{\log^+M_u(\sigma, F)}{e^{\sigma\rho}} = T, \; \; \; \; \liminf\limits_{\sigma\rightarrow+\infty}\frac{\log^+M_u(\sigma, F)}{e^{\sigma\rho}} = \tau. $

根据文献[7, 10, 24, 26],我们容易得到L-S变换(1.4)的级,下级,型与下型的相关结果.

定理B^[10]  若L-S变换$ F(s)\in L_\infty $,其级为$ \rho (0<\rho<\infty) $,则

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

定理C^[24]  若L-S变换$ F(s)\in L_\infty $,其级为$ \rho (0<\rho<\infty) $,型为$ T $,则

$ T = \limsup\limits_{n\rightarrow+\infty}\frac{\lambda_n}{\rho e}(A_n^*)^{\frac{\rho}{\lambda_n}}. $

进一步,若$ F(s) $的下级为$ \mu $,下型为$ \tau $,且$ \lambda_n\sim\lambda_{n+1} $,函数

$ \psi(n) = \frac{\log A^*_n-\log A^*_{n+1}}{\lambda_{n+1}-\lambda_n} $

关于$ n(>n_0) $非减,则

$ \mu = \liminf\limits_{n\rightarrow+\infty}\frac{\lambda_n\log\lambda_n}{-\log A_n^*}, \; \; \; \tau = \liminf\limits_{n\rightarrow+\infty}\frac{\lambda_n}{\rho e}(A_n^*)^{\frac{\rho}{\lambda_n}}. $

此外,文献[11]与[25]讨论了慢增长L-S变换(1.4)的增长性,证明了变换的$ h $ -级,对数级,对数型与$ A_n^* $, $ \lambda_n $的几个关系定理.然而,较少有涉及无限级L-S变换(1.4)的文献.本文第一个目的是通过引入$ q $ -级,下$ q $ -级, $ q $ -型,下$ q $ -型以及双下$ q $ -型等概念,讨论无限级L-S变换(1.4)的增长性.

定义1.3  令$ q\geq 1 $, $ q\in {\Bbb N}_+ $,定义L-S变换$ F(s)\in L_\infty $的$ q $ -级$ \rho_q $为

$ \limsup\limits_{\sigma\rightarrow+\infty}\frac{\log_{q+1} M_u(\sigma, F)}{\sigma} = \rho_q, \; \; \rho_q\in [0, +\infty]. $

如果$ \rho_q\in (0, +\infty) $,则称$ F(s) $为具有有限$ q $ -级的整函数.此外,定义$ F(s) $的下$ q $ -级为

$ \liminf\limits_{\sigma\rightarrow+\infty}\frac{\log_{q+1}M_u(\sigma, F)}{\sigma} = \mu_q. $

注1.2  显然,若$ q = 1 $,有$ \rho_q = \rho $, $ \mu_q = \mu $;若$ q\geq 2 $, $ \rho_q\in (0, +\infty) $,则$ \rho = \infty $.如无特别说明,本文约定$ q\geq 2 $.

注1.3  若$ \rho_q = \mu_q $,则称$ F(s) $具有$ q $ -正规增长,否则,称$ F(s) $为$ q $ -非正规增长.

定义1.4  若L-S变换$ F(s)\in L_\infty $,具有有限$ q $ -级$ \rho_q (0<\rho_q<\infty) $,定义$ F(s) $的$ q $ -型$ T_q $,下$ q $ -型$ \tau_q $分别为

$ \limsup\limits_{\sigma\rightarrow+\infty}\frac{\log_{q}M_u(\sigma, F)}{e^{\sigma\rho_q}} = T_q, \; \; \; \; \liminf\limits_{\sigma\rightarrow+\infty}\frac{\log_{q}M_u(\sigma, F)}{e^{\sigma\rho_q}} = \tau_q. $

类似定理B与定理C,对于有限$ q $ -级L-S变换(1.4),这里不加证明的给出下列结果.

定理1.1  若L-S变换$ F(s)\in L_\infty $,具有有限$ q $ -级$ \rho_q (0<\rho_q<\infty) $与$ q $ -型$ T_q $,则

$ \rho_q = \limsup\limits_{n\rightarrow+\infty}\frac{\lambda_n\log_q\lambda_n}{-\log A_n^*}, \; \; \; T_q = \limsup\limits_{n\rightarrow+\infty}\log_{q-1}\left[\frac{\lambda_n}{e\rho_q}\right](A_n^*)^{\frac{\rho_q}{\lambda_n}}. $

进一步,若$ F(s) $具有下$ q $ -级$ \mu_q $与下$ q $ -型$ \tau_q $,且函数$ \psi(n) $关于$ n(>n_0) $非减,以及当$ n\rightarrow+\infty $时, $ \log_{q-1}\lambda_n\sim\log_{q-1}\lambda_{n+1} $,则

$ \mu_q = \liminf\limits_{n\rightarrow+\infty}\frac{\lambda_n\log_q\lambda_n}{-\log A_n^*}, \; \; \; \tau_q = \liminf\limits_{n\rightarrow+\infty}\log_{q-1}\left[\frac{\lambda_n}{e\rho_q }\right](A_n^*)^{\frac{\rho_q}{\lambda_n}}. $

问题  注意下$ q $ -型定义,若$ e^{\sigma\rho_q} $被$ e^{\mu_q\sigma} $替换,情况会怎样呢?因此,先给出如下定义:

定义1.5  若L-S变换$ F(s)\in L_\infty $,具有有限$ q $ -级$ \rho_q (0<\rho_q<\infty) $与下$ q $ -级$ \mu_q(0<\mu_q<\infty) $,且$ \mu_q\neq \rho_q $,定义$ F(s) $的双下$ q $ -型$ \tau_{\mu_q} $为

$ \liminf\limits_{\sigma\rightarrow+\infty}\frac{\log_qM_u(\sigma, F)}{e^{\sigma\mu_q}} = \tau_{\mu_q}. $

注1.4  显然, $ \tau_{\mu_q}\geq \tau_q $.不过,很难确定是否$ \tau_{\mu_q}\geq T_q $或$ \tau_{\mu_q}\leq T_q $.

对L-S变换的$ q $ -非正规增长,我们得到:

定理1.2  若L-S变换$ F(s)\in L_\infty $,具有有限$ q $ -级$ \rho_q $,下$ q $ -级$ \mu_q $,且$ 0\leq \mu_q\neq \rho_q<\infty $,则

(1.6) $ \begin{equation} \tau_q = \liminf\limits_{\sigma\rightarrow+\infty}\frac{\log_q M(\sigma, F)}{e^{\rho_q\sigma}} = \liminf\limits_{\sigma\rightarrow+\infty}\frac{\log_q\mu(\sigma, F)}{e^{\rho_q\sigma}} = \liminf\limits_{\sigma\rightarrow+\infty}\frac{\log_{q-1}N(\sigma, F)}{e^{\rho_q\sigma}} = 0. \end{equation} $

定理1.3  若L-S变换$ F(s)\in L_\infty $,具有有限$ q $ -级$ \rho_q $,下$ q $ -级$ \mu_q $,且$ 0\leq \mu_q\neq \rho_q<\infty $,如果当$ n\rightarrow+\infty $时, $ \log_q\lambda_n\sim\log_q\lambda_{n+1} $,则

(1.7) $ \begin{equation} \tau_{\mu_q}\geq \liminf\limits_{n\rightarrow\infty}(A^*_n)^{\frac{\mu_q}{\lambda_n}}\log_{q-1}\left(\frac{\lambda_n}{e\mu_q}\right), \; \; \; \; (0\leq \tau_{\mu_q}\leq \infty). \end{equation} $

进一步,存在正整数$ n_0 $使得

$ \psi(n) = \frac{\log A^*_n-\log A^*_{n+1}}{\lambda_{n+1}-\lambda_n} $

关于$ n(>n_0) $非减,则

(1.8) $ \begin{equation} \tau_{\mu_q} = \liminf\limits_{n\rightarrow\infty}(A^*_n)^{\frac{\mu_q}{\lambda_n}}\log_{q-1}\left(\frac{\lambda_n}{e\mu_q}\right), \; \; \; \; (0\leq \tau_{\mu_q}\leq \infty). \end{equation} $

2017年, Singhal-Srivastava^[18]讨论了有限级L-S变换(1.4)的逼近,得到

定理D^[18]  若L-S变换$ F(s)\in L_\infty $,具有有限级$ \rho (0<\rho<\infty) $与型$ T $,则对任意实数$ \beta(-\infty<\beta<+\infty) $,有

$ \rho = \limsup\limits_{n\rightarrow+\infty}\frac{\lambda_n\log\lambda_n}{-\log E_{n-1}(F, \beta)\exp(-\beta\lambda_n)} = \limsup\limits_{n\rightarrow+\infty}\frac{\lambda_n\log\lambda_n}{-\log E_{n-1}(F, \beta)} $

与

$ T = \limsup\limits_{n\rightarrow+\infty}\frac{\lambda_n}{\rho e}(E_{n-1}(F, \beta)\exp(-\beta\lambda_n))^{\frac{\rho}{\lambda_n}} = \limsup\limits_{n\rightarrow+\infty}\frac{\lambda_n}{\rho\exp(\rho\beta+1)}(E_{n-1}(F, \beta))^{\frac{\rho}{\lambda_n}}. $

进一步,若$ F(s) $具有下级$ \mu $与下型$ \tau $,且$ \lambda_n\sim\lambda_{n+1} $以及函数$ \psi(n) $关于$ n(>n_0) $非减,则

$ \mu = \liminf\limits_{n\rightarrow+\infty}\frac{\lambda_n\log\lambda_n}{-\log E_{n-1}(F, \beta)}, \; \; \; \tau = \liminf\limits_{n\rightarrow+\infty}\frac{\lambda_n}{\rho \exp(\rho\beta+1)}(E_{n-1}(F, \beta))^{\frac{\rho}{\lambda_n}}. $

注1.5  

$ E_n(F, \beta) = \inf\limits_{p\in\Pi_n}\parallel F-p\parallel_\beta, n = 1, 2, \cdots , $

其中

$ \parallel F-p\parallel_\beta = \max\limits_{-\infty<t<+\infty}|F(\beta+{\rm i}t)-p(\beta+{\rm i}t)|, $

$ \Pi_k = \left\{\sum\limits_{j = 1}^k b_je^{\lambda_js}: (b_1, b_2, \cdots , b_k)\in {\Bbb C}^k\right\}. $

注1.6  若L-S变换(1.4)满足$ n\geq k+1 $时, $ A^*_n = 0 $且$ A^*_k\neq0 $,则称$ F(s) $为阶为$ k $的指数型多项式,记为$ p_k $,即, $ p_k(s) = \int^{\lambda_k}_0\exp(sy){\rm d}\alpha(y) $.若选取适合的$ \alpha(y) $,函数$ p_k(s) $可简化为$ \exp(s\lambda_i) $项的多项式,即, $ \sum\limits_{i = 1}^kb_i\exp(s\lambda_i) $.

受文献[18, 24-25]启发,本文将讨论逼近算子$ E_n(F, \beta) $与双下$ q $ -型$ \tau_{\lambda_q} $的内在关系,证明了下述结果.

定理1.4  若L-S变换$ F(s)\in L_\infty $,具有下$ q $ -级$ \lambda_q (0\leq \mu_q\neq \rho_q<\infty) $,若$ \log_q\lambda_n\sim\log_q\lambda_{n+1} $,则对任意的实数$ \beta(-\infty<\beta<+\infty) $,有

(1.9) $ \begin{equation} \tau_{\mu_q}\geq \liminf\limits_{n\rightarrow\infty}(E_{n-1}(F, \beta)\exp(-\beta\lambda_n))^{\frac{\mu_q}{\lambda_n}}\log_{q-1}\left(\frac{\lambda_n}{e\mu_q}\right), \; \; \; \; (0\leq \tau_{\mu_q}\leq \infty). \end{equation} $

进一步,若函数$ \psi(n) $关于$ n(>n_0) $非减,则

$ \tau_{\mu_q} = \liminf\limits_{n\rightarrow\infty}(E_{n-1}(F, \beta)\exp(-\beta\lambda_n))^{\frac{\mu_q}{\lambda_n}}\log_{q-1}\left(\frac{\lambda_n}{e\mu_q}\right), \; \; \; \; (0\leq \tau_{\mu_q}\leq \infty), $

即

(1.10) $ \begin{equation} \exp(\beta\mu_q)\tau_{\mu_q} = \liminf\limits_{n\rightarrow\infty}\log_{q-1}\left(\frac{\lambda_n}{e\mu_q}\right)(E_{n-1}(F, \beta))^{\frac{\mu_q}{\lambda_n}}. \end{equation} $

为证明定理1.2与1.3,需要以下两引理.

引理2.1^[10]  若L-S变换$ F(s)\in L_\infty $,则对任意的$ \sigma(-\infty<\sigma<+\infty) $和$ \varepsilon(>0) $,有

$ \frac{1}{2}\mu(\sigma, F)\leq M_u(\sigma, F)\leq K\mu((1+2\varepsilon)\sigma, F), $

这里$ K $为常数.

引理2.2^[11]  若L-S变换$ F(s)\in L_\infty $,则对$ \sigma_0>0 $,有

$ \log \mu(\sigma, F) = \log \mu(\sigma_0, F)+\int_{\sigma_0}^\sigma N(t, F){\rm d}t. $

定理1.2的证明  由$ F(s) $的下$ q $ -级为$ \mu_q $且$ \rho_q>\mu_q>0 $,则对任意小的$ \varepsilon\ (0<\varepsilon<\rho_q-\mu_q) $,存在常数$ \sigma_0 $,使得当$ \sigma>\sigma_0 $时,有

(2.1) $ \begin{equation} \log_q M_u(\sigma, F)>\exp\{(\mu_q-\varepsilon)\sigma\}, \end{equation} $

以及趋于无穷的一序列$ \{\sigma_k\} $,使得

(2.2) $ \begin{equation} \log_q M_u(\sigma_k, F)<\exp\{(\mu_q+\varepsilon)\sigma_k\}. \end{equation} $

那么,当$ k\rightarrow+\infty $时,有

$ \exp\{(\mu_q-\varepsilon-\rho_q)\sigma_k\}\leq \frac{\log_q M_u(\sigma, F)}{\exp(\rho_q\sigma_k)} \leq \exp\{(\mu_q+\varepsilon-\rho_q)\sigma_k\}. $

结合$ 0<\varepsilon<\rho_q-\mu_q $,可得

(2.3) $ \begin{equation} \liminf\limits_{\sigma\rightarrow+\infty}\frac{\log_q M_u(\sigma, F)}{\exp(\rho\sigma)} = 0. \end{equation} $

再根据引理2.1与引理2.2,有

$ \rho_q = \limsup\limits_{\sigma\rightarrow+\infty}\frac{\log_{q+1} M_u(\sigma, F)}{\sigma} = \limsup\limits_{\sigma\rightarrow+\infty}\frac{\log_{q+1} \mu(\sigma, F)}{\sigma} = \limsup\limits_{\sigma\rightarrow+\infty}\frac{\log_{q} N(\sigma, F)}{\sigma} $

与

$ \mu_q = \liminf\limits_{\sigma\rightarrow+\infty}\frac{\log_{q+1} M_u(\sigma, F)}{\sigma} = \liminf\limits_{\sigma\rightarrow+\infty}\frac{\log_{q+1} \mu(\sigma, F)}{\sigma} = \liminf\limits_{\sigma\rightarrow+\infty}\frac{\log_{q} N(\sigma, F)}{\sigma}. $

类似于(2.3)式的证明,易得

$ \liminf\limits_{\sigma\rightarrow+\infty}\frac{\log_q \mu(\sigma, F)}{\exp(\rho_q\sigma)} = \liminf\limits_{\sigma\rightarrow+\infty}\frac{\log_{q-1}N(\sigma, F)}{\exp(\rho_q\sigma)} = 0. $

故定理1.2证毕.

定理1.3的证明  分两情形证明定理1.3.

情形2.1   $ q = 2 $.令

$ \vartheta = \liminf\limits_{n\rightarrow+\infty}(A_n^*)^{\frac{\mu_2}{\lambda_n}}\log \frac{\lambda_n}{e\mu_2}, \; \; \; (0<\vartheta<+\infty). $

于是,对任意的$ \varepsilon>0 $,存在整数$ n_0(\varepsilon) $使得当$ n>n_0(\varepsilon) $时,有

(2.4) $ \begin{equation} \log A_n^*>\frac{\lambda_n}{\mu_2}\log\left\{(\vartheta-\varepsilon)\left(\log \frac{\lambda_n}{e\mu_2}\right)^{-1}\right\}. \end{equation} $

依据引理2.1与(2.4)式,对$ n>n_0(\varepsilon) $,可得

(2.5) $ \begin{eqnarray} \frac{\log_2 M_u(\sigma, F)}{e^{\mu_2\sigma}}&\geq& \frac{\log(\log A_n^*+\lambda_n\sigma-\log 2)}{e^{\mu_2\sigma}} \nonumber\\ & >&e^{-\mu_2\sigma}\log\left\{\frac{\lambda_n}{\mu_2}\log\left\{(\vartheta-\varepsilon)\left(\log \frac{\lambda_n}{e\mu_2}\right)^{-1}\right\}+\lambda_n\sigma-\log2\right\}. \end{eqnarray} $

令

$ \frac{1}{\mu_2}\left[\log\frac{\log\frac{\lambda_n}{e\mu_2}}{\vartheta} \right] \leq \sigma<\frac{1}{\mu_2}\left[\log\frac{\log\frac{\lambda_{n+1}}{e\mu_2}}{\vartheta} +\frac{1}{\log\frac{\lambda_n}{e\mu_2}}\right], $

选取

$ \sigma = \frac{1}{\mu_2}\left[\log \frac{\log\left(\frac{\lambda_n}{e\mu_2}\right)}{\vartheta} +\frac{1}{\log\frac{\lambda_n}{e\mu_2}}\right]. $

于是,根据(2.5)式得

(2.6) $ \begin{equation} \frac{\log_2 M_u(\sigma, F)}{e^{\mu_2\sigma}} \geq\frac{\vartheta\log\left\{\frac{\lambda_n}{\mu_2}\left[\log\frac{\vartheta-\varepsilon}{\vartheta} +\frac{1}{\log\frac{\lambda_n}{e\mu_2}}\right]-\log2\right\}} {\log\left(\frac{\lambda_{n+1}}{e\mu_2}\right)\exp\left(\frac{1}{\log\frac{\lambda_n}{e\mu_2}}\right)}. \end{equation} $

又因为当$ n\rightarrow+\infty $时, $ \lambda_n\sim\lambda_{n+1} $和$ \lambda_n\rightarrow+\infty $,那么

(2.7) $ \begin{equation} \log\frac{\lambda_{n+1}}{\log\frac{\lambda_n}{e\mu_2}} \sim\log\left(\frac{\lambda_n}{\mu_2}\right), \; \; \; e^{\frac{1}{\log\frac{\lambda_n}{e\mu_2}}}\rightarrow1, \; \; n\rightarrow+\infty. \end{equation} $

让$ \varepsilon\rightarrow0 $和$ n\rightarrow+\infty $,由(2.6)与(2.7)式,易得$ \tau_{\mu_2}\geq \vartheta. $

若$ \vartheta = 0 $,显然有$ \tau_{\mu_q}\geq \vartheta $;若$ \vartheta = \infty $,类似上述讨论可得$ \tau_{\mu_2}\geq \vartheta $.于是, (1.7)式成立.

接下来证明(0.8)式对$ q = 2 $成立.令$ \mu(\sigma, F) $为$ \Re s = \sigma, -\infty<t<+\infty $时$ F(s) $的最大项.由

$ \psi(n) = \frac{\log A^*_n-\log A^*_{n+1}}{\lambda_{n+1}-\lambda_n} $

关于$ n(>n_0) $非减,若$ \psi(n-1)\leq \sigma<\psi(n) $,有

$ \log \mu(\sigma, F) = \log A_n^*+\lambda_n\sigma. $

由$ \tau_{\mu_2}<\infty $,根据定理1.2的证明,对任意小的$ \varepsilon>0 $,以及$ \sigma>\sigma_0 $,对满足$ \psi(n-1)\leq \sigma<\psi(n) $的所有$ n $,有

(2.8) $ \begin{equation} \log \mu(\sigma, F) = \log A_n^*+\lambda_n\sigma\geq (\tau_{\mu_2}-\varepsilon)\exp(\mu_2\sigma). \end{equation} $

令$ A_{n_1}^*\exp(\sigma\lambda_{n_1}) $和$ A_{n_2}^*\exp(\sigma\lambda_{n_2}) $为$ \Re s = \sigma(>\sigma_0) $的两个连续最大项,这里$ n_1>n_0, $$ \psi(n-1)>\sigma_0 $以及$ n_2-1\geq n_1 $.这样,由(1.8)式,对于所有的$ \sigma>\sigma_0 $且满足$ \psi(n_2-1)\leq \sigma<\psi(n_2) $,有

$ \log(\log A_{n_2}^*+\lambda_{n_2}\sigma)\geq (\tau_{\mu_2}-\varepsilon)\exp(\mu_2\sigma); $

对$ n_1\leq n\leq n_2-1 $,有

$ \psi(n_1) = \psi(n_1+1) = \cdots = \psi(n) = \cdots = \psi(n_2-1), $

以及当$ \sigma = \psi(n) $时,有$ A_n^*\exp(\lambda_n\sigma) = A_{n_2}^*\exp(\lambda_{n_2}\sigma) $.因此,存在正整数$ n_1' $使得当$ n>n_1' = \max\{n_0, n_1\} $, $ \sigma>\sigma_0 $时,有

$ A_n^*>\exp\left\{\exp\left\{(\tau_{\mu_2}-\varepsilon)e^{\mu_2\sigma}\right\}-\lambda_n\sigma\right\}. $

于是

(2.9) $ \begin{equation} (A_n^*)^{\frac{\mu_2}{\lambda_n}} >\exp\left\{\frac{\mu_2}{\lambda_n} \exp\exp\left[\log(\tau_{\mu_2}-\varepsilon)+\mu_2\sigma\right]-\mu_2\sigma\right\}. \end{equation} $

取

$ \sigma = \frac{1}{\mu_2}\left[\log\frac{\log(\frac{\lambda_n}{e\mu_2})}{\tau_{\mu_2}}+o(\frac{1}{\log\lambda_n})\right], $

那么

(2.10) $ \begin{eqnarray} (A_n^*)^{\frac{\mu_2}{\lambda_n}} &>&\exp\left\{\frac{\mu_2}{\lambda_n}\exp\exp\left[\log\frac{\tau_{\mu_2}-\varepsilon}{\tau_{\mu_2}} +\log\log(\frac{\lambda_n}{e\mu_2})+o(\frac{1}{\log\lambda_n})\right]\right.\\ & &\left.-\log\log(\frac{\lambda_n}{e\mu_2})-o(\frac{1}{\log\lambda_n})\right\}\\ &>& \exp\left\{\frac{1}{e}\exp\exp\left[\log\frac{\tau_{\mu_2}-\varepsilon}{\tau_{\mu_2}}+o(\frac{1}{\log\lambda_n})\right] +\log\frac{\tau_{\mu_2}}{\log(\frac{\lambda_n}{e\mu_2})}-o(\frac{1}{\log\lambda_n})\right\}. \end{eqnarray} $

上式让$ \varepsilon\rightarrow0 $和$ n\rightarrow+\infty $,结合对任意$ x>0 $有$ e^x\geq ex $成立,则

(2.11) $ \begin{equation} \vartheta = \liminf\limits_{n\rightarrow\infty}(A_n^*)^{\frac{\mu_2}{\lambda_n}}\log\left(\frac{\lambda_n}{e\mu_2}\right)\geq \tau_{\mu_2}. \end{equation} $

这样,结合(0.7)与(1.11)式,即(0.8)式当$ q = 2 $时成立.

情形2.2   $ q\geq 3 $.置

$ \vartheta_1 = \liminf\limits_{n\rightarrow+\infty}(A_n^*)^{\frac{\mu_q}{\lambda_n}}\log_{q-1}\left(\frac{\lambda_n}{e\mu_q}\right), \; \; \; (0<\vartheta<+\infty). $

于是,对任意的$ \varepsilon>0 $,存在正整数$ n_0(\varepsilon) $,使得当$ n>n_0(\varepsilon) $时,有

(2.12) $ \begin{equation} \log A_n^*>\frac{\lambda_n}{\mu_q}\log\left[ (\vartheta_1-\varepsilon)\left(\log_{q-1}\left(\frac{\lambda_n}{e\mu_q}\right)\right)^{-1}\right]. \end{equation} $

那么,对$ n>n_0(\varepsilon) $,由(1.12)式与引理2.1,有

(2.13) $ \begin{eqnarray} \frac{\log_{q} M_u(\sigma, F)}{e^{\mu_q\sigma}} & \geq &\frac{\log_{q-1}(\log A_n^*+\lambda_n\sigma-\log 2)}{e^{\mu_q\sigma}} \\ &>&e^{-\mu_q\sigma}\log_{q-1}\left\{\frac{\lambda_n}{\mu_q}\log\left\{(\vartheta_1-\varepsilon)\left(\log_{q-1} \frac{\lambda_n}{e\mu_q}\right)^{-1}\right\}+\lambda_n\sigma-\log2\right\}.\end{eqnarray} $

令

$ \frac{1}{\mu_q}\left[\log\frac{\log_{q-1}\frac{\lambda_n}{e\mu_q}}{\vartheta_1} \right] \leq \sigma<\frac{1}{\mu_q}\left[\log\frac{\log_{q-1}\frac{\lambda_{n+1}}{e\mu_q}}{\vartheta_1} +\frac{1}{\log_{q-1}\lambda_n}\right], $

取

$ \sigma = \frac{1}{\mu_q}\left[\log \frac{\log_{q-1}\left(\frac{\lambda_n}{e\mu_q}\right)}{\vartheta} +\frac{1}{\log_{q-1}\lambda_n}\right]. $

于是

$ \frac{\log_{q} M_u(\sigma, F)}{e^{\mu_q\sigma}} \geq\frac{\vartheta_1\log\left\{\frac{\lambda_n}{\mu_q}\left[\log\frac{\vartheta_1-\varepsilon}{\vartheta_1} +\frac{1}{\log_{q-1}\lambda_n}\right]-\log2\right\}} {\log_{q-1}\left(\frac{\lambda_{n+1}}{e\mu_q}\right)\exp\left(\frac{1}{\log_{q-1}\lambda_n}\right)}. $

类似情形2.1可得$ \tau_{\mu_q}\geq \vartheta_1 $.

另一方面,类似情形2.1,存在正整数$ n_3 $,使得当$ n>n_3 $, $ \sigma>\sigma_0 $时,有

$ A_n^*>\exp\left\{\exp_{p-1}\left\{(\tau_{\mu_q}-\varepsilon)e^{\mu_q\sigma}\right\}-\lambda_n\sigma\right\}, $

这里$ \exp_0(x) = x, \exp_1x = e^x $, $ \exp_p(x) = \exp(\exp_{p-1}(x)) $.于是,有

(2.14) $ \begin{equation} (A_n^*)^{\frac{\mu_q}{\lambda_n}} >\exp\left\{\frac{\mu_q}{\lambda_n} \exp_q\left[\log(\tau_{\mu_q}-\varepsilon)+\mu_q\sigma\right]-\mu_q\sigma\right\}. \end{equation} $

取

$ \sigma = \frac{1}{\mu_q}\left[\log\frac{\log_{q-1}(\frac{\lambda_n}{e\mu_q})}{\tau_{\mu_q}}+o(\frac{1}{\log_{q-1}\lambda_n})\right], $

类似(2.10)式的讨论易得$ \vartheta_1\geq \mu_q $,即(0.8)式对$ q\geq 3 $时也成立.

定理1.3证毕.

分两种情形证明定理1.4.

情形3.1   $ q = 2 $.置

$ \vartheta_2 = \liminf\limits_{n\rightarrow\infty}\log\left(\frac{\lambda_n}{e\mu_2}\right)(E_{n-1}(F, \beta)\exp(-\beta\lambda_n))^{\frac{\mu_2}{\lambda_n}} , \; \; \; \; (0<\vartheta_1<+\infty). $

由上式知,对任意小的$ \varepsilon>0 $,存在正整数$ n_0(\varepsilon) $,使得当$ n>n_0(\varepsilon) $时,有

(3.1) $ \begin{equation} \log (E_{n-1}(F, \beta)\exp(-\beta\lambda_n))>\frac{\lambda_n}{\mu_2}\log\left[ (\vartheta_2-\varepsilon)\left(\log\left(\frac{\lambda_n}{e\mu_2}\right)\right)^{-1}\right]. \end{equation} $

因为$ F(s)\in L_\infty $,即,对$ \beta(-\infty<\beta<+\infty) $, $ F(s)\in \overline{L}_\beta $.于是

(3.2) $ \begin{eqnarray} E_n(F, \beta) &\leq&\parallel F-p_n\parallel_\beta\leq |F(\beta+{\rm i}t)-p_n(\beta+{\rm i}t)|\\ &\leq &\left|\int_0^{+\infty}\exp\{(\beta+{\rm i}t)y\}{\rm d}\alpha(y)-\int_0^{\lambda_n}\exp\{(\beta+{\rm i}t)y\}{\rm d}\alpha(y)\right|\\ & = &\left|\int_{\lambda_n}^\infty\exp\{(\beta+{\rm i}t)y\}{\rm d}\alpha(y)\right|, \end{eqnarray} $

$ \left|\int_{\lambda_{k}}^\infty\exp\{(\beta+{\rm i}t)y\}{\rm d}\alpha(y)\right| = \lim\limits_{b\rightarrow+\infty}\left|\int_{\lambda_{k}}^b\exp\{(\beta+{\rm i}t)y\}{\rm d}\alpha(y)\right|. $

记$ I_{j+k}(b;{\rm i}t) = \int_{\lambda_{j+k}}^b\exp\{{\rm i}ty\}{\rm d}\alpha(y) $, $ \lambda_{j+k}<b\leq \lambda_{j+k+1} $,那么$ |I_{j+k}(b;{\rm i}t)|\leq A_{j+k}^* $与

$ \begin{eqnarray*} & &\left|\int_{\lambda_{k}}^b\exp\{(\beta+{\rm i}t)y\}{\rm d}\alpha(y)\right|\\ & = &\left|\sum\limits_{j = k}^{n+k-1}\int_{\lambda_j}^{\lambda_{j+1}}\exp\{\beta y\}{\rm d}_yI_j(y;{\rm i}t)+ \int_{\lambda_{n+k}}^b\exp\{\beta y\}{\rm d}_yI_{n+k}(y;{\rm i}t)\right|\\ & = &\left|\left[\sum\limits_{j = k}^{n+k-1}e^{\lambda_{j+1}\beta}I_j(\lambda_{j+1};{\rm i}t)-\beta\int_{\lambda_j}^{\lambda_{j+1}}e^{\beta y}I_j(y;{\rm i}t){\rm d}y\right]\right.\\ & &\left.+e^{\beta b}I_{n+k}(b;{\rm i}t)-\beta\int_{\lambda_{n+k}}^{b}e^{\beta y}I_j(y;{\rm i}t){\rm d}y\right|\\ &\leq&\sum\limits_{j = k}^{n+k-1}\left[A_j^*e^{\lambda_{j+1}\beta}+A_j^*(e^{\lambda_{j+1}\beta}-e^{\lambda_{j}\beta})\right] +2e^{\beta\lambda_{n+k+1}}A_{n+k}^*-e^{\beta\lambda_{n+k}}A_{n+k}^*\\ & \leq&2\sum\limits_{j = k}^{n+k} A_n^*e^{\beta\lambda_{n+1}}. \end{eqnarray*} $

上式$ b\rightarrow+\infty $时,得

(3.3) $ \begin{equation} \left|\int_{\lambda_{k}}^\infty\exp\{(\beta+{\rm i}t)y\}{\rm d}\alpha(y)\right|\leq 2\sum\limits_{n = k}^{+\infty}A_n^*\exp\{\beta\lambda_{n+1}\}. \end{equation} $

又因为对任意的$ \sigma(\beta<\sigma<+\infty) $, $ A_n^*\leq 2 M_u(\sigma, F)e^{-\sigma\lambda_n} $.于是,由(3.2)与(3.3)式,得

(3.4) $ \begin{equation} E_n(F, \beta)\leq 2\sum\limits_{k = n+1}^\infty A_{k-1}^*\exp\{\beta\lambda_k\}\leq 4M_u(\sigma, F)\sum\limits_{k = n+1}^\infty\exp\{(\beta-\sigma)\lambda_k\}. \end{equation} $

选取$ h'(0<h'<h) $,使得对$ n\geq0 $, $ (\lambda_{n+1} -\lambda_n)\geq h' $,再结合(3.4)式,对$ \sigma\geq \beta+1 $,可得

$ \begin{eqnarray*} E_n(F, \beta) &\leq& 4M_u(\sigma, F)\exp\{\lambda_{n+1}(\beta-\sigma)\}\sum\limits_{k = n+1}^\infty \exp\{(\lambda_k-\lambda_{n+1})(\beta-\sigma)\}\\ &\leq& 4M_u(\sigma, F)\exp\{\lambda_{n+1}(\beta-\sigma)\}\exp\{h'(n+1)\}\sum\limits_{k = n+1}^\infty (\exp\{-h'k\})\\ & = &4M_u(\sigma, F)\exp\{\lambda_{n+1}(\beta-\sigma)\}\left(1-\exp\{h'\}\right)^{-1}, \end{eqnarray*} $

即

(3.5) $ \begin{equation} E_{n-1}(F, \beta)\leq KM_u(\sigma, F)\exp\{\lambda_{n}(\beta-\sigma)\}, \end{equation} $

这里$ K $是常数.令

(3.6) $ \begin{equation} \gamma_n = E_{n-1}(F, \beta)\exp(-\beta\lambda_n), \; n = 1, 2, \cdots. \end{equation} $

由(3.1), (3.5)和(3.6)式,对$ n>n_0(\varepsilon) $,有

$ \begin{eqnarray*} \nonumber \frac{\log_2 M_u(\sigma, F)}{e^{\mu_2\sigma}}&\geq& \frac{\log(\log \gamma_n+\lambda_n\sigma-\log K)}{e^{\mu_2\sigma}} \\ & >&e^{-\mu_2\sigma}\log\left\{\frac{\lambda_n}{\mu_2}\log\left\{(\vartheta_2-\varepsilon)\left(\log \frac{\lambda_n}{e\mu_2}\right)^{-1}\right\}+\lambda_n\sigma-\log K\right\}. \end{eqnarray*} $

取

$ \sigma = \frac{1}{\mu_2}\left[\log \frac{\log\left(\frac{\lambda_n}{e\mu_2}\right)}{\vartheta_2} +\frac{1}{\log\frac{\lambda_n}{e\mu_2}}\right], $

类似情形2.1易得$ \tau_{\mu_2}\geq \vartheta_2 $.

另一方面,由

$ \psi(n) = \frac{\log A^*_n-\log A^*_{n+1}}{\lambda_{n+1}-\lambda_n} $

关于$ n(>n_0) $非减,类似定理1.3,存在正整数$ n_4 $,使得当$ n>n_4 $, $ \sigma>\sigma_0 $时,有

(3.7) $ \begin{equation} \log A_n^*>\exp\left\{(\tau_{\mu_2}-\varepsilon)e^{\mu_2\sigma}\right\}-\lambda_n\sigma. \end{equation} $

又因为对$ \beta<+\infty $,有

$ \begin{eqnarray*} \nonumber A_n^*\exp\{\beta\lambda_n\}& = &\sup\limits_{\lambda_n<x\leq\lambda_{n+1}, -\infty<t<+\infty}\left|\int_{\lambda_n}^x\exp\{{\rm i}ty\}{\rm d}\alpha(y)\right|\exp\{\beta\lambda_n\}\\ &\leq& \sup\limits_{\lambda_n<x\leq\lambda_{n+1}, -\infty<t<+\infty}\left|\int_{\lambda_n}^x\exp\{(\beta+{\rm i}t)y\}{\rm d}\alpha(y)\right| \\ &\leq &\sup\limits_{-\infty<t<+\infty}\left|\int_{\lambda_n}^\infty\exp\{(\beta+{\rm i}t)y\}{\rm d}\alpha(y)\right|, \end{eqnarray*} $

以及由$ \Pi_{n-1} $定义知,存在$ p_1\in \Pi_{n-1} $使得

(3.8) $ \begin{equation} \|F-p_1\|_\beta\leq 2E_{n-1}(F, \beta). \end{equation} $

那么,对任意的$ p\in \Pi_{n-1} $,有

(3.9) $ \begin{equation} A_n^*\exp\{\beta\lambda_n\}\leq |F(\beta+{\rm i}t)-p(\beta+{\rm i}t)|\leq \|F-p\|_\beta, \end{equation} $

结合(3.8)式可得

$ A_n^*\exp\{\beta\lambda_n\}\leq 2E_{n-1}(F, \beta). $

这样,对任意$ \beta<+\infty $与$ F(s)\in L_\infty $,再结合$ \gamma_n $的定义,得

(3.10) $ \begin{equation} \gamma_n\geq \frac{A_n^*}{2}. \end{equation} $

结合(3.7)与(3.10)式得

(3.11) $ \begin{equation} (\gamma_n)^{\frac{\mu_2}{\lambda_n}} >\exp\left\{\frac{\mu_2}{\lambda_n} \exp\exp\left[\log(\tau_{\mu_2}-\varepsilon)+\mu_2\sigma\right]-\mu_2\sigma-\frac{\mu_2\log 2}{\lambda_n}\right\}. \end{equation} $

取

$ \sigma = \frac{1}{\mu_2}\left[\log\frac{\log(\frac{\lambda_n}{e\mu_2})}{\tau_{\mu_2}}+o(\frac{1}{\log\lambda_n})\right], $

又由$ x>0 $, $ e^x\geq ex $成立.这样,类似情形2.1易得, $ \tau_{\mu_2}\leq \vartheta_2 $.

这样,对$ q = 2 $, $ \tau_{\mu_q} = \vartheta_2 $成立.

情形3.2   $ q\geq3 $.类似情形2.2与情形3.1,有

$ \tau_{\mu_q} = \liminf\limits_{n\rightarrow\infty}(E_{n-1}(F, \beta)\exp(-\beta\lambda_n))^{\frac{\mu_q} {\lambda_n}}\log_{q-1}\left(\frac{\lambda_n}{e\mu_q}\right). $

又由$ [\exp(-\beta\lambda_n)]^{\frac{\mu_q}{\lambda_n}} = \exp(-\beta\mu_q) $,故(0.10)式得证.

定理1.4证毕.