How do I show that $u \in W^{k,1}(U)$ where $u(x)=|x|^{-\alpha}$? Let $U=\{x \in \mathbb{R}^n | |x|<1\}$, $\alpha$ be positive and $u(x)=|x|^{-\alpha}$. Show that $u \in W^{k,1}$ if $k\geq 0 $ is an integer satisfying $k+ \alpha <n$.
My idea: In this question, firstly I tried to show that $u \in L_{loc}^{1}(U)$ and $D^{\alpha}(U) \in L^{1}(U)$ for $|\alpha|\leq k$.
For first part: $\int_{U}|u(x)|dx=\int_{U}|x|^{-\alpha}dx=\int_{0}^{1} r^{-\alpha+n-1} dr$.
This integral is finite only if $\alpha <n$.
I am confused for second part, how do I find $D^{\alpha}(u)$. I know for $\alpha =1$.
I am studying Sobolev space first time, anyone can please suggest me some hints for second part?

In fact
\begin{eqnarray}
\int_U|D^ku|dx&=&\int_{U}\bigg|\sum_{k_1+\cdots+k_n=k}\frac{\partial^k}{\partial x_1^{k_1}\cdots\partial x_n^{k_n}}|x|^{-\alpha}\bigg|dx\\
&\le&\sum_{k_1+\cdots+k_n=k}\int_{U}\bigg|\frac{\partial^k}{\partial x_1^{k_1}\cdots\partial x_n^{k_n}}|x_1^2+\cdots+x_n^2|^{-\alpha/2}\bigg|dx\\
&\le&\sum_{k_1+\cdots+k_n=k}\int_{U}\bigg|\frac{\partial^k}{\partial x_1^{k_1}\cdots\partial x_n^{k_n}}|x_1^2+\cdots+x_n^2|^{-\alpha/2}\bigg|dx\\
&\le&\sum_{k_1+\cdots+k_n=k}C\int_{U}|x|^{-\alpha-k}dx\\
&\le&\sum_{k_1+\cdots+k_n=k}C\int_0^1r^{-\alpha-k+n}\omega_{n-1}dr\\
&<&\infty.
\end{eqnarray}
You already showed that
$$ \int_{U}|u|dx<\infty $$
and hence $u=|x|^{-\alpha}\in W^{k,1}(U)$.
