Take $U= B^{0}(0,1)$ to be the unit ball in $\mathbb R^n$ and consider $u(x)=|x|^{-\alpha}$ ($x \in U$, $x \neq 0$).

Question

Take $U= B^{0}(0,1)$ to be the unit ball in $\mathbb R^n$ and consider $u(x)=|x|^{-\alpha}$ ($x \in U$, $x \neq 0$).

For which values of $\alpha>0,~n,~p$ does $u~ \text{belong to} W^{1,p}(U)$?

This is common question in Sobolev spaces. My question is after we obtain $Du=\frac{\alpha}{|x|^{\alpha +1}}$, we have (forget $\alpha$ for a second)$\|Du\|_{p}^{p}=\int \frac{1}{|x|^{p\alpha + p}}dx=\int \frac{1}{r^{p\alpha + p}} r^{n-1}dx=\int \frac{1}{r^{p\alpha + p-n+1}}dx$. So we want $p\alpha + p -n +1 >1$, right? It seems not since I got the opposite direction of the inequality previously. I appreciate if anyone could point out where I get wrong.

Answer 1

You need $p\alpha+p-n+1<1$ instead, so that

$$\frac{1}{\alpha}\vert\vert Du\vert\vert_p^p=C_n\int_0^1\frac{1}{r^{p\alpha+p-n+1}}dr=C_n\int_0^1r^{-(p\alpha+p-n+1)}dr<\infty,$$

which comes from ensuring the antiderivative of $r^{-(p\alpha+p-n+1)}$ doesn't blow up at $0$.

Note your expression for $Du$ is correct but with bars around it.