Lesbesgue theorem for differentiation of indefinite integrals Let $ 1<p<\infty$ and suppose $f \in L^p(\mathbb{R}^n)$ . Prove that also $Mf \in L^p(\mathbb{R}^n)$ and 
$$ \|Mf\|_p \leq 2(3^np')^{1/p}\|f\|_p.$$
My ideas: I was thinking of using Hardy-Littlewood theorem in the following way, but could not get success, any help will be appreciated. 
$$ \| Mf\|_p^p = p\int_{0}^{\infty} \lambda(\left\lbrace x\mid Mf(x)>t\right\rbrace)t^{p-1} \, dt$$
then maybe somehow try to apply Hardy-Littlewood theorem and then maybe apply Fubini's theorem to get the answer. 
Note :- $M$ is the Hardy-Littlewood Maximal function 
Note :- M is the hardy littlewood Maximal function m defined in the following way , 
Assume that $f \in L^1_{loc}(R^n)$. THen the Hardy-Littlewood maximal function for f is the function Mf defined on R^n by 
$Mf(x) = sup_{0<r<\infty} 1/(\lambda(B(x,r)) \int_{B(x,r)} |f(y)|dy$ 
Hardy Littlewood theorem : Let $f \in L^1(R^n) $. Then 
$\lambda({x| Mf(x)>t}) \leq 3^n ||f||_1/(t) $ for $0<t<\infty$
 A: For $t>0$ define
$$
f_{t}\left(  x\right)  :=\left\{
\begin{array}
[c]{ll}%
f\left(  x\right)  & \text{if }\left\vert f\left(  x\right)  \right\vert
>\frac{t}{2},\\
0 & \text{otherwise.}%
\end{array}
\right.
$$
Then $f_{t}\in L^{1}(  \mathbb{R}^{n})  $. Indeed,
\begin{align*}
\int_{\mathbb{R}^{N}}\left\vert f_{t}\right\vert \,dx &  =\int_{\{
\vert f\vert >\frac{t}
{2}\}  }\left\vert f\right\vert \,dx\\
&  \leq\left(  \frac{2}{t}\right)  ^{p-1}\int_{\{
\vert f\vert >\frac{t}
{2}
}\left\vert f\right\vert ^{p}\,dx<\infty.
\end{align*}
Moreover, since $\left\vert f\right\vert \leq\left\vert f_{t}\right\vert
+\frac{t}{2}$ we have that $\operatorname*{M}\left(  f\right)  \leq
\operatorname*{M}\left(  f_{t}\right)  +\frac{t}{2}$, and so
$$
\left\{  x\in\mathbb{R}^{n}:\,\operatorname*{M}\left(
f\right)  \left(  x\right)  >t\right\}  \subseteq\left\{  x\in\mathbb{R}%
^{n}:\,\operatorname*{M}\left(  f_{t}\right)  \left(  x\right)
>\frac{t}{2}\right\}  .
$$
By the Hardy Littlewood theorem applied to $f_{t}\in L^{1}\left(  \mathbb{R}^{n}\right)  $ 
\begin{align}
\lambda \left(\{ {M}(f)>t\}  \right)    \leq\frac{3^n
}{t}\int_{\mathbb{R}^{n}}\left\vert f_{t}\right\vert \,dx  =\frac{3^n}{t}\int_{\{\vert f\vert >\frac{t}{2}\}  }\left\vert f\right\vert
\,dx.
\end{align}
Hence, using Fubini's theorem, we obtain
\begin{align*}
\int_{\mathbb{R}^{n}}\left(  \operatorname*{M}\left(  f\right)  \right)
^{p}\,dx &  =p\int_{0}^{\infty}t^{p-1}\lambda\left(  \left\{  x\in\mathbb{R}^{n}:\,\operatorname*{M}\left(  f\right)  \left(  x\right)  >t\right\}
\right)  \,dt\\
&  \leq2\ell p\int_{0}^{\infty}t^{p-2}\int_{\left\{  x\in\mathbb{R}%
^{n}:\,\left\vert f\left(  x\right)  \right\vert >\frac{t}{2}\right\}
}\left\vert f\left(  y\right)  \right\vert \,dy  \,dt\\
&  =3^n p\int_{\mathbb{R}^{N}}\left\vert f\left(  y\right)  \right\vert
\left(  \int_{0}^{2\left\vert f\left(  y\right)  \right\vert }t^{p-2}%
\,dt\right)  \,dy \\
&  =\frac{3^n p2^{p}}{p-1}\int_{\mathbb{R}^{n}}\left\vert f\left(  y\right)
\right\vert ^{p}\,d y .
\end{align*}
A: Here's the answer I'll give. The Hardy Littlewood Maximal Inequality $$|\{x : (Mf)(x) > \lambda\}| \le \frac{3^n}{\lambda}||f||_1$$ gives what is known as a weak $(1,1)$ estimate. It's also easy to see that $$||Mf||_\infty \le ||f||_\infty$$ which is the weak $(\infty,\infty)$ estimate. The fact that there exists a constant $C$, depending only on $n$ and $p$, so that $$||Mf||_p \le C||f||_p$$ then follows from what is known as the Marcinkiewicz interpolation theorem. The Marcinkiewicz interpolation theorem does give you the explicit constants, so I'll let you track those down. 
