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Assume $\mathbb{‎‎H}=W^{2,2}(\Omega) \cap W^{1,2}_0(\Omega)$ with the norm induced from inner product $$\langle u,v\rangle_{‎\mathbb{‎‎H}}=\int_\Omega \Delta u \,\Delta v\, dx$$ for any $u \in W^{2,2}(\Omega) \cap W^{1,2}_0(\Omega)$ it is well-known that we have inequality (Rellich inequality) $$ ‎\Lambda_N ‎\int_{\Omega}‎\frac{u^2}{|x|^4}\mathrm{d}x ‎\leq ‎\int_\Omega ‎|\Delta u|^2 \, ‎\mathrm{d}x‎ $$ where $‎\Lambda_N=(‎\frac{N^2(N-4)^2}{16})‎$ is optimal constant and also it is known that $$\|u\|^2=‎\int_\Omega ‎‎\Big(|\Delta u|^2-\Lambda_N \frac{u^2}{|x|^4}\Big) \, \mathrm{d}x $$ ‎ Defines an other norm on $W^{2,2}(\Omega) \cap W^{1,2}_0(\Omega)$.

My question is this that is these two norm equivalent? To see an improved case of this inequality with reminder term see Corollary 1 of Paper.

I know if I could show that $W^{2,2}(\Omega) \cap W^{1,2}_0(\Omega)$ is a hilbert space with the new inner product then this two norms must be equivalent due to Open mapping theorem.

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  • $\begingroup$ Notice the difference between $||a||||b||$ and $\|a\|\|b\|$. I set $\|u\|^2$ in the latter format, which is standard, where it had been in the former. ${}\qquad{}$ $\endgroup$ – Michael Hardy Aug 18 '15 at 13:16

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