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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.

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

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