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By Hardy's inequality on bounded domain which contain the origin, we have for $u\in H^1_0(\Omega)$, $$ \int_\Omega\frac{u^2(x)}{|x|^2}dx\leq \left(\frac{n-2}{2}\right)^2\int_\Omega|\nabla u|^2dx $$ Can we derive that, for $u\in H^1_0(\Omega)$, $$ \int_\Omega\frac{u^2(x)}{|x-x_0|^2}dx\leq \left(\frac{n-2}{2}\right)^2\int_\Omega|\nabla u|^2dx,\quad\forall x_0\in\Omega $$

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For any $x_0\in\Omega$, take $y=x-x_0$ which gives $$ \int_\Omega\frac{u^2(x)}{|x-x_0|^2}dx=\int_{\Omega-x_0}\frac{u^2(y+x_0)}{|y|^2}dy $$ Then $g(y):=u(y+x_0)\in H_0^1(\Omega-x_0)$ and $0\in\Omega-x_0$ which implies $$ \int_{\Omega-x_0}\frac{u^2(y+x_0)}{|y|^2}dy\leq \left(\frac{n-2}{2}\right)^2\int_{\Omega-x_0}|\nabla g(y)|^2dy $$ Since $$ \int_{\Omega-x_0}|\nabla g(y)|^2dy=\int_{\Omega}|\nabla u(x)|^2dx $$ then $$ \int_\Omega\frac{u^2(x)}{|x-x_0|^2}dx\leq \left(\frac{n-2}{2}\right)^2\int_\Omega|\nabla u|^2dx,\quad\forall x_0\in\Omega. $$

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