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


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