2
$\begingroup$

I am trying to prove that for a measurable function $f$, the Hardy-Littlewood maximal function is not in $L^1$ unless $f$ is $0$ a.e.

I was able to udnerstand the proof of the inequality here with $1$ replaced by an arbitrary constant $a>0$ : A question about the Hardy-Littlewood maximal function.

However I want to conclude by saying that Mf is not integrable since $1/|x|^n$ is not integrable on $B(0,a)^c$. Is this true? Am I thinking in the right direction?

EDIT: I found this in Frank Jones' measure theory book: https://imgur.com/a/tct1vI2. So it seems to be true, I just have no clue why it should be true.

$\endgroup$
3
$\begingroup$

If you know that $Mf\geq c\|x\|^{-n}$ for $\|x\|\geq1$, then (with $b$ the volume of $B_1(0)$) \begin{align*} \int_{\mathbb R^n} Mf&\geq\int_{\|x\|\geq 1}\frac{c}{\|x\|^n}\,dx =c\,\int_{\|x\|\geq 1}\int_{\|x\|}^\infty \frac{n}{s^{n+1}}\,ds\,dx\\ \ \\ &=cn\,\int_1^\infty\int_{1\leq\|x\|\leq s}\frac1{s^{n+1}}\,dx \,ds =cn\,\int_1^\infty\frac{m(B_s(0)\setminus B_1(0))}{s^{n+1}} \,ds\\ \ \\ &=-bc+cbn\,\int_1^\infty \frac1s\,ds=\infty. \end{align*}

$\endgroup$
2
  • $\begingroup$ I'm wondering if the term after the second equality should be replaced with $cn\int_{1}^\infty \int_{1\leq \| x\| \leq s}\frac{1}{s^{n+1}} dx ds$. After all, $\|x\|\geq 1$ $\endgroup$
    – Eric
    Dec 27 '20 at 12:32
  • $\begingroup$ Indeed. Edited. $\endgroup$ Dec 27 '20 at 13:05
0
$\begingroup$

It is true and it follows immediately if you use polar coordinates in $\mathbb R^{n}$. Look for 'polar coordinates' in the index of Rudin's RCA.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.