我们的第一个主要结果如下。
定理1
让\(0<p<1/2)和\(f\在H_{p}(G_{m})中\).那么存在一个绝对常数\(c{p}\),仅取决于第页,这样的不平等
$$\sum^{\infty}_{n=1}\frac{\log^{p}n\VertR^{\psi}_{n} (f)\Vert_{H_{p}(G_{m})}^{p}}{n ^{2-2p}}\leq c{p}\Vert f\Vert_{H_{p}(G_{m})}^{p}$$
(3.1)
持有,哪里\(磅/平方英寸)_{n} (f)\)表示关于Vilenkin–Fourier级数的第n个Riesz对数平均值(f).
为了证明定理1我们将使用以下引理。
引理1
(请参见[38])
让\(x\在I_{N}(x_{k} e(电子)_{k} +x_{l} e(电子)_{l} )\),\(1\leqx{k}\leqm_{k} -1个\),\(1\leqx{l}\leqm_{l} -1个\),\(k=0,\ldots,N-2\),\(l=k+1,\ldots,N-1).然后
$$\int_{I_{N}}\bigl\vert K^{psi}_{N}(x-t)\bigr\vert\,d\mu(t)\leq\frac{cM_{l} M(M)_{k} }{nM_{N}},当}N \geq M_{N}时,\quad\textit{$$
让\(x\在I_{N}(x_{k} e(电子)_{k} )\),\(1\leqx{k}\leqm_{k} -1个\),\(k=0,\ldots,N-1).然后
$$\int_{I_{N}}\bigl\vert K^{psi}_{N}(x-t)\bigr\vert\,d\mu(t)\leq\frac{cM_{K}}{M_{N{}},\quad\textit{when}N\geqM_{N}$$
引理2
(请参见[39])
让\(x\在I_{N}(x_{k} e(电子)_{k} +x_{l} e(电子)_{l} )\),\(1\leqx{k}\leqm_{k} -1个\),\(1\leqx{l}\leqm_{l} -1个\),\(k=0,\ldots,N-2\),\(l=k+1,\ldots,N-1).然后
$$\int_{I{N}}\sum_{j=M_{N}+1}^{N}\frac{vertK^{psi}_{j}(x-t)\vert}{j+1}\,d\mu(t)\leq\frac}cM_{k} M(M)_{l} {M_{N}^{2}}$$
让\(x\在I_{N}(x_{k} e(电子)_{k} )\),\(1\leqx{k}\leqm_{k} -1个\),\(k=0,\ldots,N-1).然后
$$\int_{I{N}}\sum_{j=M_{N}+1}^{N}\frac{vertK^{psi}_{j}(x-t)\vert}{j+1}\,d\mu(t)\leq\frac}{cM_{K}}{M_{N}}l_{N}$$
证明
通过使用Abel变换,Riesz对数均值的核可以重写为(另请参见[39])
$$L^{\psi}_{n}=\frac{1}{L{n}}\sum^{n-1}_{j=1}\压裂{K^{psi}_{j}}{j+1}+\压裂{K^{psis}_{n}}{l_{n{}}$$
(3.2)
因此,根据(2.2)我们得到了
$$\sup_{n\in\mathbb{n}}\int_{G_{m}}\bigl\vert L^{alpha}_{n}\bigr\vert\,d\mu\leq c<\infty,\quad\text{其中}\alpha=w\text{或}\psi$$
因此\(R^{psi}_{n}\)以为界\(L_{\infty}\)到\(L_{\infty}\).按命题2定理的证明1将完成,如果我们显示
$$\sum^{\infty}_{n=1}\frac{\log^{p} n个\int_{\bar{I}}\vert R^{\psi}_{n} 一个\vert^{p}\,d\mu}{n^{2-2p}}\leqc{p}<\infty,\quad\text{for}0<p<1/2$$
(3.3)
对于每个第页-原子一,其中我表示原子的支持。
让一是具有支撑的任意p原子我和\(\mu(I)=M_{N}^{-1}\)。我们可以假设\(I=I_{N}\)很容易看出\(R^{\psi}_{n}a=\sigma^{\psi}_{n}(a)=0\),何时\(n\leq M_{n})因此,我们认为\(n>M_{n}\).
自\(\垂直a\垂直{\infty}\leq cM_{N}^{2}\)如果我们申请(3.2),那么我们可以得出结论
$$\开始{对齐}&\bigl\vert R^{\psi}_{n} 一个(x)\bigr\vert\\&\quad=\int_{I_{N}}\bigl\vert a(t)\biger\vert\bigl\vert L^{psi}_{N}(x-t)\bigr\vert,d\mu(t q\frac{cM_{N}^{1/p}}{L_{N}}\int_{I{N}}\sum_{j=M_{N}+1}^{N-1}\frac}\vertK^{psi}_{j}(x-t)\vert}{j+1}\,d\mu(t)\\&\qquad{}+\frac{cM_{N}^{1/p}}{l_{N}}\int_{I{N}}\bigl\vert K^{psi}{N}(x-t)\bigr\vert,d\ mu(t)。\结束{对齐}$$
(3.4)
让\(x\在I_{N}(x_{k} e(电子)_{k} +x_{l} e(电子)_{l} )\),\(1\leqx{k}\leqm_{k} -1个\),\(1\leqx{l}\leqm_{l} -1个\),\(k=0,\ldots,N-2\),\(l=k+1,\ldots,N-1).来自引理1和2因此
$$\bigl\vert R^{\psi}_{n} 一个(x)\bigr\vert\leq\frac{cM_{l} M(M)_{k} M(M)_{N} ^{1/p-2}}{\log(N+1)}$$
(3.5)
让\(x\在I_{N}(x_{k} e(电子)_{k} )\),\(1\leqx{k}\leqm_{k} -1个\),\(k=0,\ldots,N-1).应用引理1和2我们可以得出这样的结论
$$\bigl\vert R^{\psi}_{n} 一个(x)\bigr\vert\leq{M_{N}^{1/p-1}M_{k}}$$
(3.6)
通过组合(2.1)和(3.4)–(3.6)我们获得
$$\开始{对齐}&\int_{上划线{I_{N}}}\bigl\vert R^{\psi}_{n} 一个(x)\bigr\vert^{p}\,d\mu(x)\\&\quad=\sum^{N-2}_{k=0}\sum^{N-1}_{l=k+1}\sum_{x_{j}=0,j\in\{l+1,\ldots,N-1}^{m_{j-1}}\int_{I_{N}^{k,l}}\bigl\vertR^{psi}_{n} 一个\bigr\vert^{p}\,d\mu+\sum^{N-1}_{k=0}\int_{I_{N}^{k,N}}\bigl\vertR^{psi}_{n} 一个\bigr\vert^{p}\,d\mu\\&\quad\leq c\sum^{N-2}_{k=0}\sum^{N-1}_{l=k+1}\分形{m_{l+1}\点m_{N-1}}{m_{N}}\分形_{l} M(M)_{k} )^{p} M(M)_{N} ^{1-2p}}{{\log}^{p}(N+1)}+\sum^{N-1}_{k=0}\压裂{1}{M_{N}}M_{k}^{p} M(M)_{N} ^{1-p}\\&\quad\leq\frac{cM_{N}^{1-2p}}{{log^{p}(N+1)}}\sum^{N-2}_{k=0}\sum^{N-1}_{l=k+1}\压裂{(M_{l} M(M)_{k} )^{p}}{M_{l}}+\sum^{N-1}_{k=0}\压裂{M_{k}^{p}}{M_}N}^{p2}}\\&\quad\leq\frac{cM_{N}^{1-2p}}}{log^{p{(N+1)}+c_{p}。\结束{对齐}$$
(3.7)
很容易看出这一点
$$\sum^{\infty}_{n=M_{n}+1}\frac{1}{n^{2-2p}}\leq\frac}c}{M_{n}^{1-2p}},\quad\text{for}0<p<1/2$$
(3.8)
通过组合(3.7)和(3.8)我们得到了
$$开始{aligned}&\sum^{infty}_{n=M_{n}+1}\frac{log^{p}n\int_{overline{I{n}}\vert R_{n} 一个\vert^{p}\,d\mu}{n^{2-2p}}\\&\quad\leq\sum^{infty}_{n=M_{n}+1}\biggl(\frac{c_{p} M(M)_{N} ^{1-2p}}{N^{2-p}}+\压裂{c{p}}}{N ^{2-p}}\biggr)+c{p{}\\&\四\leq c_{p} M(M)_{N} ^{1-2p}\sum^{\infty}_{N=M_{N}+1}\frac{1}{N^{2-2p}}+\sum^{\infoty}_{N=M_{N}+1}\ frac{1\N^{2-p}}+c{p}\leq c_{p}<\infty。\结束{对齐}$$
这意味着(3.3)成立,证明完整。 □
我们的下一个主要结果特别表明了定理中的不等式1至少在沃尔什-傅里叶级数(参见问题2(在下一节中)。
定理2
让\(0<p<1/2)和\(\varPhi:\mathbb{N}\rightarrow{}[1,\infty)\)是任何非递减函数,满足条件
$$\lim_{n\rightarrow\infty}\varPhi(n)=+\infty$$
(3.9)
那么存在一个鞅\(f\在H_{p}(G_{2})中\)这样的话
$$\sum_{n=1}^{\infty}\frac{\log^{p}n\Vert R^{w}_{n} (f)\Vert_{p}^{p}\varPhi(n)}{n^{2-2p}}=\infty$$
(3.10)
哪里\(右^{w}_{n} (f)\)表示关于f的Walsh–Fourier级数的第n个Riesz对数平均值.
证明
很明显,如果我们假设\(\varPhi(n)\geq cn\),其中c(c)那么是某个正常数
$$\frac{\log^{p}n\varPhi(n)}{n^{2-2p}}\geqn^{1-2p}\log^}n\rightarrow\infty,\quad\text{as}n\rightarrow\infty$$
还有(3.10)持有。因此,在不损失一般性的情况下,我们可以假设存在正整数的递增序列\({\alpha^{\prime}_{k}:k\in\mathbb{N}\}\)这样的话
$$\varPhi\bigl(\alpha^{\prime}_{k}\bigr)=o\bigl(\alpha^{\prime}_{k}\bigr),\quad\text{as}k\rightarrow\infty$$
(3.11)
让\({\alpha_{k}:k\in\mathbb{N}\}\subseteq\{\alfa^{prime}_{k{:k\in \mathbb{N}\}\)是正整数的递增序列,这样\(\alpha_{0}\geq2\)和
$$\开始{对齐}&\sum_{k=0}^{\infty}\frac{1}{\varPhi^{1/2}(2^{2\alpha_{k})}<\infty,\end{aligned}$$
(3.12)
$$开始{对齐}和\sum_{\ta=0}^{k-1}\frac{2^{2\alpha_{eta}/p}}{\varPhi^{1/2p}(2^{2\alpha_}\eta}})}\leq\frac}2^{1\alpha_1}/p+1}{\valPhi^}{1/2p/}$$
(3.13)
$$开始{aligned}&\frac{2^{2\alpha_{k-1}/p+1}{\varPhi^{1/2p}(2^{2\alpha_}k-1})}\leq\frac}1}{128\alpha_{k}}\ frac{2\alpha_{k}。\结束{对齐}$$
(3.14)
我们注意到,在条件(3.11)我们可以得出这样的结论
$$\frac{2^{2\alpha_{\eta}/p}}{\varPhi^{1/2p}(2^{2\alpha_}\ta}})}\geq{\biggl(\frac}2^{1\alpha_2\eta{}}{\ varPhi(2^}2\alfa_{\ta}{})第y天$$
紧接着就是这样一个递增的序列\({\alpha_{k}:k\in\mathbb{N}\}\),满足条件(3.12)–(3.14),可以被构造。
让
$$f^{(A)}(x):=\sum_{{k;2\alpha_{k}<A\}}\lambda_{k} 一个_{k} $$
哪里
$$\lambda{k}=\frac{1}{\varPhi^{1/2p}(2^{2\alpha_{k})}$$
和
$$a{k}={2^{2\alpha{k}(1/p-1)}}$$
发件人(3.12)和引理1我们可以得出这样的结论\(f=(f^{(n)},n在mathbb{n}中)在H_{p}(G_{2})中.
很容易证明
$$\widehat{f}^{w}(j)=\textstyle\begin{cases}\frac{2^{2\alpha_{k}(1/p-1)}}{\varPhi^{1/2p}(2^{2\alpha_{k})},&&text{if}j\in\{2^{2\alpha_{k}},\ldots,2^{2\alpha_{k}+1}-1\},k\in\mathbb{N},\\0,&&text{if}j\notin\bigcup_{k=1}^{infty}\{2^{2\alpha_{k}}},ldots,2^{2\alpha_{k}+1}-1\}。\结束{cases}$$
(3.15)
对于\(n=\sum_{i=1}^{s}2^{n_{i}}\),\(n{1}<n{2}<\cdots<n{s}\)我们表示
$$\mathbb美元{答}_{0,2}:=\Biggl\{n\in\mathbb{n}:n=2^{0}+2^{2}+\sum_{i=3}^{s_{n}}2^{n_{i}}\Biggr\}$$
让\(2^{2\α{k}}\leqj\leq2^{2\alpha{k}+1}-1)和\(j\in\mathbb{答}_{0,2}\).然后
$$\开始{对齐}R^{w}_{j} (f)=&\压裂{1}{l_{j}}\sum_{n=1}^{2^{2\alpha_{k}}-1}\frac{S_{n} (f)}{n} +\压裂{1}{l_{j}}\sum_{n=2^{2\alpha_{k}}}^{j}\压裂{S_{n} (f)}{n} :=I+\mathit{II}。\结束{对齐}$$
(3.16)
让\(n<2^{2\alpha_{k}})。然后从(3.13), (3.14)和(3.15)我们有
$$\开始{aligned}\bigl\vert S^{w}_{n} (f)(x)\bigr\vert\leq&\sum_{\eta=0}^{k-1}\sum_{v=2^{2\alpha_{\eta}}}^{2^{2\\alpha_}\eta}+1}-1}\bigl\vert\widehat{f}^{w}2\α{\eta}+1}-1}\frac{2^{2\β{(1/p-1)}}{\varPhi^{1/2p}(2^{2\alpha{\eta})}\\leq&\sum{\eta=0}^{k-1}\压裂{2^{2\alpha_{\eta}/p}}{\varPhi^{1/2p}(1/p-2)}}{\varPhi^{1/2p}(2^{2\alpha_{k})}。\结束{对齐}$$
因此,
$$开始{对齐}\vert I\vert\leq&\frac{1}{l_{j}}\sum_{n=1}^{2^{2\alpha_{k}}-1}\frac}\vertS^{w}_{n} f(x)\vert}{n}\\leq&\frac{1}{l{2^{2\alpha{k}}}\frac}1}{128\alpha_k}}\frac{2^2\alfa_k}(1/p-2)}}{varPhi^{1/2p}c{1}{n}\leq\frac{1}}{128\alpha_{k}}\frac{2^{2\alpha_}k}(1/p-2)}}{varPhi^{1/2p}(2^{2\alpha{k})}。\结束{对齐}$$
(3.17)
让\(2^{2\α_{k}}\leqn\leq2^{2\alpha_{k{+1}-1)。然后我们有以下内容:
$$\开始{对齐}S^{w}_{n} (f)=&\sum_{\eta=0}^{k-1}\sum_{v=2^{2\alpha_{\eta}}^{2^{2\\alpha_}+1}-1}\widehat{f}^{w}(v)w{v}+\sum_}v=2^}2\alfa_{k}}^}{n-1}\wide hat{f}^{w}(v)w_{v}0}^{k-1}\frac{2^{{2\alpha_{eta}}(1/p-1)}}{\varPhi^{1/2p}(2^{2\ alpha_}})}\bigl(D^{w}_{2^{2\alpha_{\eta}+1}}-D^{w}_{2^{2\alpha_{\eta}}}\bigr)+\frac{2^}{2\alpha_{k}}(1/p-1)}}{\varPhi^{1/2p}(2^{2\alpha_}})}\bigl(D^{w}_{n} -D^{w}_{2^{2\alpha_{k}}\biger)。\结束{对齐}$$
这给了
$$开始{对齐}\mathit{II}&=\frac{1}{l_{j}}\sum_{n=2^{2\alpha_{k}}}^{2^{2\\alpha_}k}+1}}\frac}{1}}{n}\Biggl(\sum_{eta=0}^{k-1}\frac:2^{2\alpha_{eta}(1/p-1)}{varPhi^{1/2p}(2^{2\alpha_{\eta}})}\bigl(D^{w}_{2^{2\alpha_{\eta}+1}}-D^{w}_{2^{2\alpha_{\eta}}}}\bigr)\Biggr)\\&&\quad{}+\frac{1}{l_{j}}\frac{2^{2\alpha_{k}(1/p-1)}}{\varPhi ^{1/2p}(2^{2\alpha_{k}})}\sum_{n=2^{2\alpha_{k}}}}^{j}\frac{(D^{w}_{n} -D^{w}_{2^{2\alpha_{k}})}{n}\\&:=\mathit{二}_{1} +\mathit{二}_{2}. \结束{对齐}$$
(3.18)
让\(x\在I{2}(e_{0}+e_{1})\在I_{0}\反斜杠I{1}\)我们对Dirichlet核使用了众所周知的等式(有关详细信息,请参见例如[17]和[29]):回想一下
$$D美元^{w}_{2^{n}}(x)=\textstyle\begin{cases}2^{n},&\text{if}x\在I_{n}中,\\0,&\text{if}x\ notin I_{n},\end{cases{$$
(3.19)
和
$$D美元^{w}_{n} =w_{n}\sum^{infty}_{k=0}n_{k} 第页_{k} D类^{w}_{2^{k}}=w_{n}\sum^{infty}_{k=0}n_{k}\bigl(D^{w}_{2^{k+1}}-D^{w}_{2^{k}}\biger),\quad\text{表示}n=\sum^{\infty}_{i=0}n_{i} 2个^{i} $$
(3.20)
所以我们可以得出结论
$$D美元^{w}_{n} (x)=\textstyle\begin{cases}w_{n},&&\text{if}n\text{是奇数,}\\0,&&\text{if}n\text{是偶数。}\end{cases}$$
自\(\alpha_{0}\geq2\),\(k\in\mathbb{N}\)我们获得\(2α{k}geq 4),对于所有人\(k\in\mathbb{N}\)如果我们申请(3.19)我们得到了
$$\mathit美元{二}_{1}=0 $$
(3.21)
和
$$\begin{aligned}&&mathit开始{二}_{2} =\压裂{1}{l_{j}}\压裂{2^{2\alpha_{k}(1/p-1)}}{\varPhi^{1/2p}(2^{2\alpha_}})}\总和{n=2^{2\\alpha_{k} -1个}}^{(j-1)/2}\压裂{w{2n+1}}{2n+1}=\压裂{1}{l{j}}\压裂}2^{2\alpha{k}(1/p-1)}r{1}}{varPhi^{1/2p}(2^{2\alpha_k}})}\sum{n=2^{2\_{k} -1个}}^{(j-1)/2}\frac{w_{2n}}{2n+1}。\结束{对齐}$$
让\(x\在I_{2}(e_{0}+e_{1})中\)然后,根据沃尔什函数的定义,我们得到
$$w_{4n+2}=r_{1} 周_{4n}=-w{4n{$$
和
$$\开始{aligned}\vert\mathit{二}_{2} \vert=&\frac{1}{l_{j}}\frac}2^{2\alpha_{k}(1/p-1)}}{\varPhi^{1/2p}(2^{2\alpha_}})}\Biggl\vert\sum{n=2^{2\\alpha_{k} -1个}}^{(j-1)/2}\frac{w_2n}}{2n+1}\Biggr\vert\\=&\frac}1}{l_{j}}\frac{2^{2\alpha_{k}(1/p-1)}}{varPhi^{1/2p}_{k} -2个}+1} ^{(j-1)/4}\biggl(\frac{w_{4n-4}}{4n-3}+\frac}w_{4-2}}{4-1}\biggr)\biggr\vert\\=&\frac[1}{l_{j}}\frac[2\alpha_{k}(1/p-1)}}{\varPhi^{1/2p}(2^{2\alpha_k}})}\biggl\vert\vert裂缝{w{j-1}}{j}+sum{n=2^{2\alpha_{k} -2个}+1} ^{(j-1)/4}\biggl(压裂{w_{4n-4}}{4n-3}-\压裂{w_{4n-2}}{4n-1}\biggr)\biggr\vert\\\geq&\frac{c}{log({2^{2\alpha_{k}+1}})}\frac}2^{2\alpha_}k}(1/p-1)}}{varPhi^{1/2p}}\biggr\vert-\sum_{n=2^{2\alpha_{k} -2个}+1} ^{(j-1)/4}\vert{w_{4n-4}}\vert\biggl(\frac{1}{4n-3}-\frac{1}{4n-1}\biggr)\biggr)\\geq&\frac{1}{4\alpha_{k}}\frac{2^{2\alpha_{k}(1/p-1)}}{\varPhi^{1/2p}(2^{2\alpha_{k})}\Biggl(\frac{1}{j}-\和{n=2^{2\alpha_{k} -2个}+1} ^{(j-1)/4}\biggl(\frac{1}{4n-3}-\frac{1}{4n-1}\biggr)\biggr)。\结束{对齐}$$
(3.22)
通过简单的计算,我们可以得出以下结论
$$\开始{aligned}&\sum_{n=2^{2\alpha_{k} -2个}+1} ^{(j-1)/4}\biggl(\frac{1}{4n-3}-\frac{1}{4n-1}\biggr)\\&\quad=\sum_{n=2^{2\alpha_{k} -2个}+1} ^{(j-1)/4}\压裂{2}{(4n-3)(4n-1)}\\&\quad\leq\sum_{n=2^{2\alpha_{k} -2个}+1} ^{(j-1)/4}\压裂{2}{(4n-4)(4n-2)}=\压裂{1}{2}\sum_{n=2^{2\alpha_{k} -2个}+1} ^{(j-1)/4}\frac{1}{(2n-2)(2n-1)}\\&&quad\leq\frac{1}{2}\sum_{n=2^{2\alpha_{k} -2个}+1} ^{(j-1)/4}分形{1}{(2n-2)(2n-2)}=frac{1}}{8}\sum_{n=2^{2\alpha_{k} -2个}+1} ^{(j-1)/4}分形{1}{(n-1)(n-1_{k} -2个}+1} ^{(j-1)/4}压裂{1}{(n-1)(n-2)}=\压裂{1{8}\sum_{l=2^{2\alpha_{k} -2个}+1} ^{(j-1)/4}\biggl(\frac{1}{n-2}-\压裂{1}{n-1}\biggr)\\&\quad\leq\frac{1}}{8}\bigl(压裂{1{2^{2\alpha_{k} -2个}-1} -\压裂{4}{j-5}\biggr)\leq\frac{1}{8}\bigl(\压裂{1}{2^{2\alpha_{k} -2个}-1} -\压裂{4}{j}\biggr)。\结束{对齐}$$
自\(2^{2\α{k}}\leqj\leq2^{2\alpha{k}+1}-1),其中\(α{k}\geq 2),我们获得
$$\压裂{2}{2^{2\alpha_{k}}-4}\leq\压裂{2]{2^{4}-4}=\压裂{1}{6}$$
和
$$\开始{aligned}\vert\mathit{二}_{2} \vert\geq&\frac{1}{4\alpha_{k}}\ frac{2^{2\alpha_}k}(1/p-1)}}{\varPhi^{1/2p}(2^{2\alpha_c}})}\biggl(\frac}1}{j}-\frac{1}{8}\biggl(\frac{1'{2^{2\alpha_{k} -2个}-1} -\压裂{4}{j}\biggr)\大gr)\\geq&\压裂{1}{4\alpha_{k}}\压裂{2^{2\alpha_}k}(1/p-1)}{\varPhi^{1/2p}(2^{2\alpha_{k})}\大gl(\frac{3}{2j}-\压裂{1}{2^{2\alpha_{k}+1}-8}\biggr)\\geq&\压裂{1}{4\alpha_}k}\frac{2^}2\alfa_{k{(1/p-1)}}{\varPhi^{1/2p}-\压裂{1}{2}\压裂{1}{2^{2\α{k}}-4}\biggr)\\geq&\压裂{1'{4\α{k{}\压裂}2^{1\α_k}(1/p-1)}}{varPhi^{1/2p}(2^{2\α_k{})}\bigl(\frac{1}{4} \压裂{1}{2^{2\alpha_{k}}}+\压裂{1\{2}\frac{1}}{2\压裂{k}{}-\压裂{1}{2{2\alpha_k}}-4}\biggr)varPhi^{1/2p}}\压裂{2^{2\α_{k}(1/p-1)}}{\varPhi^{1/2p}(2^{2\alpha_{k{})}\biggl(\压裂{1}{4}\frac{1}}{2^}2\ alpha_}k}}}-\压裂{1}{6}\frac{1}{2^α_}k{}}}}\bigr)\\geq&\压裂{1'{48α{k}}\压裂{2^{2\alpha{k}(1/p-2)}}{varPhi^{1/2p}(2^{2\alpha_k}})}\geq\frac{1}{64\alpha_ k}}\frac}2^{1\alpha_ k}(1/2)}}}{(2^{2\alpha{k})}。\结束{对齐}$$
(3.23)
通过组合(3.14), (3.16)–(3.23)的\(\在I_{2}(e_{0}+e_{1})中\)和\(0<p<1/2)我们发现了
$$\开始{aligned}\bigl\vert R^{w}_{j} f(x)\bigr\vert\geq&&\ vert\mathit{二}_{2} \vert-\vert\mathit{二}_{1} \vert-\vert I\vert\\geq&\frac{1}{64\alpha_{k}}\frac{2^{2\alpha_}k}(1/p-2)}}{varPhi^{1/2p}\alpha_{k})}=\frac{1}{128\alpha_a{k}}\frac}2^{2\alpha_k}(1/p-2)}}{\varPhi^{1/2p}(2^{2\alpha_c}}){。\结束{对齐}$$
因此,
$$\开始{aligned}&\bigl\Vert R^{w}_{j} (f)\bigr\Vert_{\text{弱-}L_{p} (G{2})}^{p}\\&\quad\geq\frac{1}{128\alpha_{k}^{p{}\\frac{2^{2\alpha_}k}(1-2p)}}{varPhi^{1/2}(2^{2\alpha_c})^{w}_{j} (f)\bigr\vert\geq\frac{1}{128\alpha_{k}}\frac}2^{2\alpha_}k}(1/p-2)}}{\varPhi^{1/2p}{\varPhi^{1/2}(2^{2\alpha_{k}})}\mu\biggl\{x\在I_{2}中(e_{0}+e_{1}):\bigl\vert R^{w}_{j} (f)\bigr\vert\geq\frac{1}{128\alpha_{k}}\frac{2^{2\alpha_}k}(1/p-2)}}{\varPhi^{1/2p}1/2}(2^{2\alpha_{k})}\bigl}}{\varPhi^{1/2}(2^{2\alpha_{k})}。\结束{对齐}$$
(3.24)
此外,
$$\开始{对齐}&\sum_{j=1}^{\infty}\frac{\VertR^{w}_{j} (f)\垂直_{\text{弱-}L_{p} (G{2})}^{p}\log^{p{{{(j)}\varPhi(j){j^{2-2p}}\\&\quad\geq\sum_{{j\in\mathbb{答}_{0,2}:2^{2\alpha_{k}}_{j} (f)\垂直_{\text{弱-}L_{p} }^{p}\log^{p{{p}{(j)}\varPhi(j){j^{2-2p}}\\&\quad\geq\frac{c}{\alpha_{k}^{p}}\frac}{2^{2\alpha_}k}(1-2p)}}{\varPhi^{p/2}(2^{2\alpha_{k})}}\sum_{j\in\mathbb{答}_{0,2}:2^{2\alpha_{k}}<j\leq2^{2\alpha_}k}+1}-1}}\frac{\log^{p}{(j)}\varPhi(j){j^{2-2p}}\\&\quad\geq\frac}c\varPhi(2^{1\alpha_c}})^{p}}\frac{2^{2\alpha_{k}(1-2p)}}{varPhi^{1/2}(2^{2\alpha_}})}\sum_{j\in\mathbb{答}_{0,2}:2^{2\alpha_{k}}<j\leq2^{2\alpha_}k}+1}-1}}\frac{1}{j^{2-2p}}\\&\quad\geq\varPhi^{1/2}\bigl(2^{1\alpha_c}}\bigr)\rightarrow\infty,\quad\\text{as}k\rightarrow\infty。\结束{对齐}$$
证据是完整的。 □