Let $A$ be a self-adjoint operator on a Hilbert space $H=\big(L^2(\Bbb{R}^2),||.||\big)$, assumed to be unbounded. We have the following characterization of its spectrum $\sigma(A)$:
$\lambda\in\sigma(A)$ if and only if $\exists (u_n)\in D(A), \|u_n\|=1,$ such that $\lim_n \|(A-\lambda)u_n\|=0;$
$\lambda$ is an eigenvalue of $A$ with infinite multiplicity if and only if there exists a sequence of linearly independent vectors $u_i\in D(A)$ such that $(A-\lambda)u_i=0;$
of course $2$ is the form of $1$.
Now we give another characterization of the essential spectrum of $A,$ denoted $\sigma_{ess}.$$\lambda\in\sigma_{ess}(A)$ if and only if $\exists (u_n)\in D(A), \|u_n\|=1,$ $u_n\to 0$ weakly, such that $\lim_n \|(A-\lambda)u_n\|=0.$
I have self-adjoint operators $T=-\Delta_{\Bbb{R}^2}-(x^2+y^2)$ and $R=i(x\partial_y-y\partial_x)$ such that
(1) $\:\sigma(T)=\sigma_{ess}(T)=\Bbb{R}$ and the point spectrum of $T$ is empty; i.e., $\sigma_p(T)=\emptyset.$
(2) $\:\sigma(R)=\sigma_{ess}(R)=\sigma_p(R)=\Bbb{Z}$ (here any eigenvalue of $R$ has infinite multiplicity.)
(3)Note that $RT=TR$.
My question: Let $a\in \sigma(T)=\sigma_{ess}(T)=\Bbb{R}$ and $m\in \sigma(R)=\sigma_{ess}(R)=\sigma_p(R)=\Bbb{Z}.$ Can I find a sequence $(u_n)\in D(R)\cap D(T),\quad\|u_n\|=1,\quad u_n\to 0$ weakly, such that $\lim_n \|(T-a)u_n\|=\lim_n \|(R-m)u_n\|=0?$. I think there is no such sequence.
Thanks in advance.