I want to generalize the fact that the weak derivative of $|x|$ is $sgn(x)$.
The expected result should be $\frac{d|x|^{\alpha}}{dx_i} = \alpha x_i |x|^{\alpha - 2}$. A way to do this is found in page 50 of these notes.
However, I don't have the tools to believe that:
$|x|^{\alpha}$ is weakly differentiable provided that the pointwise derivative, which is defined almost everywhere, is locally integrable
So I would like to do it without using this fact.
The domain
I think weak derivatives have a dependence on the domain so I specify that I want the derivative of $|x|^{\alpha}$ in $\mathbb{R}^N \setminus \{0\}$
The computation
$\int |x|^{\alpha} \frac{d\phi}{dx_i} d(x_1,\ldots,x_N)$, $\phi \in \mathcal{C}_0^{\infty}$, by Fubini's theorem (?) I can write:
$\int_{\mathbb{R}^{N-1} \setminus \{0\}} (\int_{\mathbb{R} \setminus \{0\}} |x|^{\alpha} \frac{d\phi}{dx_i} dx_i) d(x_1,\ldots,x_{i-1},x_{i+1},x_N)$
integrating by parts $\int_{\mathbb{R} \setminus \{0\}} |x|^{\alpha} \frac{d\phi}{dx_i} dx_i$ and using that $\phi \in \mathcal{C}_0^{\infty}$:
$\int_{-\infty}^0 |x|^{\alpha} \frac{d\phi}{dx_i} dx_i = -\int_{-\infty}^0 \alpha x_i |x|^{\alpha-2} \phi dx_i$, analogously, $\int_{0}^\infty |x|^{\alpha} \frac{d\phi}{dx_i} dx_i = -\int_{0}^\infty \alpha x_i |x|^{\alpha-2} \phi dx_i$
so that I get the result equal to:
$-\int_{\mathbb{R}^N \setminus \{0\}} \alpha x_i|x|^{\alpha}\phi dx_i$
The only thing to finish the proof would be to see where I missed the - sign.