TITLE: Show that there exist $f \in L^1$ but $f^*$ is not integrable over the unit ball .
QUESTION [6 upvotes]: In Stein's functional analysis, for some $0\lt t\le 1$ he defined
$$
f(x)=
\begin{cases}
|x|^{-d}\log\left(\dfrac{1}{|x|}\right)^{-1-t}&\text{whenever }0\le |x| \le \frac{1}{2}\\ 
\\
f(x)=0 &\text{ otherwise}
\end{cases}
$$
and he state that 
$$
f^*(x)\ge c|x|^{-d}\log\left(\frac{1}{|x|}\right)^{-t},
$$ 
then $f^*(x)$ is not integrable over the unit ball. However, I can prove this only when $d=1$.  
My attempt:
When $d=1$, 
$$\int\limits_0^{\frac12} \frac cx \log\left(\frac1x\right)^{\!-t-1} \mathrm{d}x =c \frac{(\log 2) ^{-t}}{t}$$ so $f \in L^1$, and also we have 
$$
f^*(y) \ge \frac1y\int\limits_0^y \frac cx \log\left(\frac1x\right)^{\!-t-1} \mathrm{d}x=\frac{c'}{y} \log\left(\frac1y\right)^{\!-t}
$$ 
whenever $y \in \left(0,\frac12\right)$ and RHS is not integrable over the unit ball .
However, when $d \neq 1$, we can not use the relation
$$
\int \frac{f(x)}{x} \mathrm{d}x =\int f(x)\, \mathrm{d} \log(x),
$$ 
so how to deduce the desired conclusion?

REPLY [2 votes]: Hint: Polar coordinates in $\mathbb R^d$ gives
$$\int_{\mathbb R^d} f\,dv = nv(B)\int_0^\infty r^{d-1}\int_S f(r\zeta)\,d\sigma(\zeta)\,dr.$$
Here $f$ is a measurable and nonnegative, $v$ is volume measure, $B$ is the unit ball, $S$ is the unit sphere, and $\sigma$ is normalized surface measure on $S.$
This will reduce the $d>1$ case to the $d=1$ case.