# Extending the Hardy Littlewood Maximal Function Estimate to the Case $1<p<\infty$

This question pertains to the extension of the weak-type estimate of the maximal function in $L^{1}$ to a "sharp" estimate in $L^{p}$ for $1<p<\infty$ (the case $p=\infty$ is evidently trivial).

If $f$ is integrable on $\mathbb{R}^{d}$, then the Hardy-Littlewood Maximal function $f$ is defined to be: \begin{equation*} f^{*}(x)=\sup_{x\in B}\frac{1}{m(B)}\int_{B}|f(y)|dy, \end{equation*} where the supremum is taken over all balls containing $x$ (it can be interpreted as a sort of maximal average of $f$).

The well-known weak-type estimate for $f^{*}$ in $L^{1}$ is given by \begin{equation*} m(\{x:f^{*}(x)>\alpha\})\leq\frac{A}{\alpha}||f||_{L^{1}(\mathbb{R}^{d})} \end{equation*} where $A$ depends only on $d$ (taking $A=3^{d}$ is sufficient for the proof).

It is well-known that this is the best estimate one can hope for, for despite the estimate showing $f^{*}$ is not too much bigger than $f$, it is nevertheless true that in general $f\in L^{1}$ does not imply $f^{*}\in L^{1}$.

Anyway, from this exercise, apparently we can modify the situation slightly to obtain a positive result on the integrability of $f^{*}$.

Let $f$ be a measurable function on $\mathbb{R}^{d}$ such that: \begin{equation*} ||f||_{L^{p}(\mathbb{R}^{d})}=\int_{\mathbb{R}^{d}}|f(x)|^{p}dx<\infty. \end{equation*} Then the task is to prove the following inequality: \begin{equation*} ||f^{*}||_{L^{p}(\mathbb{R}^{d})}\leq C||f||_{L^{p}(\mathbb{R}^{d})}, \end{equation*} where $C$ depends only on $p$ and $d$.

There are some preliminary results which I proved which make it possible to prove this without any advanced "machinery" in functional analysis:

\begin{equation*} \text{(1) } \int_{\mathbb{R}^{d}}|f(x)|^{p}dx=\int\limits_{0}^{\infty}p\alpha^{p-1}m(\alpha)d\alpha, \end{equation*}

\begin{equation*} \text{(2) } m^{*}(\alpha)=m\left(\{x:f^{*}(x)>\alpha\}\right)\leq\frac{2A}{\alpha}\int_{\{x:|f(x)>\frac{\alpha}{2}\}}|f(x)dx. \end{equation*}

In the above results $A$ is a constant depending on only $d$, $1<p<\infty$, $\alpha>0$ and $m(\alpha)$ is the same as in LHS of $(2)$ except for $f$.

These results I've proved, and using them I compute the following to prove the statement: \begin{align*} ||f^{\star}||_{L^{p}(\mathbb{R}^{d})} &=\int_{\mathbb{R}^{d}}|f^{*}(x)|^{p}dx\\ &=\int_{\mathbb{R}^{d}}\int_{0}^{|f^{*}(x)|^{p}}p\alpha^{p-1}d\alpha dx\\ &=\int_{0}^{\infty}p\alpha^{p-1}m^{\star}(\alpha)d\alpha\\ &\leq\int_{0}^{\infty}p\alpha^{p-1}\left(\frac{2A}{\alpha}\int_{\{x:|f(x)|>\frac{\alpha}{2}\}}|f(x)|dx\right)d\alpha\\ &=2Ap\int_{0}^{\infty}\alpha^{p-2}\left(\int_{\{x:|f(x)|>\frac{\alpha}{2}\}}|f(x)|dx\right)d\alpha\\ &=2Ap\int_{0}^{\infty}\int_{\{x:|f(x)|>\frac{\alpha}{2}\}}\alpha^{p-2}|f(x)|dxd\alpha\\ &=\ldots\\ &=C\int_{\mathbb{R}^{d}}|f(x)|^{p}dx\\ &=C||f||_{L^{p}(\mathbb{R}^{d})}. \end{align*}

Unfortunately, I am having trouble filling in the $\ldots$ to obtain the conclusion. I've tried messing rewriting expressions in terms of the previous results and working backwards, but I can't get things to work out.

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Why not just integrate over $\alpha$ first, from $0$ to $2|f(x)|$? – user31373 Jun 16 '12 at 21:07
I'm not sure I follow your suggestion. At what point would I integrate the $\alpha$ iterate from $0$ to $2|f(x)|$? Since I have $\alpha$ integration from $0\to\infty$. – Taylor Martin Jun 17 '12 at 0:13

\begin{align}\int_{0}^{\infty}\int_{\{x\colon |f(x)|>\frac{\alpha}{2}\}}\alpha^{p-2}|f(x)|\,dx\,d\alpha &=\int_{\mathbb R^d}\int_{0}^{2|f(x)|}\alpha^{p-2}|f(x)|\,d\alpha\,dx \\&=\int_{\mathbb R^d}\frac{1}{p-1}(2|f(x)|)^{p-1}|f(x)|\,dx \\&=\frac{2^{p-1}}{p-1}\int_{\mathbb R^d}|f(x)|^{p}\,dx\end{align}