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I would like to know a proof of a variant of Hardy's inequality below. Could anyone introduce me a reference or give me a proof? Thank you very much for your assistance.

Set $0<r<\frac{2(n-s)}{n-2}$, $0<s<2$ and $n\ge 3$. Let $\Omega \subset \mathbb{R}^n$ be a bounded domain with smooth boundary. Them there exists a constant $C>0$ such that $$ \int_{\Omega}\frac{|u|^r}{|x|^s}dx\le C\int_{\Omega}|\nabla u|^2dx $$ for any $u \in H^1_0(\Omega)$.

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    $\begingroup$ s=0, r=2n/(n-2) is Sobolev embedding. s=2, r=2 is standard Hardy. By interpolation you get all couples (s,r) in the segment joining these two points, and by boundedness of the domain you get the smaller values of r $\endgroup$
    – Piero D'Ancona
    Commented Mar 28, 2013 at 21:35
  • $\begingroup$ @Higgs: I think the idea is arithmetic geometric inequality: $\sqrt{ab} \le (a+b)/2$, which implies $\frac{|u|^{\alpha r_1 + (1-\alpha) r_2}}{|x|^{\alpha s_1 + (1- \alpha) s_2}} \le \frac{|u|^{r_1}}{|x|^{s_1}} + \frac{|u|^{r_2}}{|x|^{s_2}}$. $\endgroup$
    – John Jiang
    Commented Mar 30, 2013 at 13:48

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