# Norm of Hardy-Littlewood maximal operator

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 $f$.

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 $\alpha\to0$.
• Aguirre. Thank you very much. I have another question: Since the proof in Stein's book depend on the decomposition of $\mathbb{R}^n$, could we estend the inequality to \begin{equation} |\{x\in B : Mf(x)>\alpha\}| \ge \frac{2^{-d}}{\alpha}\int_{x\in B:|f(x)|>\alpha} |f(x)| \ dx ? \end{equation} Here, $B$ denotes the ball on $\mathbb{R}^n$. – beginner Oct 9 '14 at 22:12