Is this reasoning is correct:
Let $R$ be a self adjoint differential operator, acting on $L^2(\Bbb{R}^{2d}), d\in \Bbb{N}^*$, such that $R$ is diagonalizable in an orthonormal basis $(\Phi_{\alpha,\beta})_{(\alpha,\beta)\in \Bbb{N}^{d}}$ with
$$ R \Phi_{\alpha,\beta}= (|\alpha|-|\beta|)\Phi_{\alpha,\beta}$$
where $\alpha=(\alpha_1,\dots,\alpha_d)$ and $|\alpha|=|\alpha_1|+\dots+|\alpha_d|$.
Let $m\in \Bbb{Z}$ . I want to show that if there is a sequence $(u_{n})$ (depends on $m$) $L^2(\Bbb{R}^{2d})$ and $R(u_n)\in L^2(\Bbb{R}^{2d})$ such that
1]- $u_{n}\overset{Weakly}{\longrightarrow} 0$ i.e., for $f\in L^2(\Bbb{R}^{2d})$ we have $<f,u_{n}>\to 0$ as $n\to\infty$.
2]- $\big(R- m) u_{n} \big) \overset{L^2(\Bbb{R}^{2d})}{\longrightarrow} 0$ as $n\to\infty$.
3]- $||u_n||=1$ for all $n\in \Bbb{N}$.
Then $u_{n}=c\Phi_{\alpha,\beta}$ with $|c|=1$ and $\alpha-\beta=m$.
Here is my reasoning:
Since $u_{n}\in L^2(\Bbb{R}^{2d})$, then $u_{n}=\sum_{\alpha',\beta'} c^{n}_{\alpha',\beta'}\Phi_{\alpha',\beta'}$ (The convergence of this series being in $L^2(\Bbb{R}^{2d})$) .
So $$\big(R- m) u_{n} =\sum_{\alpha',\beta'} c^{n}_{\alpha',\beta'}\big((|\alpha'|-|\beta'|) - m\big) \Phi_{\alpha',\beta'}$$
(because $ R \Phi_{\alpha',\beta'}= (|\alpha'|-|\beta'|)\Phi_{\alpha',\beta'}$ et $(\sum g_k)'= (\sum (g_k)')$.
Hence $$||\big(R- m) u_{n}||^2=\sum_{\alpha',\beta'} \Big|c^{n}_{\alpha',\beta'}\big((|\alpha'|-|\beta'|) - m \big) \Big|^2$$.
Now by hypothesis $1$, for $f=\Phi_{\alpha',\beta'}$ (Here, we point out that $u_n$ depends on $ m=|\alpha|-|\beta|$, so it is necessary that $(\alpha',\beta')\not=(\alpha,\beta)$).
Therefore :$$<u_{n},\Phi_{\alpha',\beta'}>=c^{n}_{\alpha',\beta'} \to 0$$
as $n\to\infty$.
But $||u_{n}||^2=\sum_{\alpha',\beta'}| c^{n}_{\alpha',\beta'}|^2=1$
I take $n\to\infty$ for both members, and consequently necessarily we get:
$|c^{n}_{\alpha,\beta}|=1$
$ c^{n}_{\alpha',\beta'}=0$ avec $(\alpha',\beta')\not=(\alpha,\beta)$.
Finaly: $u_{n}=c\Phi_{\alpha,\beta}$ with $|c|=1$ and $\alpha-\beta=m$.
Thanks in advance.