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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.

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  • $\begingroup$ Sorry, why can't we have $k(\varphi)=\|T\varphi\|$, and $\varphi_n=(1-\frac{1}{2^n})\varphi$? $\endgroup$ – fierydemon May 25 '20 at 23:26
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    $\begingroup$ @fierydemon: $\varphi$ is not assumed to be in the domain of $T$. $\endgroup$ – Nate Eldredge May 26 '20 at 0:06
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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.

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