$\gdef\leq{\leqslant}$ $\gdef\geq{\geqslant}$ $\gdef\lrb#1{\lbrace#1\rbrace}$ $\gdef\norm#1{\Vert#1\Vert}$ $\gdef\F{\mathcal{F}}$ $\gdef\T{\mathcal{T}}$ $\gdef\intrn{\int_{\R^n}}$ $\gdef\intr{\int_{\R}}$ $\gdef\ip{2\pi i}$ $\gdef\qed{~\tag*{\Large□}}$ $\gdef\C{C^\infty}$ $\gdef\bc{\mathbb{C}}$ $\gdef\S{\mathscr{S}}$ $\gdef\D{\mathscr{D}}$ $\gdef\H{\mathscr{H}}$ $\gdef\act#1{\lt#1\gt}$
$Q(x,r):$以$x$为心，$r$为边长的方体. $B(x,r)\subset Q(x,r)\subset B(x,\sqrt{n}r)$
Hardy-Littlewood极大算子$\bar Mf(x)=\sup_{r>0}\frac{1}{Q(x,y)}\int_{Q(x,r)}|f(y)|dy,\forall f\in L^1_{loc}(\R^n)$
断言$\exist c>0,c=c(n)~s.t.~c^{-1}\bar Mf(x)\leq Mf(x)\leq c\bar Mf(x),\forall x\in\R^n,\forall f\in L^1_{loc}(\R^n)$
$\R^n$中的二进方体(dyadic cube)$$
[2^k_{m_1},2^k_{(m_1+1)}]\times[2^k_{m_2},2^k_{(m_2+1)}]\times\cdots\times[2^k_{m_n},2^k_{(m_n+1)}]
$$ 其中$k,m_1,...,m_n\in\Z$
引理(C-Z分解定理) 设$f\in L^1(\R^n)$，给定$\alpha>0$，则存在$\R^n$的分解：$R^n=\Omega\cup F$且$\Omega\cap F=\empty$，还满足
(1) $\Omega=\cup_{k\in\Z}Q_k,\mathring{Q_j}\cup\mathring{Q_k}=\empty~(\forall j\neq k)$，其中$Q_k(\forall k\in\Z)$是二进方体.
(2)$\alpha<\frac{1}{|Q_k|}\int_{Q_k}|f(y)|dy\leq2^n\alpha,(\forall k\in\Z)$
(3) $|f(x)|\leq\alpha~a.e.~x\in F$
(4) $|\Omega|=\sum_{k\in\Z}|Q_k|\leq\frac{1}{\alpha}\norm{f}_{L^1}$
(5) $f=g+b$满足$\norm{g}_{L^\infty}\leq2^n\alpha,\norm{g}_{L^1}\leq\norm{f}_{L^1}\quad(5.1)$
$b(x)=\sum_{j\in\Z}b_j(x)\quad(5.2)$且每个$b_j$支撑在$Q_j$上
$\int_{Q_j}b_j(x)dx=0\quad(5.3)$
$\norm{b_j}_{L^1(\R^n)}=\norm{b_j}_{L^1(Q_j)}\leq2^{n+1}\alpha|Q_j|\quad(5.4)$
证明：$1^\circ$ 设$f\in L^1(\R^n),\alpha>0$给定，$\exist r>0$(充分大) $s.t.~\frac1{|Q|}\int_Q|f(y)|dy<\alpha\quad(3.1),Q=Q(x,r)$的任何方体
事实上，$(3.1)\lrArr|Q|>\frac{1}{\alpha}\int_Q|f(y)|dy\quad(3.1)'$，只要找到$r>0~s.t.~|Q|>\frac{1}{\alpha}\norm{f}_{L^1}\quad(3.2)$，则$(3.2)\rArr(3.1)'\rArr(3.1)$
$|Q|=(2r)^n=2^nr^n$
分割$\R^n$为可数个二进方体的并，取$k_0=1+[log_2r]$，都满足$|Q|\geq\frac{1}{\alpha}\norm{f}_{L^1}$，这些是第0代的方体
$2^\circ$ 分解第0代的每个方体为$2^n$等分(通过边长)，得到一个新的二进方体的网，称为第一代
设$\H^{(1)}=\lbrace Q|Q$为第一代$\rbrace$，从$\H^{(1)}$中选一个方体$Q$，若$\frac{1}{|Q|}\int_Q|f(y)|dx>\alpha\quad(3.3)$
设$\S^{(1)}=\lbrace Q\in\H^{(1)}|$且$Q$满足$(3.3)\rbrace,\S^{(1)}\subset\H^{(1)}$
$\forall Q\in\H^{(1)}\backslash\S^{(1)}$，有$\frac{1}{|Q|}\int_Q|f(x)|dx\leq\alpha\quad(3.4)$
通过边长$2^n$等分方体$Q\in\H^{(1)}\backslash\S^{(1)}$，称这些方体为第二代
设$\H^{(2)}=\lbrace Q|Q$为第二代$\rbrace$
$\forall Q\in\H^{(1)}\backslash S^{(1)},\tilde{Q}\in\H^{(2)}$. 若$2^n\tilde{Q}=Q$，称$Q$为$\tilde{Q}$的母亲
设$\S^{(2)}=\lrb{Q\in\H^{(2)}|\frac{1}{|Q|}\int_Q|f(y)|dy\gt\alpha}$
对$\forall Q\in\H^{(2)}\backslash\S^{(2)}~i.e.~Q$满足(3.4)，对这些$Q$继续分下去到$m$代
$3^\circ~\Omega\overset{def}{=}\bigcup^\infty_{m=1}\lrb{Q|Q\in\S^{(m)}}=\bigcup_{k\in\Z}Q_K$
$\forall Q_k,\exist m_0\in\N~s.t.~Q_k\in\S^{(m_0)}$，但它的母亲$\hat{Q}\in\H^{m_0-1}\backslash\S^{(m_0-1)}$
且$\frac{1}{|Q|}\int_{\hat Q}|f(y)|dy\leq\alpha$.定义$F=\R^n\backslash\Omega$
现在来验证(1)-(5)
(1)显然
(2)事实上，$\forall Q_k~(\Omega=\bigcup_{k\in\Z}Q_k)\exist m_0\in\N~.s.t~Q_k\in\S^{(m_0)}$，且其母亲$\widehat{Q_k}\notin\S^{(m_0-1)},\widehat{Q_k}\in\H^{(m_0-1)}\backslash\S^{(m_0-1)}$
$\rArr\widehat{Q_k}$满足(3.4),$\frac{1}{|\widehat{Q_k}|}\int_{\widehat{Q_k}}|f(y)|dy\leq\alpha$
$\rArr\int_{Q_k}|f(y)|dy\leq\int_{\widehat{Q_k}}|f(y)|dy\leq\alpha|\widehat{Q_k}|=\alpha\cdot2^n|Q_k|$
$\rArr\frac{1}{|Q_k|}|f(y)|dy\leq\alpha\cdot2^n$
而$Q_k\in\S^{(m_0)}$，有$\frac{1}{|Q_k|}\int_{Q_k}|f(y)|dy\gt\alpha$
(3) 事实上，$\forall x\in F=\R^n\backslash\Omega\rArr x\notin Q_k~(\forall k\in\Z)$
$\exist!$没有被选到的包含$x$的二进方体$Q_x^{(k)},k=0,1,...$指第k代.
则$\forall k\in\N$有$\frac{1}{|Q^{(k)}_x|}\int_{Q_x^{(k)}}f(y)dy\leq\frac{1}{|Q^{(k)}_x|}\int_{Q_x^{(k)}}|f(y)|dy\leq\alpha$
$\lim\limits_{k\rarr\infty}\frac{1}{|Q^{(k)}_x|}\int_{Q_x^{(k)}}|f(y)|dy=|f(x)|~a.e.~x\in F\rArr|f(x)|\leq\alpha~.a.e.~x\in F$（Lebsgue微分定理）
(4) 事实上，$|\Omega|=|\bigcup_{k\in\Z}Q_k|=\sum_{k\in\Z}|Q_k|$
而$\forall Q_k,Q_k\in\S^{(m_0)}\rArr\frac{1}{|Q_k|}\int_{Q_k}|f(y)|dy\gt\alpha$
$\rArr|Q_k|\leq\frac{1}{\alpha}\int_{Q_k}|f(y)|dy$
$$
\begin{aligned}
\rArr\sum_{k\in\Z}|Q_k|&\leq\frac{1}{\alpha}\sum_{k\in\Z}\int_{Q_k}|f(y)|dy\\
&=\frac{1}{\alpha}\int_{\bigcup_{k\in\Z}Q_k}|f(y)|dy\\
&\leq\frac{1}{\alpha}\intrn|f(y)|dy=\frac{1}{\alpha}\norm{f}_{L^1}
\end{aligned}
$$ (5)对$\forall j\in\Z$，定义$b_j(x)=(f(x)-\frac{1}{|Q_j|}\int_{Q_j}f(y)dy)\Chi_{Q_j}(x)$,
$b(x)=\sum_{j\in\Z}b_j(x),g(x)=f(x)-b(x)$
(5.2)由定义立即得到
(5.3)事实上，$$
\begin{aligned}
\int_{Q_j}b_j(x)dx&=\int_{Q_j}(f(x)-\frac{1}{|Q_j|}\int_{Q_j}f(y)dy)dx\\
&=\int_{Q_j}f(x)dx-\frac{1}{|Q_j|}\int_{Q_j}f(y)dy\int_{Q_j}1dx\\
&=0
\end{aligned}
$$
(5.4)$$
\begin{aligned}
&\int_{Q_j}|b_j(x)|dx\\
&=\int_{Q_j}|f(x)-\frac{1}{|Q_j|}\int_{Q_j}f(y)dy|dx\\
&\leq\int_{Q_j}|f(x)|dx+\frac{1}{|Q_j|}|\int_{Q_j}f(y)dy|\int_{Q_j}1dx\\
&=\int_{Q_j}|f(x)|dx+|\int_{Q_j}f(y)dy|\\
&\leq2\int_Q|f(y)|dy\\
&\leq2\cdot2^n\alpha\cdot|Q_j|=2^{n+1}\alpha|Q_j|
\end{aligned}
$$ 对(5.1) $g(x)=f(x)-b(x)=f(x)-\sum_{j\in\Z}(f(x)-\frac{1}{Q_j}\int_{Q_j}f(y)dy)\Chi_{Q_j}(x)$
$\forall x\in\R^n=\Omega\cup F$
(I)$x\in F,g(x)=f(x)\rArr\norm{g(x)}_{L^\infty(F)}\leq\alpha$
(II) $x\in\Omega\rArr x\in Q_{j_0}$（最多有$2^n$个$Q_{j_0}$）
若$x\in\mathring{Q_{j_0}},g(x)=f(x)-(f(x)-\frac{1}{|Q_{j_0}|}\int_{Q_{j_0}}f(y)dy)\Chi_{Q_{j_0}}(x)=\frac{1}{|Q_{j_0}|}\int_{Q_{j_0}}|f(y)|dy\leq2^n\alpha$
若$x\notin\mathring{Q_{j_0}},m\lrb{x\in\Omega|x\notin\mathring{Q_{j_0}},\forall j\in\Z}=0$
$\rArr\norm{g}_{L^\infty(\R^n)}\leq max{\alpha,2^n\alpha}=2^n\alpha$ $$\qed$$
定理(Benedek-Calders$\acute{o}$n-Panzone定理)
设$T$是次线性算子且满足
(i)$T$是弱(p,p)型$(1\lt p\lt\infty),~T:L^p\rarr L^{p,\infty},\norm{Tf}_{L^{p,\infty}}\leq c\norm{f}_{L^p}$
$i.e.~d_{Tf}(\alpha)\leq\frac{c_1}{x^p}\norm{f}^p_{L^p},\forall f\in L^p(\R^n)$
(ii) $\exist c_2,c_3>1~s.t.~\forall f:Supp(f)\subset B(x_0,r),\int_{B(x_0,r)}f(y)dy=0$，有$\int_{\R^n\backslash B(x_0,c_2r)}|Tf(x)|dx\leq c_3\norm{f}_{L^1}$
则$T$为弱(1,1)型$~i.e.~d_{Tf}(\alpha)\leq\frac{c_4}{\alpha}\norm{f}_{L^1},\forall\alpha>0,\forall f$具有紧支撑.
证明：$f\in L^1(\R^n),\alpha>0$给定
由C-Z分解定理，$\R^n=\Omega\cup F,f=g+b,|\Omega|=\sum_k|Q_k|\leq\frac{1}{\alpha}\norm{f}_{L^1}$
$$
\begin{aligned}
\norm{g}^p_{L^p}=\intrn|g(x)|^pdx&=\sum_k\int_{Q_k}|g(x)|^{p-1}|g(x)|dx\\
&+\int_F|g(x)|^{p-1}|g(x)|dx\\
&\leq\sum_k\int_{Q_k}(2^n\alpha)^{p-1}|g(x)|dx\\
&+\int_F\alpha^{p-1}|g(x)|dx
\end{aligned}
$$ $1^\circ\forall x\in F,g(x)=f(x)$
$2^\circ\forall x\in Q_{k_0},g(x)=f(x)-\sum_kb_k(x)=f(x)-b_{k_0}(x)=f(x)-(f(x)-\frac{1}{|Q_{k_0}|}\int_{Q_{k_0}}f(y)dy)\Chi_{Q_{k_0}}=\frac{1}{|Q_{k_0}|}\int_{Q_{k_0}}f(y)dy~i.e.~g(x)=\frac{1}{|Q_{k_0}|}\int_{Q_{k_0}}|f(y)|dy$
$\rArr|g(x)|\leq2^n\alpha~a.e.~\Omega,\norm{g(x)}_{L^\infty(\Omega)}\leq2^n\alpha$ $$
\begin{aligned}
\rArr\norm{g}^p_{L^p}&\leq((2^n\alpha)^{p-1}+\alpha^{p-1})(\int_{Q_k}|g(x)|dx+\int_F|g(x)|dx)\\
&\leq((2^n\alpha)^{p-1}+\alpha^{p-1})(\int_\Omega|f(x)|dx+\int_F|f(x)|dx)
\end{aligned}
$$ $\rArr\norm{g}^p_{L^p}\leq c\alpha^{p-1}\norm{f}_{L^1}$
$b_k(x)=(f(x)-\frac{1}{|Q_k|}\int_{Q_k}f(y)dy)\Chi_{Q_k}(x)\rArr\int_{Q_k}|b_k(x)|dx\leq2\int_{Q_k}|f(x)|dx$
估计$d_{Tf}(\alpha)~(\forall\alpha>0),f=g+b$
$|T(f)|=|T(g+b)|\leq|Tg|+|Tb|$
$\rArr d_{Tf}(\alpha)\leq d_{Tg}(\frac{\alpha}{2})+d_{Tb}(\frac{\alpha}{2})$
第一步，对$d_{Tb}(\frac{\alpha}{2})$ $$
\begin{aligned}
\int_{Q_k}|b_k(x)|^pdx&\leq2^p\int_{Q_k}|f(x)|^pdx+2^p\int_{Q_k}|\frac{1}{|Q_k|}\int_{Q_k}f(y)dy|^pdx\\
&\leq2^{p+1}\int_{Q_k}|f(x)|^pdx
\end{aligned}
$$ $$
\begin{aligned}
\rArr\forall N\in\N,\norm{b-\sum^{N+1}_{k=1}b_k}^p_{L^p}=\norm{\sum^\infty_{k=N}b_k}^p_{L^p}&=\intrn|\sum^\infty_{k=N}b_k(x)|^pdx\\
&=\sum_{k\geq N}\int_{Q_k}|b_k(x)|^pdx\\
&\leq\sum_{k\geq N}2^{p+1}\int_{Q_k}|f(x)|^pdx\\
&=2^{p+1}\int_{\bigcup_{k\geq N}Q_k}|f(x)|^pdx
\end{aligned}
$$ 而$\sum^\infty_{k=1}|Q_k|\leq\alpha^{-1}\norm{f}_{L^1}\lt+\infty$
所以$\lim\limits_{N\rarr\infty}\sum^\infty_{k\geq N}|Q_k|=0\rArr\lim\limits_{N\rarr\infty}|\bigcup_{k\geq N}Q_k|=0\rArr\lim\limits_{N\rarr\infty}\norm{b-\sum_{1\leq k\leq N}b_k}_{L^p}=0$
($b=\sum^\infty_{k=1}b_k=\lim\limits_{N\rarr\infty}\sum^N_{k=1}b_k~in~L^p$)
由$T$弱(p,p)型，$T(\sum_{k\geq N}b_k(x))\overset{\mu}{\rarr}0$
$\rArr\exist N_l,T(\sum_{k\geq N_l}b_k(x))\overset{a.e.}{\rarr}0~(l\rarr\infty)$
$|Tb(x)|=|T(\sum_kb_k(x))|\leq\sum_k|Tb_k(x)|~a.e.~x\in\R^n$
取$B_k$为$Q_k$的同心圆且$r_{B_k}=2diamQ_k$，设$\hat{\Omega}=\cup_kB_k,m(B_k)\leq c_n|Q_k|$
$$
\begin{aligned}
d_{Tb}(\frac{\alpha}{2})&=m(\lrb{x\in\R^n||Tb(x)|\gt\frac{\alpha}{2}})\\
&=m(\lrb{x\in\hat{\Omega}}||Tb(x)|>\frac{\alpha}{2})+m(\lrb{x\in\R^n\backslash\hat{\Omega}}||Tb(x)|\gt\frac{\alpha}{2})\\
&\leq m(\hat{\Omega})+J
\end{aligned}
$$ 对$m(\hat{\Omega}),m(\hat{\Omega})\leq\sum_{k\in\N}|B_k|\leq c_n\sum_{k\in\N}|Q_k|\leq\frac{c}{a}\norm{f}_{L^1}$
对$J$,$$
\begin{aligned}
J&=m(\lrb{x\in\R^n\backslash\hat{\Omega}}||Tb(x)|\gt\frac{\alpha}{2})\\
&\leq\frac{c}{\alpha}\int_{\R^n\backslash\hat{\Omega}}|Tb(x)|dx\\
&\leq\frac{c}{\alpha}\int_{\R^n\backslash\hat{\Omega}}\sum_k|Tb_k(x)|dx\\
&\leq\frac{c}{\alpha}\sum_k\int_{Q_k}|b_k(x)|dx\\
&\leq\frac{c}{\alpha}\sum_k\int_{Q_k}|f(x)|dx\leq\frac{c}{\alpha}\norm{f}_{L^1}
\end{aligned}
$$ $$
\rArr d_{Tb}(\frac{\alpha}{2})\leq\frac{c}{\alpha}\norm{f}_{L^1}\rArr d_{Tf}(\alpha)\leq\frac{c}{\alpha}\norm{f}_{L^1}\qed
$$