# Approximation of vectors using self-adjoint operators

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.

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

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.