# How to show contradiction in the Hardy inequality when the singularity power is greater that 2.

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.

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 $\Lambda(\lambda^{-1}r)\to\infty$.