I am reading an article and there, author claim that $$L(.)=\Delta ^2(.)-\frac{\lambda}{|x|^4}(.): W^{2,2}(\Omega) \cap W^{1,2}_0(\Omega) \to W_0^{-2,2}(\Omega) $$ is coercive if $ 0\leq \lambda<\Lambda_N=(\frac{N^2(N-4)^2}{16})$ , because of the following inequality:
For all $ u \in W^{2,2}(\Omega) \cap W^{1,2}_0(\Omega) $ و $N>4$ $$ \Lambda_N \int_{\Omega}\frac{u^2}{|x|^4}\mathrm{d}x \leq \int_{\Omega} |\Delta u|^2 \mathrm{d}x $$ where $\Lambda_N=(\frac{N^2(N-4)^2}{16})$ is optimal constant.
My question is this that how the inequality with the optimal constant conclude that operator is coercive for $ 0\leq \lambda<\Lambda_N=(\frac{N^2(N-4)^2}{16})$.
my try:
the norm on the hilbet space $\mathbb{H}=W^{2,2}(\Omega) \cap W^{1,2}_0(\Omega)$ is $$\langle u,v\rangle_{\mathbb{H}}=\int_{\Omega} \Delta u \Delta v dx$$
I must show that $$\langle Lu,u \rangle_{\mathbb H} \geq c ||u||_{\mathbb H}^2 $$ for a positive constant $c$.
with simple calculations and using above inequality, I have showed that $$\langle Lu,u \rangle_{L^2} \geq c ||\Delta u||_{L^2}=c||u||_{\mathbb H}^2 $$ for a positive $ c $
but I must show that $$\langle Lu,u \rangle_{\mathbb H} \geq c||u||_{\mathbb H}^2 $$ .
It appears that you are using the notation $\langle \cdot, \cdot \rangle_{\mathbb{H}}$ for the inner product on you Hilbert space $W^{2,2}(\Omega)\cap W^{1,2}_0(\Omega)$ and $\langle \cdot,\cdot\rangle_{L^2}$ for the dual pairing with the space $W^{-2,2}_0(\Omega)$, which is the co-domain of the operator $L$. The term $\langle Lu,u\rangle_{\mathbb{H}}$ then does not make sense unless $u$ has more differentiability and even then will not be coercive.
The appropriate bi-linear form for coercivity, i.e. for existence of solutions via the Lax-Milgram theorem, is $\langle Lu,u\rangle_{L^2}$ which you have already shown is coercive.