本文恒设$\{\Omega, \Im, P\}$是完备的概率空间, 随机变量都定义在$\{\Omega, \Im, P\}$上.自从许宝禄先生和Robbins^[1]提出完全收敛的概念以来, 许多学者进行了研究, 得到了很多深刻的结果, 参见考文献[2-9]等.特别地, 李德立等^[2]得到如下结论:

定理A  设$\{X, X_n, n\ge 1\}$是独立同分布的随机变量序列, $\beta>-1, $ $ \{a_{ni}=c_{ni} (i/n )^\beta (1/n ), $ $1\le i\le n, n\ge 1\}$是正常数阵列且$\sum\limits_{i=1}^n a_{ni}=1, \forall n\ge 1$及$0<c\le c_{ni}\le C<\infty, $ $\forall~1\le i\le n, $ $n\ge 1$, 其中$c$和$C$是常数.若

$\left\{ \begin{array}{*{35}{l}} E|X{{|}^{1/(1+\beta )}}<\infty, & -1<\beta <-1/2, \\ E|X{{|}^{2}}\log (1+|X|)<\infty, ~~ & \beta =-1/2, \\ E|X{{|}^{2}}<\infty, & \beta >-1/2, \\ \end{array} \right.$

(1.1)

则

$\sum\limits_{n=1}^\infty P\left(\left|\sum\limits_{i=1}^n a_{ni}X_i -EX\right|>\varepsilon\right)<\infty, ~\forall~\varepsilon>0.$

(1.2)

反之, 若存在$\varepsilon>0$使 (1.2) 式成立, 则 (1.1) 式成立.

梁汉营^[3]不但把定理A推广到NA列情形, 而且作了改进, 得到了更一般的结果:

定理B  设$r>1, \{X_n, n\ge 1\}$是均值为零的NA随机变量序列满足

$P(|X_n|>x)\le DP(|X|>x), \, \forall\, x\ge 0, \, \forall \, n\ge 1, $

(1.3)

其中$D$是正常数. $\{a_{ni}, 1\le i\le n, n\ge 1\}$如定理A所述.若

$\left\{\begin{array}{ll} E|X|^{(r-1)/(1+\beta)}<\infty, & -1<\beta<-1/r, \\ E|X|^r\log (1+|X|)<\infty, ~~& \beta=-1/r, \\ E|X|^{r}<\infty, & \beta>-1/r, \end{array}\right.$

(1.4)

则

$\sum\limits_{n = 1}^\infty {{n^{r - 2}}} P\left( {\mathop {\max }\limits_{1 \le k \le n} \left| {\sum\limits_{i = 1}^k {{a_{ni}}} {X_i}} \right| > \varepsilon } \right) < \infty ,\forall \varepsilon > 0.$

(1.5)

反之, 若$r\ge 2, \{X, X_n, n\ge 1\}$是同分布的NA随机变量序列, 若 (1.5) 式成立, 则 (1.4) 式成立且$EX=0.$

令$\alpha>-1$,

$ A_0^\alpha=1, ~~A_n^\alpha=\frac{(\alpha+1)(\alpha+2)\cdots (\alpha+n)}{n!}, n=1, 2, \cdots . $

Gut^[4]对独立同分布列得到如下结果:

定理C  设$ r>1, \{X, X_n, n\ge 0\}$是独立同分布的随机变量序列, 若$EX=0$及

$\left\{\begin{array}{ll} E|X|^{(r-1)/(1+\beta)}<\infty, & -1<\beta<-1/r, \\ E|X|^r\log (1+|X|)<\infty, ~~& \beta=-1/r, \\ E|X|^r<\infty, & -1/r<\beta\le 0, \end{array}\right.$

(1.6)

则

$\sum\limits_{n=1}^\infty n^{r-2}P\left(\left|\sum\limits_{i=0}^n A_{n-i}^{\beta}X_i \right|>A_n ^{1+\beta} \varepsilon\right)<\infty, ~\forall~\varepsilon>0.$

(1.7)

反之, 若 (1.7) 式成立, 则 (1.6) 式成立且$EX=0$.

文献[3, 定理2.2]把定理C推广到NA随机变量情形, 文献[5, 定理3.2]进一步推广和改进了文献[3, 定理2.2], 得到

定理D  设$r>1, -1<\beta\le 0, p>0, $当$-1/2<\beta\le 0$时进一步假设$p<2(1+\beta)/(1+2\beta).$ $\{X, X_{n}, n\ge 0\}$是同分布的NA随机变量序列.若$EX=0$及

$\left\{\begin{array}{ll} E|X|^{p(r-1)/(1+\beta)}<\infty, & -1<\beta<-1/[p(r-1)+1], \\ E|X|^{p(r-1)+1}\log (1+|X|)<\infty, ~~& \beta=-1/[p(r-1)+1], \\ E|X|^{pr/(1+\beta-p\beta)}<\infty, & -1/[p(r-1)+1]<\beta\le 0, \end{array}\right.$

(1.8)

则

$\sum\limits_{n = 1}^\infty {{n^{r - 2}}} P\left( {\mathop {\max }\limits_{0 \le k \le n} \left| {\frac{1}{{A_n^{1 + \beta }}}\sum\limits_{i = 0}^k {A_{n - i}^\beta } {X_i}} \right| > \varepsilon {n^{ - (1 + \beta )(p - 1)/p}}} \right) < \infty ,\forall \varepsilon > 0.$

(1.9)

反之, 如果 (1.9) 式成立且当$1<r<2$时进一步假设$\{X, X_n, n\ge 1\}$是严平稳NA随机变量序列, 则 (1.8) 式成立且当$p\ge 1$时有$EX=0.$

邱德华等^[9]把上述结果推广到$\tilde{\rho}$混合随机变量序列. Chow ^[10]首先提出矩完全收敛概念, 它是完全收敛的深化. Chow^[10]还研究了独立同分布的随机变量序列部分和的矩完全收敛性, 得到

定理E  设$1\le p <2, r>1, \{X_{n}, n\ge 1\}$是均值为零的独立同分布的随机变量序列.如果$E\{|X_1|^{rp}+|X_1|\log (1+|X_1|)\}<\infty$, 则

$ \sum\limits_{n=1}^\infty n^{r-2-1/p}E\left\{\left|\sum\limits_{k=1}^n X_k\right |-\epsilon n^{1/p}\right \}_+<\infty, ~~\forall \epsilon>0, $

其中$x_+=\max\{0, x\}.$

定理E被许多学者进行了推广和改进, 见参考文献[11-17]等等.本文的目的是:利用$\tilde{\rho}$混合随机变量 (定义见后面) 的Rosenthal型最大值不等式不但把定理B$\sim$定理E推广到$\tilde{\rho}$混合随机变量序列, 得到了$\tilde{\rho}$混合随机变量序列加权和的矩完全收敛性定理, 而且条件和结论都有所改进.本文所用的证明方法与以前证明矩完收敛的方法还不同, 以前一般都是先证明完全收敛, 再利用完全收敛来证明矩完全收敛, 本文利用下面的引理2.3来证明矩完全收敛性, 这方法较以前的要简便且具有广泛性.从本文的证明来看, 本文的结论对成立Rosenthal型最大值不等式的随机变量 (如NA随机变量, $\varphi$混合随机变量等) 都成立.

设$\{\xi_{n}, n\ge 1\}$是随机变量序列, $\Im_{S}=\sigma(\xi_n:n \in S \subset N)(N\mbox{是自然数集}).$给定子$\sigma$-代数$\Im_1, \Im_2 \subset \Im$, 令

$ \rho{(\Im_1, \Im_2)}=\sup\{|corr(\zeta, \eta)|, \zeta \in L_2(\Im_1) , \eta \in L_2(\Im_2)\}, $

其中

$ corr(\zeta, \eta)=\frac{E(\zeta\eta)-E\zeta E\eta}{\sqrt{Var(\zeta)Var(\eta)}}. $

Stein在文献[18]中定义了如下的相关系数:

$ \tilde{\rho}(k)=\sup\{\rho{(\Im_S, \Im_T)}: \mbox{有限子集}\ S, T\subset N \mbox{满足 }\ dist(S, T)\ge k\}, k\ge 0. $

显然, $ 0\le \tilde{\rho}(k+1)\le\tilde{\rho}(k)\le 1, \tilde{\rho}(0)=1.$

定义1.1  称随机变量序列$\{\xi_{n}, n\ge 1\}$为$\tilde{\rho}$混合的, 如果存在$k\in N $使$\tilde{\rho}(k)<1.$

$\tilde{\rho}$混合与通常的$\rho$混合有一定的类似, 但不完全相同.很多学者对$\tilde{\rho}$混合随机变量序列进行了研究, 如文献[7-9]研究了加权和的完全收敛性, Utev和Peligrad^[19]研究了极大型不等式和不变原理, Bryc和Smolenski^[20]得到了矩不等式和几乎处处收敛性, Bradley^[21-22]研究了中心极限定理, 杨善朝^[23]得到了矩不等式和强大数律, Peligrad和Gut^[24], 吴群英^[25-26]以及甘师信^[27]研究了收敛性, 邱德华和甘师信^[28]讨论了$\tilde{\rho}$混合阵列的完全收敛性和大数律, 陈平炎^[29]得到了最大值不等式和$\tilde{\rho}$混合随机变量序列部分和的算术平均值的矩存在的充要条件, 等等.

下面陈述本文的主要结果, 其证明放在下一节.

定理1.1  设$r>1, \theta>0, \tau>0, \beta>\max\{-1, -1/\theta\}, p>\max\{0, (1-\theta)/(r-1)\}.$当$\beta>-1/2$时进一步假设$p<2(1+\theta\beta)/(1+2\beta)$且$ r>\max\{1, (3-\theta+2\beta)/[2(1+\theta\beta)]\}.$ $\{X_{n}, n\ge 1\}$是均值为零的$\tilde{\rho}$混合随机变量序列且满足 (1.3) 式, $\{a_{ni}=c_{ni}i^\beta n^{-(1+\theta\beta)/p}, $ $ 1\le i\le n, $ $n\ge 1\}$是常数阵列且$ |c_{ni}|\le C, \forall 1\le i\le n, n\ge 1$, 其中$C$是正常数.若

$\tau<\left\{\begin{array}{ll} p(r-1)/(1+\theta\beta), & \max\{-1, -1/\theta\}<\beta<-1/\left[p(r-1)+\theta\right], \\ p(r-1)+\theta, & \beta=-1/\left[p(r-1)+\theta\right], \\ pr/(1+\theta\beta-p\beta), & \beta>-1/\left[p(r-1)+\theta\right], \\ \end{array}\right.$

(1.10)

$\left\{\begin{array}{ll} E|X|^{p(r-1)/(1+\theta\beta)}<\infty, & \max\{-1, -1/\theta\}<\beta<-1/\left[p(r-1)+\theta\right], \\ E|X|^{p(r-1)+\theta}\log (1+|X|)<\infty, & \beta=-1/\left[p(r-1)+\theta\right], \\ E|X|^{pr/(1+\theta\beta-p\beta)}<\infty, & \beta>-1/\left[p(r-1)+\theta\right], \end{array}\right.$

(1.11)

则

$\sum\limits_{n = 1}^\infty {{n^{r - 2}}} E\left\{ {\mathop {\max }\limits_{1 \le k \le n} \left| {\sum\limits_{i = 1}^k {{a_{ni}}} {X_i}} \right| - \varepsilon } \right\}_ + ^\tau < \infty ,\forall \varepsilon > 0.$

(1.12)

反之, 设$r\ge 2, |c_{ni}|\ge c, \forall 1\le i\le n, n\ge 1$, 其中$c$是正常数, $\{X, X_n, n\ge 1\}$是同分布的$\tilde{\rho}$混合随机变量序列.若存在某$\varepsilon>0$及$\tau>0$使 (1.12) 式成立, 则 (1.11) 式成立.若进一步有$\tau$满足 (1.10) 式且$c\le |c_{ni}|\le C, \sum\limits_{i=1}^n a_{ni}=1, \forall n\ge 1$, 则还有$EX=0.$

定理1.2  设$\{a_{ni}=c_{ni}(n-i+1)^\beta n^{-(1+\theta\beta)/p}, 1\le i\le n, n\ge 1\}$, 其余的与定理1.1完全相同, 则定理1.1的结论同样成立.

定理1.3  设$r>1, -1<\beta\le 0, p>0, \tau>0, $当$-1/2<\beta\le 0$时进一步假设$p<2(1+\beta)/(1+2\beta).$ $\{X_{n}, n\ge 0\}$是均值为零的$\tilde{\rho}$混合随机变量序列且 (1.3) 式对任意$n\ge 0$都成立.若 (1.8) 式成立且

$\tau<\left\{\begin{array}{ll} p(r-1)/(1+\beta), ~~& -1<\beta<-1/[p(r-1)+1], \\ p(r-1)+1, & \beta=-1/[p(r-1)+1], \\ pr/(1+\beta-p\beta), & -1/[p(r-1)+1]<\beta\le 0, \end{array}\right.$

(1.13)

则

$\sum\limits_{n = 1}^\infty {{n^{r - 2 + \tau (1 + \beta )(p - 1)/p}}} E\left\{ {\mathop {\max }\limits_{0 \le k \le n} \left| {\frac{1}{{A_n^{1 + \beta }}}\sum\limits_{i = 0}^k {A_{n - i}^\beta } {X_i}} \right| - \varepsilon {n^{ - (1 + \beta )(p - 1)/p}}} \right\}_ + ^\tau < \infty ,\forall \varepsilon > 0.$

(1.14)

反之, 设$\{X, X_n, n> 1\}$是同分布的$\tilde{\rho}$混合随机变量序列.如果存在$\varepsilon>0$及满足 (1.13) 式的$\tau$使 (1.14) 式成立且当$1<r<2$时进一步假设$\{X, X_n, n\ge 1\}$是严平稳$\tilde{\rho}$混合随机变量序列, 则 (1.8) 式成立且当$p\ge 1$时有$EX=0.$

注1.1  在定理1.1的条件下, 我们有

(ⅰ) $\max\{-1, -1/\theta\}<-1/\left[p(r-1)+\theta\right].$

(ⅱ) $1+\theta\beta-p\beta>0.$事实上, 若$-1/\left[p(r-1)+\theta\right]<\beta\le 0$时显然有$1+\theta\beta-p\beta>0$, 若$\beta>0$时, 由$p<2(1+\theta\beta)/(1+2\beta)$可得$1+\theta\beta-p\beta>p/2>0.$

(ⅲ) 当$\beta>-1/2$时, 若$\theta\ge 1$, 则$(3-\theta+2\beta)/[2(1+\theta\beta)]\le 1$, 若$0<\theta<1$, 则$(3-\theta+2\beta)/[2(1+\theta\beta)]>1$.

(ⅳ) 当$\beta>-1/2$时, 有$(1-\theta)/(r-1)<2(1+\theta\beta)/(1+2\beta)$.事实上, 若$\theta \ge 1$时, 显然有$(1-\theta)/(r-1)<2(1+\theta\beta)/(1+2\beta)$, 若$0<\theta< 1$时, 由$r>(3-\theta+2\beta)/[2(1+\theta\beta)]$可得$(1-\theta)/(r-1)<2(1+\theta\beta)/(1+2\beta).$

注1.2  (ⅰ) 令$p=\theta=1, $ $\{X_{n}, n\ge 1\}$是均值为零的NA随机变量序列, 则由定理1.1可得到定理B且$c_{ni}$的范围有所扩大.

(ⅱ) 令$\tau=1, c_{ni}\equiv 1, \beta=0, $即$a_{ni}=n^{-1/p}$, 则由定理1.1可得到定理E.

(ⅲ) 当$\{X_{n}, n\ge 1\}$是NA随机变量序列时, 由定理1.3可得定理D.

(ⅳ) 由定理1.1与定理1.3可得文献[9]的相应结论.

本文记号$x_+^\tau=(x_+)^\tau, \tau>0.$ $a\ll b$表示存在正常数$D$使$a\le Db, $在同一式中的$D$也可能不同.

为证明本文结论, 我们需要下面引理.

引理2.1^[19]  设$J$是正整数, $0\le r<1$, $\tau\ge 2, $则存在正常数$C=C(\tau, J, r)$使下面陈述成立:

$\{X_{n}, n\ge 1 \}$是满足$\tilde{\rho}(J)\le r$和$EX_n=0$及$E|X_n|^\tau<\infty(n \ge 1 )$的$\tilde{\rho}$混合随机变量序列, 则

$ E\mathop {\max }\limits_{1 \le j \le n} |\sum\limits_{i = 1}^j {{X_i}} {|^\tau } \le C\{ \sum\limits_{i = 1}^n E |{X_i}{|^\tau } + {(\sum\limits_{i = 1}^n E |{X_i}{|^2})^{\tau /2}}\} ,\;\;\forall \;n \ge 1. $

引理2.2^[9]  设$\{X_{n}, n\ge 1 \}$是$\tilde{\rho}$混合随机变量序列, $\{a_{ni}, 1\le i\le n, n\ge 1\}$是实常数阵列, 则存在与$n$无关的正常数$D$, 对$\forall x>0$和$ \forall n\ge 1$有

$ \left(\frac{1}{2} -P(\mathop {\max }\limits_{1\le i\le n}|a_{ni}X_i|>x)\right ) \sum\limits_{i=1}^n P(|a_{ni}X_i|>x)\le \left( \frac{D}{2}+1\right)P (\mathop {\max }\limits_{1\le i\le n}|a_{ni}X_i|>x). $

引理2.3  设$\tau>0, \{a_n, n\ge 1\}$和$\{b_n, n\ge 1\}$是正数列, $\{X_{ni}\}$是随机变量阵列.若

$ \sum\limits_{n=1}^\infty b_n \int_{1}^\infty P\left(\mathop {\max }\limits_{1\le j\le n}\left|\sum\limits_{i=1}^j X_{ni}\right|>\varepsilon a_n x^{1/\tau}\right){\rm d}x<\infty, \, \forall\, \varepsilon>0, $

则

$ \sum\limits_{n=1}^\infty b_n a_n^{-\tau} E\left\{\mathop {\max }\limits_{1\le j\le n}\left|\sum\limits_{i=1}^j X_{ni}\right|- \varepsilon a_n \right\}_+^\tau<\infty, \, \forall\, \varepsilon>0. $

证  对任意给定的$\varepsilon>0$, 因为

$ \begin{array}{l} \infty > \sum\limits_{n = 1}^\infty {{b_n}} \int_1^\infty P \left( {\mathop {\max }\limits_{1 \le j \le n} \left| {\sum\limits_{i = 1}^j {{X_{ni}}} } \right| > {2^{ - 1}}\varepsilon {a_n}{x^{1/\tau }}} \right){\rm{d}}x\\ \ge \sum\limits_{n = 1}^\infty {{b_n}} \int_1^{{2^\tau }} P \left( {\mathop {\max }\limits_{1 \le j \le n} \left| {\sum\limits_{i = 1}^j {{X_{ni}}} } \right| > {2^{ - 1}}\varepsilon {a_n}{x^{1/\tau }}} \right){\rm{d}}x\\ \ge \sum\limits_{n = 1}^\infty {{b_n}} \int_1^{{2^\tau }} P \left( {\mathop {\max }\limits_{1 \le j \le n} \left| {\sum\limits_{i = 1}^j {{X_{ni}}} } \right| > \varepsilon {a_n}} \right){\rm{d}}x\\ = ({2^\tau } - 1)\sum\limits_{n = 1}^\infty {{b_n}} P\left( {\mathop {\max }\limits_{1 \le j \le n} \left| {\sum\limits_{i = 1}^j {{X_{ni}}} } \right| > \varepsilon {a_n}} \right), \end{array} $

从而

$ \sum\limits_{n=1}^\infty b_n P\left(\mathop {\max }\limits_{1\le j\le n}\left|\sum\limits_{i=1}^ j X_{ni}\right|>\varepsilon a_n\right)<\infty. $

进而

$ \begin{align} & \sum\limits_{n=1}^{\infty }{{{b}_{n}}}a_{n}^{-\tau }E\left\{ {\mathop {\max }\limits_{1\le j\le n}}\left| \sum\limits_{i=1}^{j}{{{X}_{ni}}} \right|-\varepsilon {{a}_{n}} \right\}_{+}^{\tau } \\ & =\sum\limits_{n=1}^{\infty }{{{b}_{n}}}a_{n}^{-\tau }\int_{0}^{\infty }{P}\left( {\mathop {\max }\limits_{1\le j\le n}}\left| \sum\limits_{i=1}^{j}{{{X}_{ni}}} \right|-\varepsilon {{a}_{n}}>{{x}^{1/\tau }} \right)\text{d}x \\ & =\sum\limits_{n=1}^{\infty }{{{b}_{n}}}a_{n}^{-\tau }\int_{0}^{{{(\varepsilon {{a}_{n}})}^{\tau }}}{P}\left( {\mathop {\max }\limits_{1\le j\le n}}\left| \sum\limits_{i=1}^{j}{{{X}_{ni}}} \right|-\varepsilon {{a}_{n}}>{{x}^{1/\tau }} \right)\text{d}x \\ & \quad +\sum\limits_{n=1}^{\infty }{{{b}_{n}}}a_{n}^{-\tau }\int_{{{(\varepsilon {{a}_{n}})}^{\tau }}}^{\infty }{P}\left( {\mathop {\max }\limits_{1\le j\le n}}\left| \sum\limits_{i=1}^{j}{{{X}_{ni}}} \right|-\varepsilon {{a}_{n}}>{{x}^{1/\tau }} \right)\text{d}x \\ & \le {{\varepsilon }^{\tau }}\sum\limits_{n=1}^{\infty }{{{b}_{n}}}P\left( {\mathop {\max }\limits_{1\le j\le n}}\left| \sum\limits_{i=1}^{j}{{{X}_{ni}}} \right|>\varepsilon {{a}_{n}} \right) \\ & +\sum\limits_{n=1}^{\infty }{{{b}_{n}}}a_{n}^{-\tau }\int_{{{(\varepsilon {{a}_{n}})}^{\tau }}}^{\infty }{P}\left( {\mathop {\max }\limits_{1\le j\le n}}\left| \sum\limits_{i=1}^{j}{{{X}_{ni}}} \right|>{{x}^{1/\tau }} \right)\text{d}x \\ & ={{\varepsilon }^{\tau }}\sum\limits_{n=1}^{\infty }{{{b}_{n}}}P\left( {\mathop {\max }\limits_{1\le j\le n}}\left| \sum\limits_{i=1}^{j}{{{X}_{ni}}} \right|>\varepsilon {{a}_{n}} \right) \\ & +{{\varepsilon }^{\tau }}\sum\limits_{n=1}^{\infty }{{{b}_{n}}}\int_{1}^{\infty }{P}\left( {\mathop {\max }\limits_{1\le j\le n}}\left| \sum\limits_{i=1}^{j}{{{X}_{ni}}} \right|>\varepsilon {{a}_{n}}{{x}^{1/\tau }} \right)\text{d}x \\ & <\infty . \\ \end{align} $

证毕.

定理1.1的证明  先证充分性.对$\forall~1\le i\le n, n\ge 1, t>0$, 令$X_{ni}^{(1, t)}=X_i I(|a_{ni}X_i|\le t^{1/\tau}), $ $ X_{ni}^{(2, t)}=X_i I(|a_{ni}X_i|>t^{1/\tau}), $则

$ \begin{align} & \sum\limits_{n=1}^{\infty }{{{n}^{r-2}}}\int_{1}^{\infty }{P}\left( {\mathop {\max }\limits_{1\le k\le n}}\left| \sum\limits_{i=1}^{k}{{{a}_{ni}}}{{X}_{i}} \right|>\varepsilon {{t}^{1/\tau }} \right)\text{d}t \\ & \le \sum\limits_{n=1}^{\infty }{{{n}^{r-2}}}\int_{1}^{\infty }{P}\left( {\mathop {\max }\limits_{1\le k\le n}}\left| \sum\limits_{i=1}^{k}{{{a}_{ni}}}X_{ni}^{(1, t)} \right|>\varepsilon {{t}^{1/\tau }}/2 \right)\text{d}t \\ & +\sum\limits_{n=1}^{\infty }{{{n}^{r-2}}}\int_{1}^{\infty }{P}\left( {\mathop {\max }\limits_{1\le k\le n}}\left| \sum\limits_{i=1}^{k}{{{a}_{ni}}}X_{ni}^{(2, t)} \right|>\varepsilon {{t}^{1/\tau }}/2 \right)\text{d}t \\ & :=\text{ }{{I}_{1}}+{{I}_{2}}. \\ \end{align} $

因此, 由引理2.3可知, 要证 (1.12) 式, 只需要证$I_1<\infty$和$I_2<\infty$.下证$I_2<\infty$.

$\begin{align} & {{I}_{2}}\le \sum\limits_{n=1}^{\infty }{{{n}^{r-2}}}\int_{1}^{\infty }{P}\left( \cup _{i=1}^{n}\left( |{{a}_{ni}}{{X}_{i}}|>{{t}^{1/\tau }} \right) \right)\text{d}t \\ & \le \sum\limits_{n=1}^{\infty }{{{n}^{r-2}}}\int_{1}^{\infty }{P}\left( \cup _{i=1}^{n}\left( |{{X}_{i}}|>\frac{1}{C}{{t}^{1/\tau }}{{i}^{-\beta }}{{n}^{(1+\theta \beta )/p}} \right) \right)\text{d}t \\ & \le \sum\limits_{n=1}^{\infty }{{{n}^{r-2}}}\int_{1}^{\infty }{\sum\limits_{i=1}^{n}{P}}\left( |{{X}_{i}}|>\frac{1}{C}{{t}^{1/\tau }}{{i}^{-\beta }}{{n}^{(1+\theta \beta )/p}} \right)\text{d}t \\ & \ll \int_{1}^{\infty }{\left\{ \sum\limits_{n=1}^{\infty }{{{n}^{r-2}}}\sum\limits_{i=1}^{n}{P}\left( |X|>\frac{1}{C}{{t}^{1/\tau }}{{i}^{-\beta }}{{n}^{(1+\theta \beta )/p}} \right) \right\}}\text{d}t, \\ \end{align}$

(2.1)

令$u=y^{-\beta} x^{(1+\theta\beta)/p}, v=y$, 则$x=u^{p/(1+\theta\beta)}v^{p\beta/(1+\theta\beta)}, y=v$且

$ \frac{\partial(x, y)}{\partial(u, v)}=\frac{p}{1+\theta\beta}u^{p/(1+\theta\beta)-1}v^{p\beta/(1+\theta\beta)}, $

对任意$t\ge 1$, 于是由 (1.11) 式可得

$ \begin{align} & \sum\limits_{n=1}^{\infty }{{{n}^{r-2}}}\sum\limits_{i=1}^{n}{P}\left( |X|>\frac{1}{C}{{t}^{1/\tau }}{{i}^{-\beta }}{{n}^{(1+\theta \beta )/p}} \right) \\ & \ll \int_{1}^{\infty }{\int_{1}^{x}{{{x}^{r-2}}}}P\left( |X|>\frac{1}{C}{{t}^{1/\tau }}{{y}^{-\beta }}{{x}^{(1+\theta \beta )/p}} \right)\text{d}x\text{d}y \\ & =\frac{p}{1+\theta \beta }\int_{1}^{\infty }{\text{d}}u\int_{1}^{{{u}^{p/(1+\theta \beta -p\beta )}}}{{{u}^{p(r-1)/(1+\theta \beta )-1}}}{{v}^{p\beta (r-1)/(1+\theta \beta )}}P\left( |X|>\frac{1}{C}{{t}^{1/\tau }}u \right) \\ \end{align} $

$ \ll \left\{ \begin{align} & \int_{1}^{\infty }{{{u}^{p(r-1)/(1+\theta \beta )-1}}}P\left( |X|>\frac{1}{C}{{t}^{1/\tau }}u \right)\text{d}u, \{-1, -1/\theta \}<\beta <-1/\left[p(r-1)+\theta \right], \\ & \int_{1}^{\infty }{{{u}^{p(r-1)+\theta -1}}}\log uP\left( |X|>\frac{1}{C}{{t}^{1/\tau }}u \right)\text{d}u, ~~\beta =-1/\left[p(r-1)+\theta \right], \\ & \int_{1}^{\infty }{{{u}^{pr/(1+\theta \beta -p\beta )-1}}}P\left( |X|>\frac{1}{C}{{t}^{1/\tau }}u \right)\text{d}u, \ ~~\beta >-1/\left[p(r-1)+\theta \right] \\ \end{align} \right. $

$\ll \text{ }\left\{ \begin{array}{*{35}{l}} E{{\left| \frac{X}{{{t}^{1/\tau }}} \right|}^{p(r-1)/(1+\theta \beta )}}, & \max \{-1, -1/\theta \}<\beta <-1/\left[p(r-1)+\theta \right], \\ E{{\left| \frac{X}{{{t}^{1/\tau }}} \right|}^{p(r-1)+\theta }}\log (1+|X|), & \beta =-1/\left[p(r-1)+\theta \right], \\ E{{\left| \frac{X}{{{t}^{1/\tau }}} \right|}^{pr/(1+\theta \beta -p\beta )}}, & \beta >-1/\left[p(r-1)+\theta \right]. \\ \end{array} \right.$

(2.2)

因此由 (1.10), (2.1) 和 (2.2) 式可得$I_2<\infty.$再证$I_1<\infty$.由$EX_i=0(i\ge 1)$, (1.11) 式有

$ \begin{gathered} 0 \leqslant \mathop {\sup }\limits_{t \geqslant 1} {t^{ - 1/\tau }}{\max _{1 \leqslant k \leqslant n}}\left| {\sum\limits_{i = 1}^k E {a_{ni}}X_{ni}^{(1, t)}} \right| \hfill \\ \leqslant \left\{ \begin{gathered} \sum\limits_{i = 1}^n E |{a_{ni}}{X_i}{|^{p(r - 1)/(1 + \theta \beta )}}I(|{a_{ni}}{X_i}| > 1), \hfill \\ \max \{ - 1, - 1/\theta \} < \beta < - 1/\left[{p(r-1) + \theta } \right], \hfill \\ \sum\limits_{i = 1}^n E |{a_{ni}}{X_i}{|^{p(r - 1) + \theta }}I(|{a_{ni}}{X_i}| > 1), \beta = - 1/\left[{p(r-1) + \theta } \right], \hfill \\ \sum\limits_{i = 1}^n E |{a_{ni}}{X_i}{|^{pr/(1 + \theta \beta - p\beta )}}I(|{a_{ni}}{X_i}| > 1), \hfill \\ \beta > - 1/\left[{p(r-1) + \theta } \right]\;AAA\;pr/(1 + \theta \beta - p\beta ) \geqslant 1, \hfill \\ \mathop {\sup }\limits_{t \geqslant 1} \sum\limits_{i = 1}^n E {\left| {\frac{{{a_{ni}}{X_i}}}{{{t^{1/\tau }}}}} \right|^{pr/(1 + \theta \beta - p\beta )}}I(|{a_{ni}}{X_i}| \leqslant {t^{1/\tau }}), \hfill \\ \beta > - 1/\left[{p(r-1) + \theta } \right]\;AAA\;pr/(1 + \theta \beta - p\beta ) < 1 \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} $

$ \begin{align} & \ll \left\{ \begin{array}{*{35}{l}} {{n}^{-(r-1)}}, & \max \{-1, -1/\theta \}<\beta <-1/\left[p(r-1)+\theta \right], \\ {{n}^{-(r-1)}}\log n, & \beta =-1/\left[p(r-1)+\theta \right], \\ {{n}^{-(r-1)}}, & \beta >-1/\left[p(r-1)+\theta \right] \\ \end{array} \right. \\ & \to 0, ~~~n\to \infty . \\ \end{align} $

于是要证$I_1<\infty$, 只需要证

$I_1^*:=\sum\limits_{n=1}^\infty n^{r-2}\int_1^\infty P\left(\mathop {\max }\limits_{1\le k\le n}\left|\sum\limits_{i=1}^k\left( a_{ni}X_{ni}^{(1, t)}-Ea_{ni}X_{ni}^{(1, t)}\right) \right|>\varepsilon t^{1/\tau}/4\right){\rm d}t<\infty.$

(2.3)

取$q$使

$ \begin{gathered} q > \max \{ 2, \tau, pr/(1 + \theta \beta ), pr + \theta, pr/(1 + \theta \beta - p\beta ), 2pr/(2 - \theta )I(0 < \theta < 2), \hfill \\ \qquad \quad 2pr/[-p(1 + 2\beta ) + 2(1 + \theta \beta )]\}, \hfill \\ \end{gathered} $

由Markov不等式、引理2.1、$C_r$不等式和Jensen不等式可得

$\begin{gathered} I_1^* \ll \sum\limits_{n = 1}^\infty {{n^{r - 2}}} \int_1^\infty {{t^{ - q/\tau }}} \{ \sum\limits_{i = 1}^n E {\left| {{a_{ni}}X_{ni}^{(1, t)} - E{a_{ni}}X_{ni}^{(1, t)}} \right|^q} \hfill \\ + {(\sum\limits_{i = 1}^n E {\left( {{a_{ni}}X_{ni}^{(1, t)} - E{a_{ni}}X_{ni}^{(1, t)}} \right)^2})^{q/2}}\} {\text{d}}t \hfill \\ \ll \sum\limits_{n = 1}^\infty {{n^{r - 2}}} \int_1^\infty {{t^{ - q/\tau }}} \sum\limits_{i = 1}^n E {\left| {{a_{ni}}X_{ni}^{(1, t)}} \right|^q}{\text{d}}t \hfill \\ + \sum\limits_{n = 1}^\infty {{n^{r - 2}}} \int_1^\infty {{t^{ - q/\tau }}} {(\mathop \sum \limits_{i = 1}^n E{\left( {{a_{ni}}X_{ni}^{(1, t)}} \right)^2})^{q/2}}{\text{d}}t \hfill \\ : = {I_{11}} + {I_{12}}. \hfill \\ \end{gathered} $

(2.4)

则由 (1.10), (1.11), (2.2) 式及$q$的取法有

$ \begin{gathered} {I_{11}} = q\sum\limits_{n = 1}^\infty {{n^{r - 2}}} \int_1^\infty {{t^{ - q/\tau }}} \left\{ {\sum\limits_{i = 1}^n {\int_0^\infty {{z^{q - 1}}} } P\left( {|{a_{ni}}{X_i}|I(|{a_{ni}}{X_i}| \leqslant {t^{1/\tau }}) > z} \right){\text{d}}z} \right\}{\text{d}}t \hfill \\ = q\sum\limits_{n = 1}^\infty {{n^{r - 2}}} \int_1^\infty {{t^{ - q/\tau }}} \left\{ {\sum\limits_{i = 1}^n {\int_0^{{t^{1/\tau }}} {{z^{q - 1}}} } P\left( {|{a_{ni}}{X_i}| > z} \right){\text{d}}z} \right\}{\text{d}}t \hfill \\ \ll \int_1^\infty {{t^{ - q/\tau }}} \left\{ {\int_0^{{t^{1/\tau }}} {{z^{q - 1}}} \left( {\sum\limits_{n = 1}^\infty {{n^{r - 2}}} \sum\limits_{i = 1}^n P \left( {|X| > \frac{z}{C}{i^{ - \beta }}{n^{(1 + \theta \beta )/p}}} \right)} \right){\text{d}}z} \right\}{\text{d}}t \hfill \\ \ll \left\{ \begin{gathered} \int_1^\infty {{t^{ - \frac{q}{\tau }}}} {\text{d}}t\int_0^{{t^{\frac{1}{\tau }}}} {{z^{q - 1}}} E{\left| {\frac{X}{z}} \right|^{\frac{{p(r - 1)}}{{1 + \theta \beta }}}}{\text{d}}z, \max \{ - 1, - \frac{1}{\theta }\} < \beta < \frac{{ - 1}}{{p(r - 1) + \theta }}, \hfill \\ \int_1^\infty {{t^{ - q/\tau }}} {\text{d}}t\int_0^{{t^{1/\tau }}} {{z^{q - 1}}} E{\left| {\frac{X}{z}} \right|^{p(r - 1) + \theta }}\log \left( {1 + \left| {\frac{X}{z}} \right|} \right){\text{d}}z, \hfill \\ \beta = - 1/\left[{p(r-1) + \theta } \right], \hfill \\ \int_1^\infty {{t^{ - q/\tau }}} {\text{d}}t\int_0^{{t^{1/\tau }}} {{z^{q - 1}}} E{\left| {\frac{X}{z}} \right|^{pr/(1 + \theta \beta - p\beta )}}{\text{d}}z, \hfill \\ \beta > - 1/\left[{p(r-1) + \theta } \right] \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} $

$\begin{gathered} \ll \left\{ \begin{gathered} \int_1^\infty {{t^{ - p(r - 1)/[\tau (1 + \theta \beta )]}}} {\text{d}}t, \max \{ - 1, - 1/\theta \} < \beta < - 1/\left[{p(r-1) + \theta } \right], \hfill \\ \int_1^\infty {{t^{ - [p(r-1) + \theta]/\tau }}} {\text{d}}t, \beta = - 1/\left[{p(r-1) + \theta } \right], \hfill \\ \int_1^\infty {{t^{ - pr/[\tau (1 + \theta \beta-p\beta )]}}} {\text{d}}t, \beta > - 1/\left[{p(r-1) + \theta } \right] \hfill \\ \end{gathered} \right. \hfill \\ < \infty . \hfill \\ \end{gathered} $

(2.5)

再证$I_{12}<\infty$.分三种情况:

(i) 当$\max\{-1, -1/\theta\}<\beta<-1/\left[p(r-1)+\theta\right]$时, 又分三种情况考虑:若$ p(r-1)/(1+\theta\beta)\ge 2$且$0<\theta<2$, 此时有$E|X|^2<\infty$, 故有

$\begin{gathered} \sum\limits_{i = 1}^n E {\left( {{a_{ni}}X_{ni}^{(1, t)}} \right)^2} \ll \sum\limits_{i = 1}^n {{{\left( {{a_{ni}}} \right)}^2}} \hfill \\ \ll \left\{ {\begin{array}{*{20}{l}} {{n^{ - 2(1 + \theta \beta )/p}}, }&{\max \{ - 1, - 1/\theta \} < \beta < - 1/2, } \\ {{n^{ - (2 - \theta )/p}}\log n, }&{\beta = - 1/2, } \\ {{n^{ - [-1-2\beta + 2(1 + \theta \beta )/p]}}, }&{\beta > - 1/2.} \end{array}} \right. \hfill \\ \end{gathered} $

(2.6)

若$ p(r-1)/(1+\theta\beta)\ge 2$且$\theta\ge 2$, 此时有$E|X|^2<\infty$及$\beta>-1/2$, 则有

$\sum\limits_{i=1}^n E\left(a_{ni}X_{ni}^{(1, t)}\right)^2 \ll \sum\limits_{i=1}^n \left(a_{ni}\right)^2 \ll n^{-[-1-2\beta+2(1+\theta\beta)/p]}.$

(2.7)

若$ p(r-1)/(1+\theta\beta)< 2$, 此时由$\beta<-1/\left[p(r-1)+\theta\right]$可得$p\beta(r-1)/(1+\theta\beta)<-1$, 则由 (1.11) 式有

$ \begin{gathered} {t^{ - 2/\tau }}\sum\limits_{i = 1}^n E {\left( {{a_{ni}}X_{ni}^{(1, t)}} \right)^2} \leqslant {t^{ - p(r - 1)/[(1 + \theta \beta )\tau]}}\sum\limits_{i = 1}^n E {\left| {{a_{ni}}{X_i}} \right|^{p(r - 1)/(1 + \theta \beta )}} \hfill \\ \ll {t^{ - p(r - 1)/[(1 + \theta \beta )\tau]}}\sum\limits_{i = 1}^n {{{\left| {{a_{ni}}} \right|}^{p(r - 1)/(1 + \theta \beta )}}} \hfill \\ \ll {n^{ - (r - 1)}}{t^{ - p(r - 1)/[(1 + \theta \beta )\tau]}}. \hfill \\ \end{gathered} $

(ⅱ) 当$\beta=-1/\left[p(r-1)+\theta\right]$时, 分三种情况考虑:若$ p(r-1)+\theta\ge 2$且$0<\theta<2$, 则 (2.6) 式成立, 若$ p(r-1)+\theta\ge 2$且$\theta\ge 2$, 则 (2.7) 式成立.若$ p(r-1)+\theta< 2$, 因为$\beta\left[p(r-1)+\theta\right]=-1$, 由 (1.11) 式有

$ \begin{gathered} {t^{ - 2/\tau }}\sum\limits_{i = 1}^n E {\left( {{a_{ni}}X_{ni}^{(1)}} \right)^2} \leqslant {t^{ - \left[{p(r-1) + \theta } \right]/\tau }}\sum\limits_{i = 1}^n E {\left| {{a_{ni}}{X_i}} \right|^{p(r - 1) + \theta }} \hfill \\ \ll {t^{ - \left[{p(r-1) + \theta } \right]/\tau }}\sum\limits_{i = 1}^n | {a_{ni}}{|^{p(r - 1) + \theta }} \hfill \\ \ll {t^{ - \left[{p(r-1) + \theta } \right]/\tau }}{n^{ - (r - 1)}}\log n. \hfill \\ \end{gathered} $

(ⅲ) 当$\beta >-1/\left[p(r-1)+\theta\right]$时, 分三种情况:若$ pr/\left(1+\theta\beta-p\beta\right)\ge 2$且$0<\theta<2$, 则 (2.6) 式成立, 若$ pr/\left(1+\theta\beta-p\beta\right)\ge 2$且$\theta\ge 2$, 则 (2.7) 式成立, 若$ pr/\left(1+\theta\beta-p\beta\right)< 2$, 此时由$\beta >-1/\left[p(r-1)+\theta\right]$可得$\beta pr/\left(1+\theta\beta-p\beta\right)>-1$, 由 (1.11) 式有

$ \begin{gathered} {t^{ - 2/\tau }}\sum\limits_{i = 1}^n E {\left( {{a_{ni}}X_{ni}^{(1)}} \right)^2} \leqslant {t^{ - pr/\left[{\tau (1 + \theta \beta-p\beta )} \right]}}\sum\limits_{i = 1}^n E {\left| {{a_{ni}}{X_i}} \right|^{pr/\left( {1 + \theta \beta - p\beta } \right)}} \hfill \\ \ll {t^{ - pr/\left[{\tau (1 + \theta \beta-p\beta )} \right]}}\sum\limits_{i = 1}^n | {a_{ni}}{|^{pr/\left( {1 + \theta \beta - p\beta } \right)}} \hfill \\ \ll {t^{ - pr/\left[{\tau (1 + \theta \beta-p\beta )} \right]}}{n^{ - (r - 1)}}. \hfill \\ \end{gathered} $

综上三种情形及$q$的取法有$I_{12}<\infty.$再由 (2.3)-(2.5) 式可得$I_1<\infty, $从而 (1.12) 式成立.

再证必要性.对某$\varepsilon>0$和$\tau>0$, 由 (1.12) 式可得

$\sum\limits_{n=1}^\infty n^{r-2}P\left(\mathop {\max }\limits_{1\le k\le n}\left|\sum\limits_{i=1}^k a_{ni}X_i \right|>\varepsilon \right)<\infty.$

(2.8)

从而有

$\sum\limits_{n=1}^\infty n^{r-2} P(\mathop {\max }\limits_{1\le i\le n}|a_{ni}X_i|>\varepsilon)<\infty, $

(2.9)

再由$r\ge 2$可得

$ \lim_{n\to \infty}P(\mathop {\max }\limits_{1\le i\le n}|a_{ni}X_i|>\varepsilon)=0. $

由引理2.2和上式知当$n$充分大时, 有

$ \sum\limits_{i = 1}^n P (|{a_{ni}}{X_i}| > \varepsilon ) \ll P(\mathop {\max }\limits_{_{1 \le i \le n}} |{a_{ni}}{X_i}| > \varepsilon ). $

从而由 (2.9) 式有

$ \sum\limits_{n=1}^\infty n^{r-2} \sum\limits_{i=1}^n P(|a_{ni}X_i|>\varepsilon)<\infty. $

进而有

$ \sum\limits_{n=1}^\infty n^{r-2} \sum\limits_{i=1}^n P(|X|>\frac{1}{c}i^{-\beta}n^{(1+\theta\beta)/p}\varepsilon)<\infty. $

由此可得 (1.11) 式成立.若$\tau$还满足 (1.10) 式且$c\le |c_{ni}|\le C, $$1\leq i\leq n$, $\forall n\ge 1$, 则由 (1.11) 式和充分性的证明可得

$ \sum\limits_{n=1}^\infty n^{r-2}P\left(\left|\sum\limits_{i=1}^n a_{ni}(X_i-EX_i)\right|>\varepsilon\right)<\infty, ~\forall~\varepsilon>0, $

由上式并结合 (2.8) 式及$\sum\limits_{i=1}^n a_{ni}=1$可得$EX=0.$证毕.

定理1.2的证明  由定理1.1的证明可知定理1.2成立.

定理1.3的证明  充分性:令$a_{ni}=n^{(1+\beta)(p-1)/p}A_{n-i}^\beta/A_n^{(1+\beta)}, \forall\, 0\le i\le n, n\ge 1.$注意到

$ \lim_{n\to \infty}A_n^\beta/(n^\beta/\Gamma(1+\beta))=1, \forall\, \beta>-1. $

于是存在正常数$c$和$C$, 使当$n$充分大时有

$ c(n-i)^\beta n^{-(1+\beta)/p}\le a_{ni}\le C(n-i)^\beta n^{-(1+\beta)/p}, \ \ \forall\, 0\le i\le n-1, $

$ cn^{-(1+\beta)/p}<a_{nn}<Cn^{-(1+\beta)/p}. $

于是, 令

$ a_{ni}=c_{ni}(n-i)^\beta n^{-(1+\beta)/p}, \forall\, 0\le i\le n-1, a_{nn}=c_{nn}n^{-(1+\beta)/p}, $

其中$c<c_{ni}<C, 1\le i\le n, n\ge 1.$对任意固定的$n$及$\varepsilon>0$, 由$C_r$不等式, (1.8) 式和 (1.13) 式总有

$ n^{r-2+\tau(1+\beta)(p-1)/p}E\left\{\mathop {\max }\limits_{0\le k\le n}\left|\frac{1}{A_n^{1+\beta}}\sum\limits_{i=0}^k A_{n-i}^{\beta}X_i \right|-\varepsilon n^{-(1+\beta)(p-1)/p}\right\}_+^\tau<\infty. $

因此, 要证明 (1.14) 式, 只要证下式

$ \sum\limits_{n=1}^\infty n^{r-2}E\left\{\mathop {\max }\limits_{0\le k\le n} \left|\sum\limits_{i=0}^k a_{ni}X_i \right|-\varepsilon \right\}_+^\tau<\infty. $

而

$ \begin{gathered} E\left\{ {{\mathop {\max }\limits_{0 \leqslant k \leqslant n}}\left| {\sum\limits_{i = 0}^k {{a_{ni}}} {X_i}} \right| - \varepsilon } \right\}_ + ^\tau \hfill \\ \leqslant {2^\tau }E\left\{ {{\mathop {\max }\limits_{0 \leqslant k \leqslant n - 1}}\left| {\sum\limits_{i = 0}^k {{a_{ni}}} {X_i}} \right| - \varepsilon /2} \right\}_ + ^\tau + {2^\tau }E\left\{ {\left| {{a_{nn}}{X_n}} \right| - \varepsilon /2} \right\}_ + ^\tau . \hfill \\ \end{gathered} $

因此, 由定理1.2知要证 (1.14) 式, 只要证明

$\sum\limits_{n=1}^\infty n^{r-2}E\left\{\left|a_{nn}X_n \right|-\varepsilon/2 \right\}_+^\tau<\infty.$

(2.10)

由引理2.3有

$ \begin{gathered} \sum\limits_{n = 1}^\infty {{n^{r - 2}}} \int_1^\infty P \left( {|{a_{nn}}{X_n}| > \varepsilon {t^{1/\tau }}} \right){\text{d}}t \hfill \\ \leqslant \int_1^\infty {\left( {\sum\limits_{n = 1}^\infty {{n^{r - 2}}} P\left( {|CX| > \varepsilon {n^{(1 + \beta )/p}}{t^{1/\tau }}} \right)} \right)} {\text{d}}t \hfill \\ \ll \int_1^\infty E {\left( {\frac{{|X|}}{{{t^{1/\tau }}}}} \right)^{p(r - 1)/(1 + \beta )}}{\text{d}}t. \hfill \\ \end{gathered} $

而当$\beta=-1/[p(r-1)+1]$时有$p(r-1)/(1+\beta)=p(r-1)+1$, 当$-1/[p(r-1)+1]<\beta\le 0$时有$p(r-1)/(1+\beta)< pr/(1+\beta-p\beta)$, 因此, 由 (1.13) 式和 (1.8) 式可得 (2.10) 式成立, 从而 (1.14) 式成立.

必要性:由 (1.14) 式可得

$ \sum\limits_{n=1}^\infty n^{r-2}P\left(\mathop {\max }\limits_{0\le k\le n} \left|\frac{1}{A_n^{1+\beta}}\sum\limits_{i=0}^k A_{n-i}^{\beta}X_i \right|>\varepsilon n^{-(1+\beta)(p-1)/p}\right)<\infty, \, \forall\, \varepsilon>0. $

因此由文献[9]可知结论成立.