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Assume $ \Omega \subset \mathbb{R}^N$ is a smooth bounded domain. There is well known Hardy inequality that says

For any $ u \in W_0^{1,2}(\Omega) $, $N\geq3$ we have $$ \Lambda \int_{\Omega} \frac{u^2}{|x|^2} \, \mathrm{d}x \leq \int_{\Omega} |\nabla u|^2 \, \mathrm{d}x $$ Where $ \Lambda=\frac{(N-2)^2}{4} $ is optimal constant.

Now I am asked to prove that for any $p>2$ and $u \in C_0^{\infty}(B_r(0))$ $$ \Lambda \int_{B_r(0)} \frac{u^2}{|x|^p} \, \mathrm{d}x \leq \int_{B_r(0)} |\nabla u|^2 \, \mathrm{d}x $$ contradicts with the Hardy inequality. Where $B_r(0)$ is the ball centered at zero for some radius $r$ sufficiently small.

My try: $$ \int_{B_r(0)} \frac{u^2}{|x|^p} \, \mathrm{d}x=\int_{B_r(0)} \frac{1}{|x|^{p-2}} \frac{u^2}{|x|^2} \, \mathrm{d}x $$

for some $r$ sufficiently small, since $p>2$, we can say that

$$\frac{1}{|x|^{p-2}} > \Lambda_N$$

So this contradicts with the optimality of $\Lambda_N$ in the Hardy inequality.

Can some one give another approach.

Thanks.

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Another approach would be to consider how both sides scale under the transformation $x = \lambda y$, $\lambda>0$. By the change of variables and the chain rule, it becomes $$ \Lambda \int_{\lambda^{-1}\Omega} \frac{u^2}{\lambda^p |y|^p} \, \lambda^n\mathrm{d}y \leq \int_{\lambda^{-1}\Omega} \lambda^2 |\nabla u|^2 \lambda^n\, \mathrm{d}x $$ Thus, if $\Lambda(r)$ denotes the optimal constant for $\Omega=B_r$, we have $\Lambda(\lambda^{-1}r)=\lambda^{p-2}\Lambda(r)$. As $\lambda\to \infty$, this shows that $\Lambda(\lambda^{-1}r)\to\infty$.

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