## Weak (1,1) boundedness of Hardy Littlewood maximal function

$f$ is a locally integrable function in $\mathbb{R}^d$, define the maximal function as

$\displaystyle M(f)=\sup\limits_{x\in B}\frac{1}{|B|}\int_B|f(y)|dy$

here $B$ is an  arbitrary ball.

$\textbf{Thm: }$ Maximal operator is weakly (1,1) bounded. That is $\exists C=C(d)>0$ such that for $\forall\lambda>0$,

$\displaystyle |\{x:M(f)(x)>\lambda\}|\leq\frac{C||f||_1}{\lambda}$

$\textbf{Proof: }$ Let $E_\lambda=\{x:M(f)(x)>\lambda\}$, $K\subset E_\lambda$ is a compact set.

Then $K\subset\bigcup\limits_{i=1}^N B_i$, where $B_i=B_i(x_i)$ with $\displaystyle \frac{1}{|B_i|}\int_{B_i}|f(y)|dy>\lambda$. Apply the Vitali covering lemma

$\textbf{Vitali covering lemma: }$ Given $\{B_1,\cdots,B_n\}$ a finite collection of balls in $\mathbb{R}^n$, then there exists a subcollection $B_{ij}$, $j=1,\cdots,s$ disjoint from each other and

$\bigcup\limits_{i=1}^N B_i\subset \bigcup\limits_{j=1}^s 3B_{i_j}$

We can conclude that  $K\subset \bigcup\limits_{j=1}^s 3B_{i_j}$. So

$\displaystyle |K|\leq \sum\limits_{j=1}^s 3^d|B_{i_j}|\leq \frac{3^d}{\lambda}\sum\limits_{j=1}^s\int_{B_{i_j}}|f(y)|dy\leq \frac{3^d||f||_1}{\lambda}$

Since $K$ is arbitrary, we know $|E_\lambda|$ is bounded by the right hand side.$\hfill\square$

Actually the decay of maximal function is a little faster than $\frac{1}{\lambda}$

$\textbf{Thm: }$ There exists a constant $C=C(d)>0$ such that

$\displaystyle |\{x:M(f)(x)>\lambda\}|\leq \frac{C}{\lambda}\int_{|f|>\frac{\lambda}{2}}|f(y)|dy$

$\textbf{Proof: }$ Define

$\displaystyle g=\begin{cases}\frac{\lambda}{2},\quad |f(x)|>\frac{\lambda}{2}\\ 0, \quad |f(x)|\leq\frac{\lambda}{2}\end{cases}$

Then $|f(x)|\leq |g(x)|+\frac{\lambda}{2}$. This means

$\displaystyle \{x:M(f)(x)>\lambda\}\subset \{x:M(g)(x)>\frac{\lambda}{2}\}$

So $\displaystyle |\{x:M(f)(x)>\lambda\}|\leq |\{x:M(g)(x)>\frac{\lambda}{2}\}|\leq \frac{C}{\lambda}\int_{|f|>\frac{\lambda}{2}}|f(y)|dy$

$\hfill\square$