= Solution
The <layer cake representation> with $\varphi(t)=t^p$ gives
$$
\lVert Mf\rVert_p^p
=p\int_0^\infty s^{p-1}
\lambda(\{Mf>s\})\,ds.
$$
For each $s>0$, split
$$
f=f_0+f_1,
\qquad f_0=f\mathbf1_{\{|f|>s/2\}},
\qquad f_1=f\mathbf1_{\{|f|\leq s/2\}}.
$$
Since $Mf_1\leq s/2$ and the maximal operator is sublinear,
$$
\{Mf>s\}\subseteq\{Mf_0>s/2\}.
$$
The weak $(1,1)$ <Hardy-Littlewood maximal inequality> therefore gives
$$
\lambda(\{Mf>s\})
\leq\frac{2C_1}{s}
\int_{\{|f|>s/2\}}|f(x)|\,dx.
$$
Substitution into the distribution formula and another application of <Tonelli theorem> yield
$$
\begin{aligned}
\lVert Mf\rVert_p^p
&\leq2pC_1\int_{\mathbb R^n}|f(x)|
\int_0^{2|f(x)|}s^{p-2}\,ds\,dx\\
&=\frac{2^ppC_1}{p-1}\lVert f\rVert_p^p.
\end{aligned}
$$
Thus the <Strong Lp bound for the Hardy-Littlewood maximal function> holds with
$$
\boxed{\ \lVert Mf\rVert_p\leq
\left(\frac{2^ppC_1}{p-1}\right)^{1/p}\lVert f\rVert_p.\ }
$$
Back to article page
