How do I compute the weak derivative of this function?

Question

I have read that the function $u(x)= \frac{1}{|{x}\vert^{\alpha}}$ defined over $\Bbb R^n$ is in $W^{m,p}$ as long as $\alpha +1 <n$. I’ve tried to prove this fact, but I haven’t been able to thus far. I know I must prove that $u$ has weak derivatives in $L^p$, and that the weak derivative for the multi index $s$ is defined as the function $v$ such that: $$\int{uD^s \phi} = (-1)^{|s\vert}\int{v \phi}$$ However I have not been able to compute the weak derivative or to understand why it doesn’t exist for $\alpha + 1 \geq n$. Can anyone please explain how this works out?

Answer 1

hints:

1. Make a good guess what the weak derivative would look like. Note that because the origin has measure zero, you do not need to worry about the value at the origin.
2. prove the relationship in the definition of weak derivative. Often, one needs to cut out an $\varepsilon$-ball around the origin, in order to handle the singularity.
3. A function can have a weak derviative, but the conditions for integrability for $W^{m,p}$ can fail.