This comes from a recent lecture I've had. I have questions about on specific step in the short proof. This inequality is noted as the "Subcritical Hardy Inequality on the Whole Space".
Statement: Let $1 \leq p < N$. Then, for all $u \in W^{1,p}(\mathbb{R}^{N})$:
$(\frac{N - p}{p})^{p} \int_{\mathbb{R}^{N}} \frac{|u(x)|^{p}}{|x|^{p}} \text{d}x \leq \int_{\mathbb{R}^{N}} |\frac{x}{|x|} \cdot \nabla u|^{p} \text{d}x$
Start of Proof: Let $u \in C^{\infty}_{0}.$ We use the equation: $\text{div}(\frac{x}{|x|^{p}}) = \frac{N - p}{|x|^{p}}.$
$|\int_{\mathbb{R}^{N}} \frac{|u(x)|^{p}}{|x|^{p}} \text{d}x| = | \ \frac{1}{N-p}\int_{\mathbb{R}^{N}} \text{div}(\frac{x}{|x|^{p}}) |u(x)|^{p}\text{d}x \ |$
$= | \ \frac{-1}{N-p}\int_{\mathbb{R}^{N}} \frac{x}{|x|^{p}} p |u(x)|^{p - 1} \text{sgn}(u(x)) \nabla u \ \text{d}x \ |$
The last equality is the step I do not understand. How do we go from the middle integral to the one on the right hand side?
You integrate by parts. $$ \int (\operatorname{div}v) f=-\int v\cdot\nabla f $$ for $f\in C^\infty_c$ and $v$ weakly differentiable.