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Consider the following theorem.

Theorem: Let $A$ be a bounded self-adjoint operator on a Hilbert space $H$. Then$ λ \in σ(A) $ if, and only if there exists a sequence $ ({\psi_{n}})_{n\in \mathbb{N}} $ such that $\|\psi_{n}\|=1 $ for all $ n $ and $$\lim_{n\rightarrow \infty } \|(A−\lambda)\psi_{n}\|\rightarrow 0. $$ This is a 'part' of the so-called Weyl's criterion.

Question: is this result true in the case of an unbounded and a non-self-adjoint operator?

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  • $\begingroup$ Weyl's Criterion stays true for unbounded selfadjoint operators, see for example Wikipedia. Since it is in fact usually formulated as a criterion for the essential spectrum, I don't think there is a generalization to non-selfadjoint operators. $\endgroup$
    – Benjamin
    Oct 25, 2021 at 12:34
  • $\begingroup$ It should be true for normal operators. $\endgroup$ Oct 31, 2021 at 16:05

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It's not true of a bounded, non-self-adjoint operator. For example, consider the shift operator on $\ell^2(\mathbb{N})$: $$ S\{x_1,x_2,x_3,\cdots\}=\{0,x_1,x_2,x_3,\cdots\}. $$ $0$ is in the spectrum of $S$ because $S$ is not surjective; indeed, $\{ 1,0,0,0,\cdots \}$ is not in the range of $S$ (in fact, it is orthogonal to the range.) However, there is no sequence $\{ w_n \}$ of unit vectors in $\ell^2(\mathbb{N})$ such that $\|(S-0)w_n\|\rightarrow 0$, which is obvious because $\|Sw_n\|=\|w_n\|=1$ for all $n$.

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    $\begingroup$ +1. I have a question. Is the criterion true for normal operators? I guess it is. Indeed, the operator $S$ of this example is not normal. $\endgroup$ Oct 31, 2021 at 16:08
  • $\begingroup$ @GiuseppeNegro : Post a new question concerning the case of a normal operator. $\endgroup$ Nov 4, 2021 at 4:15

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