I need prove that for $n\geq 3$ there exists a constant C such that
$\displaystyle\int_{\mathbb{R}^n}\dfrac{u^2}{|x|^2}dx\leq C\int_{\mathbb{R}^n}|\nabla u|^2dx$
for any $u\in H^1(\mathbb{R}^n)$.
The hint is: note that $0\leq \left|\nabla u+\dfrac{\lambda u}{|x|^2}x\right|$ for all $\lambda\in \mathbb{R}$.
Thanks!
Take the quantity mentioned in the hint, square it, and then integrate over $\mathbb R^n$. Multiply it out. Then do some integration by parts with the cross term: $$ \int (\nabla u) u \cdot \frac x{|x|^2} \, dx = - \int u (\nabla u) \cdot \frac x{|x|^2} \, dx - \int u^2 \nabla\cdot\left(\frac x{|x|^2}\right) \, dx $$ that is, $$ 2 \int (\nabla u) u \cdot \frac x{|x|^2} \, dx = -\int u^2 \nabla\cdot\left(\frac x{|x|^2}\right) \, dx = -(n-2) \int \frac{u^2}{|x|^2}.$$
You need to be a little bit careful with the origin, but because $n \ge 3$, we see that we could approximate $\frac x{|x|}$ with something smooth at the origin without upsetting the integrals too much.
So you have $$ 0 \le \int |\nabla u|^2 \, dx + \lambda^2 \int \frac{u^2}{|x|^2} + 2\lambda \int (\nabla u) u \cdot \frac x{|x|^2} $$ $$ = \int |\nabla u|^2 \, dx + (\lambda^2 - (n-2)\lambda)\int \frac{u^2}{|x|^2} $$ Choose $\lambda$ appropriately.
$0\leq\left|\nabla u+\dfrac{\lambda u}{|x|^2}\right|=|\nabla u|^2+2\lambda \dfrac{u}{|x|^2}\nabla u\cdot x+\lambda^2\dfrac{u^2}{|x|^2}$
with this,
$-2\lambda \displaystyle\int_{\mathbb{R}^n}\dfrac{u}{|x|^2}\nabla u\cdot x-\lambda^2\displaystyle\int_{\mathbb{R}^n}\dfrac{u^2}{|x|^2}\leq\displaystyle\int_{\mathbb{R}^n}|\nabla u|^2$
On the other hand:
$\dfrac{1}{|x|^2}-\dfrac{2x_i^2}{|x|^2}\Rightarrow \displaystyle\int_{\mathbb{R}^n}u^2\nabla\cdot\left(\dfrac{x}{|x|^2}\right)=(n-2)\int_{\mathbb{R}^n}\dfrac{u^2}{|x|^2}$
– yemino Nov 27 '13 at 14:35