# 哈代－李特爾伍德極大函數

## 定義

${\displaystyle Mf(x)=\sup _{r>0}{\frac {1}{m(B(x,r))}}\int _{B(x,r)}|f(y)|dm(y)}$

Mf(x)可能是${\displaystyle \infty }$。） 其中m${\displaystyle \mathbb {R} ^{n}}$上的勒貝格測度

## 性質

Mf(x)是下半連續函數

### 證明

${\displaystyle c_{1}:={\frac {1}{m(B(x,r))}}\int _{B(x,r)}|f(y)|dm(y)>c}$

{\displaystyle {\begin{aligned}&Mf(x')\\&\geq {\frac {1}{m(B(x',r+\delta ))}}\int _{B(x',r+\delta )}|f(y)|dm(y)\\&\geq {\frac {1}{m(B(x',r+\delta ))}}\int _{B(x,r)}|f(y)|dm(y)\\&={\frac {1}{m(B(x',r))}}\left({\frac {r}{r+\delta }}\right)^{n}\int _{B(x,r)}|f(y)|dm(y)\\&={\frac {1}{m(B(x,r))}}\left({\frac {r}{r+\delta }}\right)^{n}\int _{B(x,r)}|f(y)|dm(y)\\&>c_{1}\cdot {\frac {c}{c_{1}}}=c\end{aligned}}}

## 哈代－李特爾伍德極大不等式

${\displaystyle f\in \mathrm {L} ^{1}(\mathbb {R} ^{n})}$可積函數，對任何常數${\displaystyle c>0}$，有不等式

${\displaystyle m(\{Mf>c\})\leq {\frac {3^{n}\|f\|_{\mathrm {L} ^{1}}}{c}}}$

### 證明

${\displaystyle {\frac {1}{m(B(x,r_{x}))}}\int _{B(x,r_{x})}|f(y)|dm(y)>c}$

K${\displaystyle \{Mf>c\}}$內的緊集${\displaystyle (B(x,r_{x}))_{x\in K}}$K的一個開覆蓋。因K緊緻，存在有限子覆蓋${\displaystyle (B(x_{i},r_{i}))_{i=1}^{N}}$。（${\displaystyle r_{i}:=r_{x_{i}}}$

{\displaystyle {\begin{aligned}m(K)&\leq \sum _{i_{j}}m(B(x_{i_{j}},3r_{i_{j}}))\\&=\sum _{i_{j}}3^{n}m(B(x_{i_{j}},r_{i_{j}}))\\&<\sum _{i_{j}}{\frac {3^{n}}{c}}\int _{B(x_{i_{j}},r_{i_{j}})}|f(y)|dm(y)\\&\leq {\frac {3^{n}}{c}}\int _{K}|f(y)|dm(y)\\&\leq {\frac {3^{n}\|f\|_{\mathrm {L} ^{1}}}{c}}\end{aligned}}}

${\displaystyle m(\{Mf>c\})=\sup _{K}m(K)\leq {\frac {3^{n}\|f\|_{\mathrm {L} ^{1}}}{c}}}$

## 參考

• Rudin, Walter (1987), Real and complex analysis, International Series in Pure and Applied Mathematics (3rd ed.), McGraw-Hill.