Let $0 < r < R$ and $p>1$ and consider the function
$$u(x) = \displaystyle\frac{\displaystyle\int_{|x|}^{R} t^{-1 }dt}{\displaystyle\int_{r}^{R} t^{-1 }dt},$$ if $r < |x|< R$ , and $u=1$ if $0\leq |x| \leq r$
Show that $\displaystyle\int_{B(0,R)} |\nabla u |^{p} \ dx = \int_{B(0,1)} 1 \ dx \ \ ({ln(R/r)})^p$
the gradient is in the weak sense.
I think the gradient is given in the natural way ( imagining u as a derivable function)
My ideas: the gradient (thinking as I said before) is :
$$ \nabla u (x) = \frac{1}{ln (\frac{R}{r})}( \frac{x_1}{|x|^2},...,\frac{x_n}{|x|^2})$$ if $r < |x|< R$ and zero in the other case.
Then
$$ \displaystyle\int_{B(0,R)} |\nabla u|^p \ dx \leq \displaystyle\int_{B(0,R) - B(0,r)} \frac{1}{(|x|ln(R/r))^p} = \frac{1}{(ln(R/r))^p} \displaystyle\int_{B(0,R) - B(0,r)} \frac{1}{|x|^p} \ dx$$
I stopped here =\ ....
Soemone can give me a help?
thanks in advance