Rudin RCA exercise 7.22. $f(x)=x^{-1}(\log x)^{-2}$ is integrable but $\int_0^1 Mf = \infty$.

Question

Define $f(x)=x^{-1}(\log x)^{-2}$ if $0<x<\frac{1}{2}$, $f(x)=0$ on the rest of $R$. Then $f \in L^1(R)$. Show that $$(Mf)(x) \ge |2x \log(2x)|^{-1} \;\;(0<x<1/4)$$ so that $\int_0^1 (Mf)(x)dx=\infty.$

$$ (Mf)(x) = \sup_{0<r<\infty} \frac{1}{m(B_r)}\int_{B_r(x)}|f(y)|\mathrm{d}y $$ where $B_r(z)$ is a ball centered around $x$ with radius $r$.

I am stuck with proving this lower bound for the maximality function. I would greatly appreciate any help.

Answer 1

The other answer is the slick way to go, but the hint reduces the problem to a calculation:

Take $0<x<1/4$ and $r=x$. Then,

$Mf(x)\ge \frac{1}{B_r(x)}\int_{B_r(x)} |f|=\frac{1}{2x}\int_{0}^{2x}|f(t)|dt=$

$\frac{1}{2x}\cdot \underset{\delta\to 0^+ } \lim \int_{\delta}^{2x}|f(t)|dt=\frac{1}{2x}\cdot \underset{\delta\to 0^+ } \lim \left |\frac{1}{\ln 2x}-\frac{1}{\ln \delta}\right |=\frac{1}{2x\ln 2x}$

from which the result follows.

Answer 2

A simpler way to show $Mf\notin L^1$ is to note there is a constant $c>0$ such that $|f|>c$ on a set of positive measure $E\subset (0,1/2).$ Then for $x > 1/2,$

$$Mf(x) \ge \frac{1}{B(x,x)}\int_{B(x,x)} |f| \ge \frac{1}{2x}\cdot c\cdot m(E),$$

which is not in $L^1.$