本文恒设$ \{\Omega, {\cal F}, P\} $是概率空间, $ \{X_{n, i}, n, i\ge 1\} $都是定义在$ \{\Omega, {\cal F}, P\} $上的$ L^p (p>0) $可积随机变量阵列, $ \{{\cal F}_{n, i}, -\infty<i<\infty, n\ge1\} $为$ {\cal F} $的子$ \sigma $ -代数满足:对每一固定的$ n\ge 1 $都有$ {\cal F}_{n, i}\subset {\cal F}_{n, i+1}. $令$ \|X\|_p = (E|X|^p)^{1/p}, p>0. $

定义1.1  设$ p\ge 1 $,称$ L^p $可积阵列$ \{X_{n, i}, {\cal F}_{n, i} \} $为$ L_p $混合的($ L_p $-mixingale),如果存在非负常数阵列$ \{C_{n, i}, n, i\ge 1\} $与常数列$ \{\psi(j), j\ge 0\} $,使得$ j\to \infty $时, $ \psi(j)\downarrow 0 $,且对任意$ i\ge 1 $和$ j\ge 0, $

$ \rm(i) $$ \|E(X_{n, i}|{\cal F}_{n, i-j})\|_p\le C_{n, i}\psi(j); $

$ \rm(ii) $$ \|X_{n, i}-E(X_{n, i}|{\cal F}_{n, i+j})\|_p\le C_{n, i}\psi(j+1). $

显然,对任意$ 1\le q\le p $和常数阵列$ \{a_{n, i}\} $,如果$ \{X_{n, i}, {\cal F}_{n, i} \} $是$ L_p $混合阵列,则$ \{X_{n, i}, {\cal F}_{n, i} \} $是$ L_q $ -混合阵列, $ \{a_{n, i}X_{n, i}, {\cal F}_{n, i} \} $是$ L_p $混合阵列,此时只需将定义1.1中的$ C_{n, i} $用$ |a_{n, i}|C_{n, i} $代替.

Mcleish^[1]首先引入了$ L_2 $ -混合序列,指出$ L_2 $ -混合序列是混合序列和鞅差序列的推广,并证明了其强大数律. McLeish^[2-3]证明了其不变原理. $ L_p $混合涵盖非常广泛, Hall和Heyde^[4]指出$ L_p $混合序列包含了鞅差序列,线性过程, $ H $ -混合序列. $ L_p $混合在估计量的强(弱)相合性和核回归函数估计等都有非常重要的应用.许多学者对其进行了深入研究,如Andrews^[5], Davidson^[6], Davidson和De Jong^[7], De Jong^[8-11], Fazekas和Klesov^[12], Gan^[13], Hansen ^[14-15], Hong等^[16], Meng和Lin^[17], Qiu等^[18]等等. De Jong^[11]得到了如下关于$ L_p $ -混合阵列$ L^r $收敛的结果：

定理1.1  设$ 1\le p\le 2 $, $ \{X_{n, i}, {\cal F}_{n, i} \} $是$ L_p $ -混合阵列满足

$ \lim\limits_{n\to \infty} \bigg(a_n^{-p} \sum\limits_{i = 1}^{n} C_{n, i}^{p}\bigg)^{1/p }\sum\limits_{j = 0}^{[n^{(p-1)/p}]+1}\psi(j) = 0, $

则$ \lim\limits_{n\to \infty} \Big\|a_n^{-1}\sum\limits_{i = 1}^{n} X_{n, i}\Big\|_p = 0, $其中$ [x] $表示$ x $的整数部分.

定理1.2  设$ \{X_{n, i}, {\cal F}_{n, i} \} $是$ L_2 $ -混合阵列满足

$ \lim\limits_{n\to \infty} \bigg(a_n^{-2} \sum\limits_{i = 1}^{n} C_{n, i}^{2}\bigg)\sum\limits_{j = 0}^{n-1}\psi(j) = 0, $

则$ \lim\limits_{n\to \infty}\Big \|a_n^{-1}\sum\limits_{i = 1}^{n} X_{n, i}\Big\|_2 = 0. $

Gan^[13]得到了如下结果:

定理1.3  设$ 1<p\le 2 $, $ \{k_n, n\ge 1\} $是正整数序列且$ k_n\uparrow \infty $, $ \{X_{n, i}, {\cal F}_{n, i} \} $是$ L_p $ -混合阵列, $ \{a_{n.i}, 1\le i\le k_n, n\ge 1\} $是常数阵列满足

$ \psi(n) = O\bigg(\frac{1}{n(\log n)^s}\bigg) , $

$ \sum\limits_{i = 1}^ \infty C_{n, i}^p\le D, \, \, \, \forall \, \, n\ge 1, $

$ \lim\limits_{n\to \infty} \max\limits_{1\le i\le k_n}|a_{n, i}| = 0, $

其中$ s $与$ D $是常数且$ s>1, D>0, $则$ \lim\limits_{n\to \infty} \Big\|\sum\limits_{i = 1}^{k_n} a_{n, i}X_{n, i}\Big\|_p = 0. $

De Jong^[11]得到了如下关于$ L_p $ -混合阵列均方收敛的结果：

定理1.4  设$ \{X_{n, i}, {\cal F}_{n, i} \} $是$ L_2 $ -混合阵列满足：存在$ \eta>0 $使

$ \lim\limits_{n\to \infty} \bigg(a_n^{-2} \sum\limits_{i = 1}^{n} C_{n, i}^{2}\bigg)\sum\limits_{j = 0}^{n}\bigg(\log(j+2)\bigg)^{1+\eta} \psi(j)^2 = 0, $

则$ \lim\limits_{n\to \infty} \Big\|a_n^{-1}\sum\limits_{i = 1}^{n} X_{n, i}\Big\|_2 = 0. $

Meng和Lin^[17]得到了如下关于$ L^p $ -混合阵列$ r $ -平均收敛的结果：

定理1.5  设$ 1<p\le 2 $, $ \{k_n, n\ge 1\} $是正整数序列, $ \{X_{n, i}, {\cal F}_{n, i} \} $是$ L_p $ -混合阵列满足

$ \limsup\limits_{n\to \infty} k_n^{-1} \sum\limits_{i = 1}^{k_n} C_{n, i}<\infty $

及

$ \limsup\limits_{n\to \infty} k_n^{-2} \sum\limits_{i = 1}^{k_n} C_{n, i}^2 = 0, $

则对任意$ r\in (1, p) $, $ \lim\limits_{n\to \infty}\Big \|k_n^{-1}\sum\limits_{i = 1}^{k_n} X_{n, i}\Big\|_r = 0. $

本文的目的是推广和改进定理1.1–1.4,在条件比定理1.5稍强的条件下得到了比定理1.5结论强的结果.

以下, $ C $都表示与$ n $无关的常数,在不同的地方可表示不同的值,即使在同一式中都如此. $ p\wedge q $表示$ \min\{p, q\} $.

定理2.1  设$ p>1 $, $ \{k_n, n\ge 1\} $和$ \{d_n, n\ge 1\} $是正整数序列, $ \{Y_{n, i}, {\cal F}_{n, i}\} $是$ L_p $ -混合阵列且满足

(2.1) $ \begin{equation} k_n^{1-1/(p\wedge 2)} d_n^{-1}\le D, \forall \, \, n\ge 1, \mbox{其中$D$是正常数}, \end{equation} $

(2.2) $ \begin{equation} \lim\limits_{n\to \infty} \bigg(\sum\limits_{i = 1}^{k_n} C_{n, i}^{p \wedge 2}\bigg)^{1/(p \wedge 2)}\sum\limits_{j = 0}^{d_n}\psi(j) = 0, \end{equation} $

则

(2.3) $ \begin{equation} \lim\limits_{n\to \infty} \bigg\|\sum\limits_{i = 1}^{k_n} Y_{n, i}\bigg\|_{p\wedge 2} = 0. \end{equation} $

证  注意到

(2.4) $ \begin{eqnarray} \sum\limits_{i = 1}^{k_n} Y_{n, i} & = & \sum\limits_{i = 1}^{k_n} \left \{ Y_{n, i}-E\big(Y_{n, i}|{\cal F}_{n, i+{d_n}}\big)\right\}\\ && + \sum\limits_{i = 1}^{k_n} \sum\limits_{j = -d_n+1}^{d_n} \left\{E\big(Y_{n, i}|{\cal F}_{n, i+j}\big)-E\big(Y_{n, i}|{\cal F}_{n, i+j-1}\big)\right\} + \sum\limits_{i = 1}^{k_n} E\big(Y_{n, i}|{\cal F}_{n, i-d_n}\big) \\ &\triangleq& T_n^{(1)}+T_n^{(2)}+T_n^{(3)}. \end{eqnarray} $

由Minkowski不等式、Hölder不等式、$ \psi(j)\downarrow 0 $、(2.1)和(2.2)式可得

(2.5) $ \begin{eqnarray} \|T_n^{(1)}\|_{p\wedge 2} & \le & \sum\limits_{i = 1}^{k_n} \| \left \{ Y_{n, i}-E\big(Y_{n, i}|{\cal F}_{n, i+{d_n}}\big)\right\} \|_{p\wedge 2} \\ & \le& k_{n}^{1-1/{(p\wedge 2)}}\bigg( \sum\limits_{i = 1}^{k_n} E\left| \left \{ Y_{n, i}-E\big(Y_{n, i}|{\cal F}_{n, i+{d_n}}\big)\right\} \right|^{p\wedge 2} \bigg)^{1/(p\wedge 2)} \\ & \le & k_n^{1-1/(p\wedge 2)} \bigg( \sum\limits_{i = 1}^{k_n} \big( C_{n, i} \psi(d_n+1)\big) ^{p\wedge 2} \bigg)^{1/(p\wedge 2)} \\ & \le & k_n^{1-1/(p\wedge 2)} \bigg( \sum\limits_{i = 1}^{k_n} C_{n, i} ^{p\wedge 2} \bigg)^{1/(p\wedge 2)}\psi(d_n) \\ &\le& k_n^{1-1/(p\wedge 2)}d_n^{-1} \bigg(\sum\limits_{i = 1}^{k_n}C_{n, i}^{p\wedge 2} \bigg)^{1/(p\wedge 2)} \sum\limits_{j = 1}^{d_n}\psi(j) = o(1). \end{eqnarray} $

类似上式的证明,我们有

(2.6) $ \begin{eqnarray} \|T_n^{(3)}\|_{p\wedge 2} & \le & k_n^{1-1/(p\wedge 2)}\bigg(\sum\limits_{i = 1}^{k_n} E\left|E\big(Y_{n, i}|{\cal F}_{n, i-d_n}\big)\right|^{p\wedge 2}\bigg)^{1/(p\wedge 2)} \\ & \le& k_n^{1-1/(p\wedge 2)}\bigg( \sum\limits_{i = 1}^{k_n} \big( C_{n, i} \psi(d_n) \big)^{p\wedge 2}\bigg)^{1/(p\wedge 2)} \\ &\le& k_n^{1-1/(p\wedge 2)}d_n^{-1} \bigg(\sum\limits_{i = 1}^{k_n}C_{n, i}^{p\wedge 2} \bigg)^{1/(p\wedge 2)}\bigg( \sum\limits_{j = 1}^{d_n}\psi(j)\bigg) = o(1). \end{eqnarray} $

因为对每一固定的$ n\ge 1 $, $ \{E\big(Y_{n, i}|{\cal F}_{n, i+j}\big)-E\big(Y_{n, i}|{\cal F}_{n, i+j-1}\big), 1\le i\le k_n\} $是鞅差序列.由Minkowski不等式、Chatterji不等式$ ^{[19, \, {\rm p}101]} $及$ C_r $不等式可得

(2.7) $ \begin{eqnarray} \|T_n^{(2)}\|_{p\wedge 2} & \le& \sum\limits_{j = -d_n+1}^{d_n} \bigg \|\sum\limits_{i = 1}^{k_n} \left\{E\big(Y_{n, i}|{\cal F}_{n, i+j}\big)-E\big(Y_{n, i}|{\cal F}_{n, i+j-1}\big)\right\} \bigg\|_{p\wedge 2} \\ &\le &C \sum\limits_{j = -d_n+1}^{d_n} \bigg(\sum\limits_{i = 1}^{k_n} E \left|E\big(Y_{n, i}|{\cal F}_{n, i+j}\big) -E\big(Y_{n, i}|{\cal F}_{n, i+j-1}\big)\right|^{p\wedge 2} \bigg)^{1/(p\wedge 2)} \\ &\le &C\sum\limits_{j = 1}^{d_n} \bigg(\sum\limits_{i = 1}^{k_n} E \left|Y_{ni}-E\big(Y_{n, i}|{\cal F}_{n, i+j}\big) \right|^{p\wedge 2} \bigg)^{1/(p\wedge 2)} \\ &&+C\sum\limits_{j = 1}^{d_n} \bigg(\sum\limits_{i = 1}^{k_n} E \left|Y_{ni} -E\big(Y_{n, i}|{\cal F}_{n, i+j-1}\big)\right|^{p\wedge 2} \bigg)^{1/(p\wedge 2)} \\ & &+ C \sum\limits_{j = -d_n+1}^{0} \bigg(\sum\limits_{i = 1}^{k_n} E \left|E\big(Y_{n, i}|{\cal F}_{n, i+j}\big) -E\big(Y_{n, i}|{\cal F}_{n, i+j-1}\big)\right|^{p\wedge 2} \bigg)^{1/(p\wedge 2)} \\ &\triangleq& T_n^{(2, 1)}+T_n^{(2, 2)}+T_n^{(2, 3)}. \end{eqnarray} $

由$ \psi(j)\downarrow 0 $和(2.2)式有

(2.8) $ \begin{eqnarray} T_n^{(2, 1)}& \le & C\sum\limits_{j = 1}^{d_n}\bigg( \sum\limits_{i = 1}^{k_n} \big( C_{n, i} \psi(j+1) \big)^{p\wedge 2} \bigg)^{1/(p\wedge 2)} \\ & \le& C\bigg(\sum\limits_{i = 1}^{k_n} C_{n, i}^{p\wedge 2} \bigg)^{1/(p\wedge 2)} \sum\limits_{j = 1}^{d_n}\psi(j) = o(1) \end{eqnarray} $

及

(2.9) $ \begin{eqnarray} T_n^{(2, 2)} \le C\bigg(\sum\limits_{i = 1}^{k_n} C_{n, i}^{p\wedge 2} \bigg)^{1/(p\wedge 2)} \sum\limits_{j = 1}^{d_n}\psi(j) = o(1) . \end{eqnarray} $

再由$ C_r $不等式、$ \psi(j)\downarrow 0 $和(2.2)式可得

(2.10) $ \begin{eqnarray} T_n^{(2, 3)}& \le& C\sum\limits_{j = -d_n+1}^{0} \bigg(\sum\limits_{i = 1}^{k_n} E \left\{|E\big(Y_{n, i}|{\cal F}_{n, i+j}\big)|^{p\wedge 2} +|E\big(Y_{n, i}|{\cal F}_{n, i+j-1}\big)|^{p\wedge 2}\right\} \bigg)^{1/(p\wedge 2)} \\ & \le& C\sum\limits_{j = -d_n+1}^{0} \bigg( \sum\limits_{i = 1}^{k_n} \left\{\big(C_{n, i}\psi(-j)\big)^{p\wedge 2}+\big(C_{n, i}\psi(-j+1)\big)^{p\wedge 2} \right\} \bigg)^{1/(p\wedge 2)} \\ & \le & C\sum\limits_{j = -d_n+1}^{0} \bigg( 2\sum\limits_{i = 1}^{k_n} \big(C_{n, i}\psi(-j)\big)^{p\wedge 2} \bigg)^{1/(p\wedge 2)} \\ & \le & C \bigg( \sum\limits_{i = 1}^{k_n}C_{n, i}^{p\wedge 2} \bigg)^{1/(p\wedge 2)} \sum\limits_{j = 0}^{d_n} \psi(j) = o(1). \end{eqnarray} $

因此,由(2.4)–(2.10)式可知(2.3)式成立.

注2.1   $ \rm(1) $由定理2.1的证明可知,当$ p = 1 $时, (2.3)式仍然成立,此时,条件(2.1)可去掉.

$ \rm(2) $定理2.1推广和改进了定理1.1与定理1.2,事实上,在$ 1\le p\le 2 $时,令$ Y_{n, i} = X_{n, i}/a_n, $$ k_n = n, d_n = [n^{(p-1)/p}]+1 $,由定理2.1可得定理1.1;在$ p = 2 $时,令$ Y_{n, i} = X_{n, i}/a_n, $$ k_n = n, d_n = n-1 $由定理2.1可得定理1.2.

$ \rm(3) $定理2.1推广和改进了定理1.3.对常数阵列$ \{a_{n, i}\} $,令$ Y_{n, i} = a_{n, i}X_{n, i}, $$ d_n = k_n $,则由定理2.1可得定理1.3.

定理2.2  设$ \{k_n, n\ge 1\} $和$ \{d_n, n\ge 1\} $是正整数序列, $ \{b_n, n\ge 1\} $是递增的正常数列, $ \{Y_{n, i}, {\cal F}_{n, i}\} $是$ L_2 $ -混合阵列且满足

(2.11) $ \begin{equation} k_n d_n^{-1}\le D, \, \, \forall \, n\ge 1, \ \mbox{其中$D$是正常数}, \end{equation} $

(2.12) $ \begin{equation} \sum\limits_{n = 1}^\infty 1/b_n<\infty, \end{equation} $

(2.13) $ \begin{equation} \lim\limits_{n\to \infty} \sum\limits_{i = 1}^{k_n} C_{n, i}^2\sum\limits_{j = 1}^{d_n}\max\{b_{j+1}-b_{j-1}, 1\}\psi(j)^{ 2} = 0, \ \mbox{其中$ b_0 = 0, $} \end{equation} $

则

(2.14) $ \begin{equation} \lim\limits_{n\to \infty} \bigg\|\sum\limits_{i = 1}^{k_n}Y_{n, i}\bigg\|_2 = 0. \end{equation} $

证  沿用定理2.1的符号和证明方法,由(2.4)式可知,要证(2.14)式只需证$ E|T_n^{(k)}|^2 = o(1), $$ k = 1, 2, 3. $类似(2.5)式的证明可得

(2.15) $ \begin{equation} E|T_n^{(1)}|^2 \le k_n \sum\limits_{i = 1}^{k_n} \big( C_{n, i} \psi(d_n+1)\big) ^2 \le k_n d_n^{-1}\sum\limits_{i = 1}^{k_n}C_{n, i}^2 \sum\limits_{j = 1}^{d_n}\psi(j)^2 = o(1) \end{equation} $

及

(2.16) $ \begin{equation} E|T_n^{(3)}|^2 \le k_n \sum\limits_{i = 1}^{k_n}\big(C_{n, i} \psi(d_n) \big)^ 2 \le k_n d_n^{-1}\sum\limits_{i = 1}^{k_n}C_{n, i}^2 \sum\limits_{j = 1}^{d_n}\psi(j)^2 = o(1) . \end{equation} $

由Cauchy-Schwartz不等式、Chatterji不等式和(2.12)式,有

(2.17) $ \begin{eqnarray} E|T_n^{(2)}|^2 & = & E\bigg(\sum\limits_{j = -d_n+1}^{d_n} (\sqrt{b_{|j|}})^{-1} \cdot \sqrt{b_{|j|}} \bigg(\sum\limits_{i = 1}^{k_n} \left\{E\big(Y_{n, i}|{\cal F}_{n, i+j}\big)-E\big(Y_{n, i}|{\cal F}_{n, i+j-1}\big)\right\}\bigg)\bigg)^2 \\ &\le & \bigg( \sum\limits_{j = -d_n+1}^{d_n} \frac{1}{b_{|j|}} \bigg)\bigg( \sum\limits_{j = -d_n+1}^{d_n} b_{|j|} E\bigg(\sum\limits_{i = 1}^{k_n}\left\{ E\big(Y_{n, i}|{\cal F}_{n, i+j}\big)-E\big(Y_{n, i}|{\cal F}_{n, i+j-1}\big)\right\} \bigg)^2\bigg) \\ &\le & C \sum\limits_{j = -d_n+1}^{d_n} b_{|j|} \sum\limits_{i = 1}^{k_n}E\left\{ E\big(Y_{n, i}|{\cal F}_{n, i+j}\big) -E\big(Y_{n, i}|{\cal F}_{n, i+j-1}\big)\right\}^2\\ & = &C \sum\limits_{j = 1}^{d_n} b_{j} \sum\limits_{i = 1}^{k_n}E\left\{ E\big(Y_{n, i}|{\cal F}_{n, i+j}\big) -E\big(Y_{n, i}|{\cal F}_{n, i+j-1}\big)\right\}^2\\ &&+C \sum\limits_{j = -d_n+1}^{0} b_{-j} \sum\limits_{i = 1}^{k_n}E\left\{ E\big(Y_{n, i}|{\cal F}_{n, i+j}\big)- E\big(Y_{n, i}|{\cal F}_{n, i+j-1}\big)\right\}^2 \\ &\triangleq& T_n^{(2, 1)}+T_n^{(2, 2)}. \end{eqnarray} $

注意到

$ \begin{eqnarray*} &&E\left\{ E\big(Y_{n, i}|{\cal F}_{n, i+j}\big)-E\big(Y_{n, i}|{\cal F}_{n, i+j-1}\big)\right\}^2 \\ & = &E\big(Y_{n, i}- E\big(Y_{n, i}|{\cal F}_{n, i+j-1}\big)\big)^2-E\big(Y_{n, i}- E\big(Y_{n, i}|{\cal F}_{n, i+j}\big)\big)^2. \end{eqnarray*} $

因此,由$ \psi(j)\downarrow 0 $和(2.13)式可得

(2.18) $ \begin{eqnarray} T_n^{(2, 1)}& = & C \sum\limits_{i = 1}^{k_n} \sum\limits_{j = 1}^{d_n} b_{j} \left\{E\big(Y_{n, i}- E\big(Y_{n, i}|{\cal F}_{n, i+j-1}\big)\big)^2-E\big(Y_{n, i}- E\big(Y_{n, i}|{\cal F}_{n, i+j}\big)\big)^2\right\}\\ &\le & C \sum\limits_{i = 1}^{k_n} b_{1} \psi(0)^2 C_{n, i}^2+C \sum\limits_{i = 1}^{k_n}\sum\limits_{j = 1}^{d_n-1}( b_{j+1}-b_{j})E \big(Y_{n, i}- E\big(Y_{n, i}|{\cal F}_{n, i+j}\big)\big)^2\\ & \le & Cb_{1} \psi(0)^2 \sum\limits_{i = 1}^{k_n} C_{n, i}^2+C \sum\limits_{i = 1}^{k_n} \sum\limits_{j = 1}^{d_n-1}( b_{j+1}-b_{j})C_{ni}^2\psi(j)^2\\ & = &o(1) \end{eqnarray} $

及

(2.19) $ \begin{eqnarray} T_n^{(2, 2)} & = & C \sum\limits_{i = 1}^{k_n}\sum\limits_{j = -d_n+1}^{0} b_{-j} \left\{E\big( E\big(Y_{n, i}|{\cal F}_{n, i+j}\big)\big)^2-E \big(E\big(Y_{n, i}|{\cal F}_{n, i+j-1}\big)\big)^2\right\} \\ & \le& C \sum\limits_{i = 1}^{k_n}\left\{ b_0E\big( E\big(Y_{n, i}|{\cal F}_{n, i}\big)\big)^2 +\sum\limits_{j = -d_n+1}^{-1}( b_{-j}-b_{-j-1})E \big(E\big(Y_{n, i}|{\cal F}_{n, i+j}\big)\big)^2 \right\}\\ & \le & Cb_{0} \psi(0)^2 \sum\limits_{i = 1}^{k_n} C_{n, i}^2+C \sum\limits_{i = 1}^{k_n}\sum\limits_{j = 1}^{d_n-1} ( b_{j}-b_{j-1})C_{ni}^2\psi(j)^2\\ & = &o(1). \end{eqnarray} $

于是,由(2.15)–(2.19)式可知(2.14)式成立.

注2.2   (1)显然,设$ \eta>0 $,令$ Y_{n, i} = X_{n, i}/a_n, k_n = d_n = n, b_n = n\big(\log(n+1)\big)^{1+\eta} $,由定理2.3可得定理1.4.

(2)设$ \eta>0 $, $ \{a_{n, i}\} $是常数阵列,令$ Y_{n, i} = a_{n, i}X_{n, i}, $$ k_n = d_n = n, $$ b_n = n\log (n+1) $$ (\log\log(n+1))^{1+\eta} $,我们得到如下结论:

设$ \{X_{n, i}, {\cal F}_{n, i}\} $是$ L_2 $ -混合阵列满足:存在$ \eta>0 $使

$ \lim\limits_{n\to \infty} \sum\limits_{i = 1}^{n} \big( a_{n, i}C_{n, i}\big)^2\sum\limits_{j = 1}^{n} \log(j+2) \big(\log\log(j+2)\big)^{1+\eta}\psi(j)^{ 2} = 0, $

则

$ \lim\limits_{n\to \infty} \bigg\|\sum\limits_{i = 1}^{n} a_{n, i}X_{n, i}\bigg\|_ 2 = 0. $

定理2.3  设$ p>1 $, $ \{k_n, n\ge 1\} $是正整数序列, $ \{Y_{n, i}, {\cal F}_{n, i}\} $是$ L_p $ -混合阵列满足

(2.20) $ \begin{equation} \sum\limits_{i = 1}^{k_n}C_{n, i}\le D, \forall \, \, n\ge 1, \mbox{其中$D$是正常数}, \end{equation} $

(2.21) $ \begin{equation} \lim\limits_{n\to \infty}\sum\limits_{i = 1}^{k_n} C_{n, i}^{p \wedge 2} = 0, \end{equation} $

则

(2.22) $ \begin{equation} \lim\limits_{n\to \infty} \bigg\|\sum\limits_{i = 1}^{k_n} Y_{n, i}\bigg\|_{p\wedge 2} = 0. \end{equation} $

证  沿用定理2.1的符号和证明方法,并用正整数$ L $代替(2.4)式中的$ d_n $.则由Minkowski不等式与$ \psi(j)\downarrow 0 $可得

(2.23) $ \begin{eqnarray} \|T_n^{(1)}\|_{p\wedge 2} \le \sum\limits_{i = 1}^{k_n}\| Y_{n, i}-E\big(Y_{n, i}|{\cal F}_{n, t+L}\big)\|_{p\wedge 2} \le \sum\limits_{i = 1}^{k_n}C_{n, i} \psi(L+1) , \end{eqnarray} $

类似地有

(2.24) $ \begin{equation} \|T_n^{(3)}\|_{p\wedge 2} \le \sum\limits_{i = 1}^{k_n}C_{n, i} \psi(L). \end{equation} $

当用正整数$ L $代替$ d_n $时(2.7)式仍成立.类似于(2.8)式的证明有

(2.25) $ \begin{eqnarray} T_n^{(2, 1)} & \le & C\sum\limits_{j = 1}^L \bigg(\sum\limits_{i = 1}^{k_n} \big( C_{n, i} \psi(j+1) \big)^{p\wedge 2} \bigg)^{1/(p\wedge 2)}\\ & \le & C(\psi(0)L )\bigg(\sum\limits_{i = 1}^{k_n} C_{n, i}^{p\wedge 2}\bigg)^{1/(p\wedge 2)} \end{eqnarray} $

和

(2.26) $ \begin{equation} T_n^{(2, 2)} \le C(\psi(0)L )\bigg(\sum\limits_{i = 1}^{k_n} C_{n, i}^{p\wedge 2}\bigg)^{1/(p\wedge 2)}. \end{equation} $

再由$ C_r $不等式和$ \psi(j)\downarrow 0 $可得

(2.27) $ \begin{eqnarray} T_n^{(2, 3)} & \le& C \sum\limits_{j = -L+1}^{0} \bigg( \sum\limits_{i = 1}^{k_n} \left\{ \big( C_{n, i} \psi(-j) \big)^{p\wedge 2} + \big( C_{n, i} \psi(-j+1) \big)^{p\wedge 2} \right\}\bigg)^{1/(p\wedge 2)} \\ & \le & C(\psi(0)L )\bigg(\sum\limits_{i = 1}^{k_n} C_{n, i}^{p\wedge 2}\bigg)^{1/(p\wedge 2)}. \end{eqnarray} $

因此,对$ \forall\, \varepsilon>0 $,由$ \psi(j)\downarrow 0 $, (2.20), (2.23)及(2.24)式,当$ L $充分大时

$ \|T_n^{(1)}\|_{p\wedge 2}+\|T_n^{(3)}\|_{p\wedge 2}<\varepsilon, $

并且由(2.21)与(2.25)–(2.27)式,当$ n $充分大时

$ \|T_n^{(2)}\|_{p\wedge 2}<\varepsilon. $

因此, (2.22)式成立.

注2.3   (1)根据定理2.3的证明可知,当$ p = 1 $时, (2.22)式仍然成立,此时,条件(2.20)可去掉.

(2)定理2.3的条件(2.21)比定理1.5的要强,但定理2.3的结论也比定理1.5的结论强.