Let $L$ an operator self-adjoint acting on $L^2(\Bbb{R}^{2})$ such that :

  1. $L(\phi_{\alpha,\beta})=(|\alpha|-|\beta|)(\phi_{\alpha,\beta})$ where $(\phi_{\alpha,\beta})$ is an orthonormal basis for $L^2(\Bbb{R}^{2})$ with $ \alpha=(\alpha_1,\alpha_2)\in\Bbb{N}^{2}$ and $|\alpha|$ its length $\alpha_1+\alpha_2$.
  2. There is a sequence $(u_n)\in L^2(\Bbb{R}^{2})$ such that :

for $p\in \Bbb{N}$ we have $(L-p)u_n\overset{L^2(\Bbb{R}^{2})}{\longrightarrow} 0$ and $ u_n\overset{weakly}{\longrightarrow} 0$.

From 1, the spectrum of the operator $L$ is the set $\sigma(L)=\{m;m\in\Bbb{Z}\}$. Now, for the above $p$ there are $\alpha, \beta$ such $p=(|\alpha|-|\beta|)$ and $L(\phi_{\alpha,\beta})=p(\phi_{\alpha,\beta})$

My question is the following: Is there a contradiction between $L(\phi_{\alpha,\beta})=p(\phi_{\alpha,\beta})$ and the fact 2.

Thanks in advance for any help.

  • $\begingroup$ I forget the condition : $ ||u_n||=1$. $\endgroup$ – A.Zoran Aug 17 '17 at 9:25

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.