在Zygmund^[1], Schipp^[2]的Walsh系统和Pál, Simon^[3]的维林肯系统中, 可以证明关于Fejér均值极大算子的弱型不等式. 该$ \sigma^* $算子的$ (H_1, L_1) $型不等式被Fujii^[4]所发现. Weisz^[5]推广了这一结果并证明当$ 1/2< p\leq 1 $时, $ \sigma^* $是从Hardy空间$ H_p $到$ L_p $有界的. Simon^[6]通过给出反例, 证明当$ 0< p <1/2 $时, $ \sigma^* $的有界性不成立. 当$ p= 1/2 $时, $ \sigma^* $的有界性不成立的反例是由Goginava^[7]给出(另见文献[8]). 当$ p= 1/2 $时, 还有如下结果. Weisz^[9]证明了$ \sigma^* $是从Hardy空间$ H_{1/2} $到弱$ L_{1/2} $空间有界的. Goginava^[10]定义极大算子

$ \tilde{\sigma}^*f=\sup\limits_{n\in {\Bbb N}}\frac{|\sigma_nf|}{\log^2(n+1)}, $

并证明该算子是从Hardy空间$ H_{1/2} $到$ L_{1/2} $有界的. 他也证明对于任意的不减函数$ \varphi:{\Bbb N}\rightarrow [1, \infty) $, 满足条件

$ \overline{\lim\limits_{n\rightarrow \infty}}\frac{\log^2(n+1)}{\varphi(n)}=+\infty $

时, 极大算子$ \sup\limits_{n\in {\Bbb N}}\frac{|\sigma_nf|}{\varphi(n)} $不是从Hardy空间$ H_{1/2} $到$ L_{1/2} $有界的. Tephnadze^[11]推广了这一结果并且证明出当$ 0 < p <1/2 $时, 极大算子$ \tilde{\sigma}_p^*f=\sup\limits_{n\in {\Bbb N}}\frac{|\sigma_nf|}{(n+1)^{1/p-2}} $是从Hardy空间$ H_p $到$ L_p $有界的, 其中$ \sigma_nf $是关于有界维林肯系统的Fejér均值.

该文的目的是把Tephnadze的结果推广到更为广泛的维林肯型系统.

该文用$ {\Bbb N} $表示非负整数的集合. 设$ m:=(m_0, m_1, \cdots, m_k, \cdots) $是满足$ m_k \geq 2 $$ (k\in {\Bbb N} ) $的序列. 对于所有的$ k\in {\Bbb N} $, 用$ Z_{m_k} $表示离散循环群. 不妨设$ Z_{m_k}=\{0, 1, \cdots, m_k-1 \} $, $ \mu_k $为$ Z_{m_k} $上的测度. 设$ G_m $表示具有$ Z_{m_k} $的乘积拓扑和乘积测度$ \mu $的完全直积, 则$ G_m $形成了哈尔测度为1的紧阿贝尔群. $ G_m $中的元素则可以表示为$ (x_0, x_1, \cdots, x_k, \cdots) $的序列, 其中对于每一个$ k\in {\Bbb N} $, 满足$ x_k\in Z_{m_k} $. 群$ G_m $的拓扑完全由

$ I_n(0):=\{ (x_0, x_1, \cdots, x_k, \cdots)\in G_m:x_k=0\ (k=0, \cdots, n-1)\} $

$ (I_0(0):=G_m) $所决定. 如果生成系统$ m $是有界的, 则称$ G_m $是有界的. 设$ I_n(x):=I_n(0)+x $$ (n\in {\Bbb N}) $和$ e_n:=(0, \cdots, 0, x_n=1, 0\cdots)\in G_m $, 则

(2.1) $ \begin{equation} G_m\backslash I_N=\bigcup\limits_{s=0}^{N-1} I_s\backslash I_{s+1}=\biggl( \bigcup\limits _{k=0}^{N-2} \bigcup\limits_{l=k+1}^{N-1} I_{N}^{k, l}\biggl) \bigcup \biggl(\bigcup\limits_{k=1}^{N-1}I_N^{k, N}\biggl), \end{equation} $

其中

$ \begin{eqnarray*} I_{N}^{k, l}=\left \{ \begin{array}{ll}I_N(0, \cdots, 0, x_k\neq 0, 0\cdots, 0, x_l\neq 0, x_{l+1}, \cdots x_{N-1}, \cdots), \\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\hskip 5cm \mbox{若}\ k<l<N, x_i\in Z_{m_i}, i\geq l+1, \\ I_N(0, \cdots, 0, x_k\neq 0, x_{k+1}=0, \cdots, x_{N-1}=0, x_N \cdots) , \\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\hskip 5cm\mbox{若}\ l=N, x_i\in Z_{m_i}, i\geq N. \end{array}\right. \end{eqnarray*} $

令$ I_N^{k, \alpha, l, \beta}:=I_N(0, \cdots, x_k=\alpha, 0, \cdots, x_l=\beta, x_{l+1}, \cdots, x_{N-1}) $, $ x_i\in Z_{m_i}, i\geq l+1 $. 显然有

(2.2) $ \begin{equation} I_{N}^{k, l}=\bigcup\limits_{\alpha=1}^{m_k-1}\bigcup\limits_{\beta=1}^{m_l-1}I_{N}^{k, \alpha, l, \beta}. \end{equation} $

设$ \Sigma_n $为集合$ I_n(x)(x\in G_m) $生成的$ \sigma $代数, $ E_n $是关于$ \Sigma_n(n\in {\Bbb N}) $的条件期望算子.空间$ L_p(G_m) $表示满足如下条件的可测函数$ f $的全体(简记为$ L_{p} $), 其中

$ \|f\|_p=\bigg(\int_{G_m}|f(x)|^p{\rm d}\mu(x)\bigg)^{1/p}<\infty, \quad(0<p<\infty). $

空间$ L_{p, \infty}(G_m) $表示满足如下条件的可测函数$ f $的全体(简记为$ L_{p, \infty} $), 其中

$ \|f\|_{L_{p, \infty}}= \sup\limits_{\lambda>0} \lambda^p \mu(|f|>\lambda)<\infty. $

用$ f=(f_n, n\in {\Bbb N}) $表示关于$ (\Sigma_n, n\in {\Bbb N}) $可测的鞅. 鞅$ f $的极大函数定义为

$ f^*=\sup\limits_{n\in {\Bbb N}}|f_n|. $

对于$ f\in L_1(G_m) $, 极大函数也可定义为

$ f^*(x) = \sup\limits_{n\in {\Bbb N}}\frac{1}{\mu(I_n(x))} \bigg|\int_{I_n(x)}f(u){\rm d}\mu(u)\bigg|. $

鞅Hardy空间$ H_p(G_m ) $表示满足如下条件的鞅的全体(简记为$ H_p $)

$ \|f\|_{H_p}:=\|f^*\|_p<\infty, \quad (0< p <\infty). $

称可测函数$ a $是一个$ p $原子, 如果存在一个区间$ I $, 使

$ \begin{eqnarray*} &(1)&{\quad} \int_I a{\rm d}\mu=0, \\ &(2)&{\quad} \|a\|_\infty \leq \mu(I)^{-\frac{1}{p}}, \\ &(3)&{\quad} \ \mbox{supp}\ (a)\subset I. \end{eqnarray*} $

令$ M_0:=1 $, $ M_{k+1}:=m_kM_k $, $ k\in {\Bbb N} $, 则每一个$ n\in {\Bbb N} $都可以唯一表示为$ n=\sum\limits_{k=0}^\infty n_kM_k, $$ 0\leq n_k<m_k, n_k\in {\Bbb N}. $数列$ (n_0, n_1, \cdots) $称为$ n $对$ m $的展开. 令

$ n^{(k)}=\sum\limits_{j=k}^\infty n_jM_j, {\qquad} |n|:=\max\{k\in {\Bbb N}:n_k\neq0\}, $

显然有$ M_{|n|}\leq n<M_{|n|+1} $.

对于$ k\in{\Bbb N} $和$ x\in G_m $, 定义Rademacher函数$ r_k $如下

$ r_k(x):=\exp(2\pi\iota \frac{x_k}{m_k})\quad (x\in G_m, \iota:=\sqrt{-1}, k\in{\Bbb N}). $

显然对于任意的$ x\in G_m, n\in {\Bbb N} $

(2.3) $ \begin{equation} \sum\limits_{i=0}^{m_n-1}r_n^i(x)=\left \{ \begin{array}{ll}0, & \mbox{若}\ x_n\neq 0, \\ m_n, {\quad} &\mbox{若}\ x_n=0. \end{array}\right. \end{equation} $

定义$ \psi_n $如下

$ \psi_n:=\prod\limits_{k=0}^\infty r_k^{n_k}\quad(n\in{\Bbb N}). $

现引入更加广泛的维林肯型(或$ \psi\alpha $)系统(另见文献[12–13]). 设函数$ \alpha_n, \alpha_j^k : G_m\rightarrow {\cal C} $$ (n, j, k\in {\Bbb N}) $满足对于任意的$ x, y\in G_m $

(1) $ \alpha_j^k $是关于$ \Sigma_j $可测的, 且$ \alpha_j^k(x+y)=\alpha_j^k(x)\alpha_j^k(y) \ \ (j, k\in {\Bbb N}); $

(2) $ |\alpha_j^k|=\alpha_j^k(0)=\alpha_0^k=\alpha_j^0=1\ \ (j, k\in {\Bbb N}); $

(3) $ \alpha_n:=\prod\limits_{j=0}^\infty \alpha_j^{n^{(j)}}\ \ (n\in {\Bbb N}). $

令$ \chi_n:=\psi_n\alpha_n\ (n\in {\Bbb N}) $. $ \chi:=\{\chi_n:n\in{\Bbb N}\} $被称作维林肯型(或$ \psi\alpha $)系统. 下面给出几个例子

1) 如果$ \forall \ k, j \in {\Bbb N} $, 有$ \alpha_j^k=1 $, $ \psi\alpha $系统就是维林肯系统.

2) 如果有$ m_j=2 $ ($ \forall j\in {\Bbb N} $) 和$ \alpha_j^{n^{(j)}}=(\beta_j)^{n_j}, $其中

$ \beta_j(x)=\exp \bigg(2\pi \iota\Big(\frac{x_{j-1}}{2^2}+\cdots +\frac{x_0}{2^{j+1}}\Big)\bigg)\quad(n, j\in{\Bbb N}, x\in G_m), $

$ \psi\alpha $系统就是二进整数群的特征系统.

3) 定义

$ \chi_n(x):=\exp\bigg(2\pi \iota \bigg (\sum\limits_{j=0}^\infty\frac{n_j}{M_{j+1}}\sum\limits_{j=0}^\infty x_jM_j\bigg)\bigg) \quad (x\in G_m, n\in{\Bbb N}), $

那么就得到了一个逼近理论中很有用的维林肯型系统.

现定义傅里叶系数, 傅里叶级数的部分和, 狄利克雷核, Fejér均值, Fejér核如下

$ \hat{f}^\chi(n)=\hat{f}(n):=\int_{G_m}f\bar{\chi}_n, \quad\hat{f}^\chi(0):=\int_{G_m}f\ \quad (f\in L_1(G_m)); $

$ S_n^{\chi}f =S_nf:=\sum\limits_{k=0}^{n-1}\hat{f}(k)\chi_k\ \quad (S^\chi_0f:=0); $

$ D^{\chi}_n(y, x)=D_n(y, x):=\sum\limits_{k=0}^{n-1}\chi_k(y)\bar{\chi}_k(x); $

$ \sigma^\chi_nf=\sigma_nf:=\frac{1}{n}\sum\limits_{k=0}^{n-1}S_kf; $

$ K^\chi_n(y, x)=K_n(y, x):=\frac{1}{n}\sum\limits_{k=0}^{n-1}D_k^\chi(y, x)\quad (y, x\in G_m). $

若$ f=(f_k, k\in {\Bbb N}) $是鞅, 上述傅里叶系数则做适当修改$ \hat{f}(n):=\lim\limits_{k\rightarrow \infty}\int_{G_m}f_k\bar{\chi}_n $. 显然有

$ S_nf(y)=\int_{G_m}f(x)D_n(y, x){\rm d}\mu(x), \quad \sigma_nf(y)=\int_{G_m}f(x)K_n(y, x){\rm d}\mu(x). $

文献[6]中, Simon证明了狄利克雷核函数满足

(2.4) $ \begin{equation} D^\chi_{M_n}(y, x)=D_{M_n}(y-x)=\left \{ \begin{array}{ll}M_n, {\quad} & \mbox{若}\ y-x\in I_n, \\ 0, &\mbox{若}\ y-x\in G_m \backslash I_n. \end{array}\right. \end{equation} $

此外对于$ y, x\in G_m $, 有

(2.5) $ \begin{equation} D_n^\chi(y, x)=\alpha_n(y)\bar{\alpha}_n(x)D^\psi_n(y-x) =\chi_n(y)\bar{\chi}_n(x)\bigg(\sum\limits_{j=0}^\infty D_{M_j}(y-x)\sum\limits_{k=m_j-n_j}^{m_j-1}r_j^k(y-x)\bigg), \end{equation} $

其中$ \psi $是维林肯系统.

由$ \alpha_j^k(x+y)=\alpha_j^k(x)\alpha_j^k(y) $和$ r_j(x+y)=r_j(x)r_j(y) $, 得

(2.6) $ \begin{eqnarray} \chi_n(y)\bar{\chi}_n(x)&=&\chi_n(y-x+x)\bar{\chi}_n(x)=\chi_n(y-x)\chi_n(x)\bar{\chi}_n(x){} \\&=&\chi_n(y-x)|\chi_n(x)|^2=\chi_n(y-x)\bar{\chi}_n(0). \end{eqnarray} $

从而

(2.7) $ \begin{equation} D_n^\chi(y, x)=D_n^\chi(y-x, 0)\ \mbox{ 和 }\ K_n^\chi(y, x)=K_n^\chi(y-x, 0). \end{equation} $

若$ f\in L_1(G_m) $, 易得部分和序列$ (S_{M_n}f, n\in{\Bbb N}) $是鞅. 接下来本文将考虑算子$ \tilde{\sigma}_p^* $的有界性, 其中

$ \tilde{\sigma}_p^*f=\sup\limits_{n\in {\Bbb N}}\frac{|\sigma_nf|}{(n+1)^{1/p-2}}. $

引理3.1   设$ A>t, t, A\in N, z\in I_t\backslash I_{t+1}. $那么

(3.1) $ \begin{equation} K_{M_A}(z, 0)=\left \{ \begin{array}{ll}0 , & \mbox{若}\ z-z_te_t\in G_m\backslash I_A, \\ { } \frac{M_t}{1-r_t(z)}, {\quad} &\mbox{若}\ z-z_te_t\in I_A. \end{array}\right. \end{equation} $

证   根据函数$ \alpha_n $和Fejér核和$ \psi\alpha $系统的定义, 得到如果$ z\in I_t\backslash I_{t+1} $ i.e. $ (z_0=\cdots=z_{t-1}=0, z_t\neq 0) $则有

(3.2) $ \begin{eqnarray} M_AK_{M_A}(z, 0)&=&\sum\limits_{k=0}^{M_A-1}D_k(z, 0){} \\ &=&\sum\limits_{k=0}^{M_A-1}(\sum\limits_{j=0}^{t-1}k_jM_j)\chi_k(z)\bar{\chi}_k(0) +\sum\limits_{k=0}^{M_A-1}M_t\sum\limits_{p=m_t-k_t}^{m_t-1}r_t^p(z-0)\chi_k(z)\bar{\chi}_k(0){} \\ &=&:\sum^1+\sum^2, {} \\ \sum^1 &=&\sum\limits_{k_{A-1}=0}^{m_{A-1}-1}\cdots \sum\limits_{k_{t+1}=0}^{m_{t+1}-1}\sum\limits_{k_{t-1}=0}^{m_{t-1}-1}\cdots \sum\limits_{k_0=0}^{m_0-1}(\sum\limits_{j=0}^{t-1}k_jM_j){} \\ &&\cdot\prod\limits_{l=0}^\infty r^{k_l}_l(z-0)\alpha_{l}^{k^{(l)}}(z)\bar{\alpha}_l^{k^{(l)}}(0)\sum\limits_{k_t=0}^{m_t-1}r_t^{k_t}(z-0){} \\ &=&\sum\limits_{k_t=0}^{m_t-1}r_t^{k_t}(z)\phi(z, 0), \end{eqnarray} $

其中函数$ \phi $不依赖于$ k_t $. 因此, $ \sum\limits^1=0. $当$ 0\leq k< M_A $时, 有$ k_j=0, \ j=A, A+1\cdots $. 另外又有

(3.3) $ \begin{eqnarray} \sum^2&=&\sum\limits_{k_{A-1}=0}^{m_{A-1}-1}\cdots \sum\limits_{k_{t+1}=0}^{m_{t+1}-1}\sum\limits_{k_{t-1}=0}^{m_{t-1}-1}\cdots\sum\limits_{k_0=0}^{m_0-1} \prod\limits_{l=0}^{A-1} r^{k_l}_l(z-0)\alpha_{l}^{k^{(l)}}(z)\bar{\alpha}_l^{k^{(l)}}(0)\sum\limits_{p=m_t-k_t}^{m_t-1}r_t^p(z-0){} \\ &=&\prod\limits_{l=0, l\neq t}^{A-1}(\sum\limits_{k_l=0}^{m_l-1}r_l^{k_l}(z)\alpha^{k^{(l)}}_l(z) \bar{\alpha}^{k^{(l)}}_l(0)) \sum\limits_{k_t=0}^{m_t-1}r_t^{k_t}(z)\alpha^{k^{(t)}}_t(z)\bar{\alpha}^{k^{(t)}}_t(0)\sum\limits_{i={m_t-k_t}}^{m_t-1} r^i_t(z). \end{eqnarray} $

由$ \alpha_n $的定义知, 知$ \alpha_n(0\leq n<M_A) $在$ I_A $上是常数. 如果$ z-z_te_t\in G_m\backslash I_A $ (即: 存在一个$ s $ ($ A>s>t $) 使$ z_s\neq 0 $), 从而得到

(3.4) $ \begin{eqnarray} \sum^2&=&\prod\limits_{l=0, l\neq s}^{A-1}(\sum\limits_{k_l=0}^{m_l-1}r_l^{k_l}(z)\alpha^{k^{(l)}}_l(z) \bar{\alpha}^{k^{(l)}}_l(0)) \sum\limits_{k_s=0}^{m_s-1}r_s^{k_s}(z)\alpha^{k^{(s)}}_t(z)\bar{\alpha}^{k^{(s)}}_t(0)\sum\limits_{i={m_t-k_t}}^{m_t-1} r^i_t(z){} \\ &=&\prod\limits_{l=0, l\neq s}^{A-1}(\sum\limits_{k_l=0}^{m_l-1}r_l^{k_l}(z)\alpha^{k^{(l)}}_l(z) \bar{\alpha}^{k^{(l)}}_l(0)) \sum\limits_{k_s=0}^{m_s-1}r_s^{k_s}(z)\sum\limits_{i={m_t-k_t}}^{m_t-1} r^i_t(z){} \\&=&0. \end{eqnarray} $

这样$ K_{M_A}(z, 0)=0 $.

另一方面, 如果$ z-z_te_t\in I_A $ (即$ z=(0, 0\cdots, 0, z_t, 0\cdots, z_{A}, \cdots)) $, 则有

(3.5) $ \begin{eqnarray} M_AK_{M_A}(z, 0) &=&M_t\prod\limits_{l=0, l\neq t}^{A-1}\bigg(\sum\limits_{k_l=0}^{m_l-1}r_l^{k_l}(z)\alpha^{k^{(l)}}_l(z) \bar{\alpha}^{k^{(l)}}_l(0)\bigg){} \\ &&\cdot\sum\limits_{k_t=0}^{m_t-1}r_t^{k_t}(z)\alpha^{k^{(t)}}_t(z)\bar{\alpha}^{k^{(t)}}_t(0)\sum\limits_{i={m_t-k_t}}^{m_t-1} r^i_t(z){} \\ &=& M_t\cdot m_0\cdots m_{t-1}\prod\limits_{l=t+1}^{A-1}\bigg(\sum\limits_{k_l=0}^{m_l-1}\alpha^{k^{(l)}}_l(z) \bar{\alpha}^{k^{(l)}}_l(0)\bigg){} \\ &&\cdot\sum\limits_{k_t=0}^{m_t-1}r_t^{k_t}(z)\alpha^{k^{(t)}}_t(z)\bar{\alpha}^{k^{(t)}}_t(0)\sum\limits_{i={m_t-k_t}}^{m_t-1}r^i_t(z){} \\ &=&M_t\cdot m_0\cdots m_{t-1} m_{t+1}\cdots m_{A-1}\sum\limits_{k_t=0}^{m_t-1}r_t^{k_t}(z)\alpha^{k^{(t)}}_t(z)\bar{\alpha}^{k^{(t)}}_t(0)\sum\limits_{i={m_t-k_t}}^{m_t-1}r^i_t(z){} \\ &=&M_t m_0\cdots m_{t-1} m_{t+1}\cdots m_{A-1}\sum\limits_{k_t=0}^{m_t-1}r_t^{k_t}(z)\sum\limits_{i={m_t-k_t}}^{m_t-1}r^i_t(z){} \\ &=&M_t M_A/m_t\sum\limits_{k_t=0}^{m_t-1}\frac{r_t^{k_t}(z)-1}{r_t(z)-1}. \end{eqnarray} $

由

$ \sum\limits_{k_t=0} ^{m_t-1}\frac{r_t^{k_t}(z)}{r_t(z)-1}=0, $

得

(3.6) $ \begin{equation} M_AK_{M_A}(z, 0)=M_tM_A\frac{1}{1-r_t(z)}. \end{equation} $

证毕.

备注3.1   设$ |n|\geq s>t, \quad n, s, t\in {\Bbb N}, z\in I_t\backslash I_{t+1}. $则有

(3.7) $ \begin{equation} K_{n^{(s+1)}, M_s}(z, 0):=\sum\limits_{k=n^{(s+1)}}^{n^{(s+1)}+M_s-1}D_k(z, 0)=\left \{ \begin{array}{ll}0 , & \mbox{若}\ z-z_te_t\in G_m\backslash I_s, \\ { } \frac{M_sM_t \chi_{n^{(s)}}(z)}{1-r_t(z)}, &\mbox{若}\ z-z_te_t\in I_s. \end{array}\right. \end{equation} $

证明类似于引理3.1, 略.

引理3.2   设$ 2< A\in {\Bbb N}_+, k\leq s < A $和$ n^*_A:=M_{2A}+M_{2A-2}+\cdots+M_2+M_0. $则有

$ n^*_{A-1}|K_{n^*_{A-1}}(z, 0)|\geq \frac{M_{2k}M_{2s}}{4}, $

对于$ z\in I_{2A}(0, \cdots, 0, z_{2k}\neq 0, 0, \cdots, 0, z_{2s}\neq 0, z_{2s+1}, \cdots, z_{2A-1}), k= 0, 1, \cdots, A-3, $$ s=k+ 2, $$ k+ 3, \cdots, A-1. $

证    设$ n\in {\Bbb N}_+ $. 由(2.5)式可知

$ D^\chi_n(y, x)=\alpha_n(y)\bar{\alpha}_n(x)D^\psi_n(y-x)=\chi_n(y)\bar{\chi}_n(x) \bigg(\sum\limits_{j=0}^\infty D_{M_j}(y-x)\sum\limits_{k=m_j-n_j}^{m_j-1}r_j^k(y-x)\bigg), $

因此

(3.8) $ \begin{equation} |D_n(z, 0)|\leq \sum\limits_{j=0}^\infty n_j D_{M_j}(z). \end{equation} $

因为$ z\in I_t\backslash I_{t+1} $

$ |D _n(z, 0)|\leq \sum\limits_{j=0}^\infty n_j D_{M_j}(z)\leq \sum\limits_{j=0}^t n_j D_{M_j}(z)\leq \sum\limits_{j=0}^t n_jM_j\leq M_{t+1}. $

因此, 当$ s\leq t $时, 得到

(3.9) $ \begin{equation} |K_{n^{(s)}, M_s}(z, 0)| =\bigg|\sum\limits_{k=n^{(s)}}^{n^{(s)}+M_s-1}D^\chi_k(z, 0)\bigg|\leq M_s M_{t+1}. \end{equation} $

由(3.2) 式得

$ K_{n^{(2t+1), M_{2t}}}(z, 0)=0 \ \mbox{ 其中 }\ t=s+1, \cdots, A-1. $

如果$ t<s\leq |n| $, $ z\in I_t\backslash I_{t+1} $和$ z-z_te_t\in I_s $, 也可得

(3.10) $ \begin{equation} \frac{1}{2}\leq \frac{|K_{n^{(s+1)}, M_s}(z, 0)|}{ M_sM_t }= \bigg|\frac{1}{1-r_t(z)}\bigg|= \bigg|\frac{1}{2\sin(\pi z_t/m_t)}\bigg|\leq\frac{m_t}{\pi} . \end{equation} $

由于$ nK_n=\sum\limits_{h=0}^{|n|}\sum\limits_{j=0}^{n_h-1}K_{n^{(h)+1}+ jM_h, M_h} $和$ (n^*_{A-1})_h=1 $假若$ 2|h $和$ h<2A $, 否则$ (n^*_{A-1})_h=0, $我们有

(3.11) $ \begin{eqnarray} n^*_{A-1}|K_{n^*_{A-1}}(z, 0)| &=& \bigg|\sum\limits_{h=0, 2|h}^{2A-2}K_{(n^*_{A-1})^{(h)+1}, M_h}(z, 0)\bigg|{} \\ &=&\bigg|\sum\limits_{t=0}^{s}K_{(n^*_{A-1})^{(2l+1)}, M_{2l}}(z, 0)\bigg| =\bigg|\sum\limits_{t=0}^{s}K_{(n^*_{A-1})^{(2l+2)}, M_{2l}}(z, 0)\bigg|{} \\ &\geq& |K_{(n^*_{A-1})^{(2s+2)}, M_{2s}}(z, 0)| -\sum\limits_{t=0}^{s-1}|K_{(n^*_{A-1})^{(2t+2)}, M_{2t}}(z, 0)|{} \\ &\geq& \frac{1}{2}M_{2s}M_{2k}-\frac{1}{\pi}\sum\limits_{t=0}^{s-1}M_{2t+1}M_{2k}. \end{eqnarray} $

很容易得到

(3.12) $ \begin{equation} \sum\limits_{t=0}^{s-1}M_{2t+1}\leq M_{2s-2}+M_{2s-1}=\frac{M_{2s}}{m_{2s-1} m_{2s-2}}+\frac{M_{2s}}{m_{2s-1}}\leq \frac{3}{4}M_{2s}. \end{equation} $

因此

(3.13) $ \begin{equation} n^*_{A-1}|K_{n^*_{A-1}}(z, 0)|\geq \Big(\frac{1}{2}-\frac{3}{4\pi}\Big)M_{2k}M_{2s}\geq \frac{M_{2k}M_{2s}}{4}. \end{equation} $

证毕.

引理3.3   设$ z\in I_{N}^{k, l}, k=0, \cdots, N-2, l=k+1, \cdots, N-1 $和$ n\geq M_N. $则有

(3.14) $ \begin{equation} \int_{I_N}|K_n(z-t, 0)|{\rm d}\mu(t)\leq \frac{cM_l M_k}{nM_N}. \end{equation} $

设$ z\in I_{N}^{k, N}, k=0, \cdots, N-1 $和$ n\geq M_N. $则有

(3.15) $ \begin{equation} \int_{I_N}|K_n(z-t, 0)|{\rm d}\mu(t)\leq \frac{c M_k}{M_N}, \end{equation} $

其中$ c $是一个常数.

证    设$ z\in I_N^{k, \alpha, l, \beta}. $应用引理3.1, 得到当$ A>l $时, $ K_{M_A}(z, 0)=0 $. 现令$ k<A\leq l. $得到

(3.16) $ \begin{equation} |K_{M_A}(z, 0)|= \bigg|\frac{M_k}{1-r_k}\bigg|= \bigg|\frac{M_k}{2\sin(\pi z_k/m_k)}\bigg|\leq \frac{m_kM_k}{\pi}. \end{equation} $

设$ z\in I_{N}^{k, l}, 0\leq k<l\leq N-1 $和$ t\in I_N $. 由$ z-t\in I_{N}^{k, l} $和$ n|K_n(z-t, 0)|\leq c\sum\limits_{A=0}^{l}M_A |K_{M_A}(z-t, 0)|, $得

(3.17) $ \begin{eqnarray} \int_{I_N}|K_n(z-t, 0)|{\rm d}\mu(t)&\leq& c\frac{1}{n}\sum\limits_{A=0}^{l}M_A \int_{I_N}|K_{M_A}(z-t, 0)|{\rm d}\mu(t){} \\ &\leq& c\frac{1}{nM_N}\sum\limits_{A=0}^{l}M_AM_k\leq c\frac{M_lM_k}{nM_N}. \end{eqnarray} $

设$ z\in I_{N}^{k, N} $, 则有

(3.18) $ \begin{eqnarray} \int_{I_N}|K_n(z-t, 0)|{\rm d}\mu(t)&\leq& c\frac{1}{n}\sum\limits_{A=0}^{|n|}M_A \int_{I_N}|K_{M_A}(z-t, 0)|{\rm d}\mu(t). \end{eqnarray} $

假设$ z=(0, \cdots, 0, z_k\neq 0, \cdots z_{N-1}=0, z_N, z_{N+1}, \cdots z_q\cdots z_{|n|-1}, \cdots) $和$ t=(0, \cdots, 0, $$ z_N, \cdots, $$ z_q-1, t_q\neq z_q, t_{q+1}, \cdots t_{|n|-1}, \cdots) $, $ q=N, \cdots |n|-1 $, 则有

(3.19) $ \begin{eqnarray} \int_{I_N}|K_n(z-t, 0)|{\rm d}\mu(t)&\leq& c\frac{1}{n}\sum\limits_{A=0}^{q-1}M_A \int_{I_N}|K_{M_A}(z-t, 0)|{\rm d}\mu(t){} \\ &\leq& c\frac{1}{nM_N}\sum\limits_{A=0}^{q-1}M_A M_k\leq c\frac{M_qM_k}{nM_N} {} \\ &\leq& c\frac{M_k}{M_N}. \end{eqnarray} $

如果假设$ z=(0, \cdots, 0, z_m\neq 0, \cdots z_{N-1}=0, z_N, z_{N+1}, \cdots z_q\cdots z_{|n|-1}, \cdots) $和$ t=(0, \cdots, 0, $$ z_N=0, \cdots, $$ z_{|n|-1}, \cdots) $, $ q=N, \cdots |n|-1 $, 也可得到

(3.20) $ \begin{eqnarray} \int_{I_N}|K_n(z-t, 0)|{\rm d}\mu(t)&\leq& c\frac{1}{n}\sum\limits_{A=0}^{|n|-1}M_A \int_{I_N}|K_{M_A}(z-t, 0)|{\rm d}\mu(t){} \\ &\leq& c\frac{1}{n}\sum\limits_{A=0}^{|n|-1}M_A \int_{I_N}M_k{\rm d}\mu(t){} \\ &\leq& c\frac{M_k}{M_N} . \end{eqnarray} $

结合(3.17), (3.19) 和(3.20) 式完成证明.

引理3.4^[14]   假设算子$ T $是线性的, 并且当$ 0<p\leq 1 $对任意$ p $原子$ a $满足

$ \int _{G_m\backslash I} |Ta|^p {\rm d}\mu\leq c_p<\infty, $

其中$ I $是原子$ a $的支撑. 如果$ T $从$ L_\infty $到$ L_\infty $有界, 则有$ \|Tf\|_{L_p(G_m)}\leq c_p \|f\|_{H_p(G_m)}. $

定理4.1   设$ 0< p <1/2 $, 则极大算子$ \tilde{\sigma}_p^* $是从鞅Hardy空间$ H_p $到$ L_p $空间有界的.

证    设$ a $是一个任意的$ p $原子, 支撑为$ I $, 且满足$ \mu(I) =M^{-1}_N $. 不妨设$ I=I_N $. 当$ n\leq M_N $时, 有$ \sigma_n(a) = 0 $. 因此, 可以假设$ n > M_N $.

由$ \|a\|_\infty\leq \mu(I)^{-\frac{1}{p}}=M_N^{\frac{1}{p}} $, 可得

(4.1) $ \begin{eqnarray} \frac{|\sigma_n(a)(z)|}{(n+1)^{1/p-2}}&\leq& \frac{1}{(n+1)^{1/p-2}}\bigg|\int_{G_m}a(t)K_n(z, t){\rm d}\mu(t)\bigg|{} \\ &=&\frac{1}{(n+1)^{1/p-2}}\bigg|\int_{I}a(t)K_n(z-t, 0){\rm d}\mu(t)\bigg|{} \\ &\leq& \frac{\|a\|_\infty }{(n+1)^{1/p-2}}\int_{I}|K_n(z-t, 0)|{\rm d}\mu(t){} \\ &\leq& \frac{M_N^{1/p} }{(n+1)^{1/p-2}}\int_{I}|K_n(z-t, 0)|{\rm d}\mu(t). \end{eqnarray} $

设$ z\in I_{N}^{k, l} $, $ 0\leq k<l\leq N $. 根据引理3.3和$ 1/p-2>0 $, 可得

(4.2) $ \begin{eqnarray} \frac{|\sigma_n(a)(z)|}{(n+1)^{1/p-2}} &\leq&c \frac{M_N^{1/p} }{M_N^{1/p-2}} \frac{M_l M_k}{nM_N} \leq c M_l M_k. \end{eqnarray} $

结合(2.1) 和(4.2)式可得

(4.3) $ \begin{eqnarray} \int_{G_m\backslash I} \Big(\frac{|\sigma_n(a)(z)|}{(n+1)^{1/p-2}}\Big)^p{\rm d}\mu(z) &=&\sum\limits_{k=0}^{N-2}\sum\limits_{l=k+1}^{ N-1}\sum\limits_{x_j=0, j\in \{l+1, \cdots, N-1\}}^{m_j-1} \int_{I_N^{k, l}} \Big(\frac{|\sigma_n(a)(z)|}{(n+1)^{1/p-2}}\Big)^p{\rm d}\mu(z){} \\ &&+\sum\limits_{k=0}^{N-1}\int_{I_N^{k, N}} \Big(\frac{|\sigma_n(a)(z)|}{(n+1)^{1/p-2}}\Big)^p{\rm d}\mu(z){} \\ &\leq & c\sum\limits_{k=0}^{N-2}\sum\limits_{l=k+1}^{ N-1}\frac{m_l\cdots m_{N-1}}{M_N}(M_lM_k)^p+c\sum\limits_{k=0}^{N-1}\frac{1}{M_N}(M_NM_k)^p{} \\ &\leq & c\sum\limits_{k=0}^{N-2}\sum\limits_{l=k+1}^{ N-1}\frac{(M_lM_k)^p}{M_l}+c\sum\limits_{k=0}^{N-1}\frac{1}{M_N}(M_NM_k)^p{} \\ &= & c\sum\limits_{k=0}^{N-2}\sum\limits_{l=k+1}^{ N-1}\frac{1}{M_l^{1-2p}}\frac{(M_lM_k)^p}{M^{2p}_l}+c\sum\limits_{k=0}^{N-1}\frac{1}{M_N^{1-2p}}\frac{(M_NM_k)^p}{M_N^{2p}}{} \\ &\leq & c\sum\limits_{k=0}^{N-2}\sum\limits_{l=k+1}^{ N-1}\frac{1}{2^{(1-2p)l}}+c\sum\limits_{k=0}^{N-1}\frac{1}{2^{N(1-2p)}}{} \\ &= & c\sum\limits_{k=0}^{N-2}\frac{1}{2^{(1-2p)k}}+c\frac{N}{2^{N(1-2p)}}\leq c. \end{eqnarray} $

证毕.

定理4.2   设$ 0<p<1/2 $, $ \varphi :{\Bbb N}\rightarrow [1, \infty) $为非减函数, 满足条件

$ \mathop{\overline{\lim}}\limits_{n\rightarrow \infty}\frac{(n+1)^{1/p-2}}{\varphi(n)}=+\infty, $

则$ \exists f\in H_p $使得

$ \sup\limits_{n\in {\Bbb N}} \bigg\|\frac{\sigma_n(f)}{\varphi(n)}\bigg\|_{L_{p, \infty}}=+\infty, $

即$ \sup\limits_{n\in {\Bbb N}}\frac{|\sigma_n(f)|}{\varphi(n)} $不是从鞅Hardy空间$ H_{p} $到$ L_{p, \infty} $有界的.

证    设$ \{\lambda_{k};k\in {\Bbb N}_{+}\} $是一个递增正整数序列, 满足

$ \lim\limits_{k\to\infty}\frac{\lambda^{1/{p-2}}_k}{\varphi(\lambda_{k})}=\infty. $

很明显对于每一个$ \lambda_{k} $, 存在一个正整数$ m' _{k} $, 使$ q_{m_{k}' }<\lambda_{k}< cq_{m_{k}' } $. 由于$ \varphi(n) $是不减函数, 可得

(4.4) $ \begin{eqnarray} \mathop{\overline{\lim}}\limits_{k\to\infty}\frac{M_{2m_{k}' }^{1/p-2}}{\varphi(q_{m_{k}' })}\geq c\mathop{\overline{\lim}}\limits_{k\to\infty}\frac{q_{m_{k}' }^{1/p-2}}{\varphi(q_{m_{k}' })} \geq c\lim\limits_{k\to\infty}\frac{\lambda_{k}^{1/p-2}}{\varphi(\lambda_{k})}=\infty. \end{eqnarray} $

选取$ \{n_{k};k\in {\Bbb N}_{+}\}\subset\{m' _{k};k\in {\Bbb N}_{+}\} $, 使

$ \lim\limits_{k\to\infty}\frac{M^{1/p-2}_{2n_k}}{\varphi(q_{n_{k}})}=\infty $

和

$ f_{n_{k}}(x)=D_{M_{2n_{k}+1}}(x, 0)-D_{M_{2n_{k}}}(x, 0), \qquad n_{k}\geq3. $

显然得到

(4.5) $ \begin{equation} \hat{f}_{n_{k}}(i)=\left\{ \begin{array}{ll}1, {\quad}&\mbox{若} \quad i = M_{2n_{k}}, \ldots, M_{2n_{k}+1}-1, \\ 0, &\mbox{其他}. \end{array}\right. \end{equation} $

进一步得到

(4.6) $ \begin{equation} S_{i}f_{n_{k}}(x)=\left\{ \begin{array}{lll}D_{i}(x, 0)-D_{M_{2n_{k}}}(x, 0), {\quad}&\mbox{若} \quad i= M_{2n_{k}}, \ldots, M_{2n_{k}+1}-1, \\ f_{n_{k}}(x), &\mbox{若} \quad i\geq M_{2n_{k}+1}, \\ 0, &\mbox{其他}. \end{array}\right. \end{equation} $

因此可得

(4.7) $ \begin{eqnarray} \|f_{n_{k}}\|_{H_{p}} &=&\parallel \sup\limits_{n\in {\Bbb N}}S_{M_{n}}f_{n_{k}}\parallel_{L_{p}}{} \\ &=&\parallel D_{M_{2n_{k}+1}}-D_{M_{2n_{k}}}\parallel_{L_{p}}{} \\ &=&\bigg(\int_{{I_{2n_{k}}}\setminus{I_{2n_{k}+1}}}M_{2n_{k}}^{p}{\rm d}\mu(x) +\int_{I_{2n_{k}+1}}(M_{2n_{k}+1}-M_{2n_{k}})^{p}{\rm d}\mu(x)\bigg)^{1/p}{} \\ &=&\bigg(\frac{m_{2n_{k}}-1}{M_{2n_{k}+1}}M^p_{2n_{k}}+\frac{(m_{2n_{k}}-1)^{p}}{M_{2n_{k}+1}}M^p_{2n_{k}}\bigg)^{1/p}{} \\ &\leq& M^{1-1/p}_{2n_{k}}. \end{eqnarray} $

由(4.6)式可得

(4.8) $ \begin{eqnarray} \frac{|\sigma_{q_{n_{k}}} f_{n_{k}}(x)|}{\varphi(q_{n_{k}})} &=&\frac{1}{\varphi(q_{n_{k}})q_{n_{k}}}\bigg|\sum\limits_{j= 0}^{q_{n_{k}}-1}S_{j}f_{n_{k}}(x)\bigg| =\frac{1}{\varphi(q_{n_{k}})q_{n_{k}}}\bigg|\sum\limits_{j= M_{2n_{k}}}^{q_{n_{k}}-1}S_{j}f_{n_{k}}(x)\bigg|{} \\ &=&\frac{1}{\varphi(q_{n_{k}})q_{n_{k}}}\bigg|\sum\limits_{j= M_{2n_{k}}}^{q_{n_{k}}-1}(D_{j}(x, 0)-D_{M_{2n_{k}}}(x, 0))\bigg|{} \\ &=&\frac{1}{\varphi(q_{n_{k}})q_{n_{k}}}\bigg|\sum\limits_{j =0}^{q_{n_{k}-1}-1}(D_{j+M_{2n_{k}}}(x, 0)-D_{M_{2n_{k}}}(x, 0))\bigg|. \end{eqnarray} $

由

$ D_{j+M_{2n_{k}}}(x, 0)-D_{M_{2n_k}}(x, 0)=\chi_{M_{2n_{k}}}(x)D_{j}(x, 0), \quad\quad j=1, 2, \ldots, M_{2n_{k}}-1, $

可得

(4.9) $ \begin{eqnarray} \frac{|\sigma_{q_{n_{k}}}f_{n_{k}}(x)|}{\varphi(q_{n_{k}})}= \frac{1}{\varphi(q_{n_{k}})q_{n_{k}}}\bigg|\sum\limits_{j=0}^{q_{n_{k}-1}-1}D_{j}(x, 0)\bigg| =\frac{1}{\varphi(q_{n_{k}})q_{n_{k}}}\frac{q_{n_{k}-1}}{q_{n_{k}}}|K_{q_{n_{k}}-1}(x, 0)|. \end{eqnarray} $

设$ x\in I^{2s, 2l}_{2n_{k}} $. 利用引理3.2可得

(4.10) $ \begin{equation} \frac{|\sigma_{q_{n_{k}}}f_{n_{k}}(x)|}{\varphi(q_{n_{k}})}\geq\frac{cM_{2s}M_{2l}}{M_{2n_{k}}\varphi(q_{n_{k}})}. \end{equation} $

从而

(4.11) $ \begin{eqnarray} &&\mu \bigg\{x\in G_{m}:\frac{\mid\sigma_{q_{n_{k}}}f_{n_{k}}(x)\mid}{\varphi(q_{n_{k}})} \geq\frac{c}{M_{2n_{k}}\varphi(q_{n_{k}})}\bigg\}{} \\ &\geq&\mu \bigg\{x\in I_{2n_{k}}^{2, 4}:\frac{\mid\sigma_{q_{n_{k}}}f_{n_{k}}(x)\mid }{\varphi(q_{n_{k}})}\geq\frac{c}{M_{2n_{k}}\varphi(q_{n_{k}})}\bigg\}\geq\mu(I_{2n_{k}}^{2, 4})>c>0. \end{eqnarray} $

因此可得

$ \begin{eqnarray*} &\nonumber&\frac{{\frac{c}{M_{2n_{k}}\varphi(q_{n_{k}})}}(\mu\{x\in G_{m}:\frac{\mid\sigma_{q_{n_{k}}}f_{n_{k}}(x)\mid}{\varphi(q_{n_{k}})}\geq\frac{c}{M_{2n_{k}}\varphi(q_{n_{k}})}\})^{1/p}}{\parallel f_{n_{k}}(x)\parallel_{H_{p}}}\\ &\geq&\frac{c}{M_{2n_{k}}\varphi(q_{n_{k}})M_{2n_{k}}^{1-1/p}}= c\frac{M_{2n_{k}}^{1/p-2}}{\varphi(q_{n_{k}})}\rightarrow \infty, \quad \mbox{当}\quad k\rightarrow \infty. \end{eqnarray*} $

证毕.