Lesbesgue theorem for differentiation of indefinite integrals

Question

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 $0<t<\infty$

Answer 1

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*}

Answer 2

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.