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
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 $d$, $1
0$ 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.