# Weyl's criterion on unbounded operators

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?

• 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. Oct 25, 2021 at 12:34
• It should be true for normal operators. Oct 31, 2021 at 16:05

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 $$n$$.

• +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. Oct 31, 2021 at 16:08
• @GiuseppeNegro : Post a new question concerning the case of a normal operator. Nov 4, 2021 at 4:15