Let $T$ be an unbounded self-adjoint operator.


Does there exist, for any $\varphi$ normalized in the Hilbert space, a constant $k(\varphi)>0$ and a sequence of normalized $(\varphi\_n)$ such that $$ \lim\_{n \rightarrow \infty} \Vert \varphi-\varphi\_n \Vert=0 $$
and $\Vert T \varphi\_n \Vert \le k(\varphi).$


Somehow this looks strange, if you think of $\varphi \notin D(T)$ as then $$\Vert T\varphi \Vert"="\infty$$


all of a sudden, on the other hand, maybe you only have to choose $k$ in a suitable way.



Approximation of vectors using self-adjoint operators
No, this only holds when $\varphi \in D(T)$.


It's enough to assume that $T$ is closed and densely defined, so that $T^{\*\*} =T$. Let $\psi \in D(T^\*)$ be arbitrary. If the hypothesis holds then we have
$$ |\langle \varphi, T^\* \psi \rangle| = \lim\_{n \to \infty} |\langle \varphi\_n, T^\* \psi \rangle| = \lim\_{n \to \infty} |\langle T \varphi\_n, \psi \rangle| \le k(\varphi) \|\psi\|$$
which is exactly the definition of $\varphi \in D(T^{\*\*}) = D(T)$. One can also see that in fact $T \varphi\_n \to T \varphi$ weakly.


Note the argument still goes through even if we only assume that $\varphi\_n \to \varphi$ weakly instead of strongly.


