Consider the function $f: \Bbb R^n \to \Bbb R$ defined by $f(x)=\log |x|$ for $x \neq 0$ and $f(0)=0$. I'm trying to prove that the Hardy-Littlewood maximal function of $f$, $Mf$, equals $\infty$. That is, for every $x \in \Bbb R^n$,
$$Mf(x) = \sup_{x \in B} \frac{1}{|B|} \int_B |f| = \infty,$$
where the supremum is taken over all balls $B$ in $\Bbb R^n$ containing $x$ and $|B|$ is the Lebesgue measure of the ball.
The reasoning I tried to use is that for any $x \in \Bbb R^n$ there is a ball $B$ containing $x$ and $0$. Since it contains $0$, $|f| \to \infty$ on $B$, and then $\frac{1}{|B|} \int_B |f| = \infty$. But I don't know if that's true, do we need that $|f|$ is exactly $\infty$ on $B$ to say that the integral on $B$ is $\infty$? In that case, would we need to define $f(0)=\infty$?
Best Answer
One may use polar coordinates.
It suffices to consider averages over balls of radii $r>1$.
\begin{align} \frac{1}{m(B_r)}\int_{B_r}\big|\log |x|\big|\,dx&=\frac{n}{r^n}\int^r_0t^{n-1}|\log t|\,dt\\ &=-\frac{n}{r^n}\int^1_0t^{n-1}\log t\,dt+\frac{n}{r^n}\int^r_1 t^{n-1}\log t\,dt \end{align}
$u=\log t$ ($e^u=t$) gives
$$ \frac{n}{r^n}\int^r_1t^{n-1}\log t\,dt=\frac{n}{r^n}\int^{\log r}_0e^{nu}u\,du\sim\frac{nr^n\log r}{nr^{n-1}}\xrightarrow{r\rightarrow\infty}\infty $$
It follows that
$$\sup_{r>0}\frac{1}{m(B_r)}\int_{B_r}\big|\log|x|\big|\,dx =\infty$$