TITLE: Norm of Hardy-Littlewood maximal operator
QUESTION [5 upvotes]: We define Hardy-Littlewood maximal operator $M$ by 
\begin{equation}
Mf(x)=\sup_{r>0} \frac{1}{|B(x,r)|} \int_{B(x,r)} |f(y)| dy
\end{equation}
where $B(x,r)$ denotes the ball centered at $x \in \mathbb{R}^n$ with radius $r>0$.
Let $1\le p<\infty$.
We define the weak Lebesgue space $wL^p(\mathbb{R}^d)$ as the set of all measurable functions $f$ on $\mathbb{R}^d$ such that
\begin{equation}
\|f\|_{wL^p}=\sup_{\gamma>0} \gamma (\{x\in \mathbb{R}^d : |f(x)|>\gamma  \})^{1/p}<\infty.
\end{equation}
It has already proven that the operator $M$ is bounded from $L^1(\mathbb{R}^n)$ to $wL^1(\mathbb{R}^n)$, that is,
\begin{equation}
\|Mf\|_{wL^{1}(\mathbb{R}^n)} \le C \|f\|_{L^1(\mathbb{R}^n)}
\end{equation} 
where $C>0$ does not depend on $f$.
My question is: Could we also get a left side norm estimate for $M$, namely 
\begin{equation}
\|Mf\|_{wL^{1}(\mathbb{R}^n)} \ge C' \|f\|_{L^1(\mathbb{R}^n)}
\end{equation}
where $C'>0$ does not depend on $f$.

REPLY [4 votes]: Yes. According to Stein's book Singular Integrals and Differentiability Properties of Functions, page 23, 5.2 (b), there is a constant $c$ such that
$$
|\{Mf(x)>c\,\alpha\}|\ge\frac{2^{-d}}{\alpha}\int_{|f|>\alpha}|f|\,dx,
$$
where $|A|$ means the measure of $A$. This gives
$$
\int_{|f|>\alpha}|f|\,dx\le \frac{2^d}{c}\bigl(c\,\alpha\,|\{Mf(x)>c\,\alpha\}|\bigr)\le\frac{2^d}{c}\|Mf\|_{wL^{1}(\mathbb{R}^d)} 
$$
Now let $\alpha\to0$.