Consider the following theorem.

Theorem: Let $A$ be a bounded self-adjoint operator on a Hilbert space $\mathcal{H}$. Then $\lambda \in \sigma(A)$ if, and only if there exists a sequence $\{\psi_{n}\}_{n\in \mathbb{N}}$ such that $||\psi_{n}|| = 1$ for all $n$ and $\lim_{n\to \infty}||(A-\lambda)\psi_{n}|| \to 0$.

This is a 'part' of the so-called Weyl's criterion. Usually, this result arises when studying spectral theory of Fredholm operators, where the essential spectrum plays a central role. In general, the above theorem comes together with some statement(s) concerning the essential spectrum of $A$ too, so that the above result is proved by using properties of the essential spectrum.

However, I'm not interested in Fredholm operators, but rather in bounded self-adjoint operators alone.

Question: How can I prove (or where can I find the proof) of the above Theorem using just the usual spectral theory of bounded self-adjoint operators and not using essential spectrum arguments?


1 Answer 1


I find it easiest to prove the contrapositive, $\lambda \in \rho(A)$ if and only if no such sequence $(\psi_n)$ exists.

First we observe that no such sequence $(\psi_n)$ exists if and only if there is a $\delta > 0$ such that $$\lVert (A - \lambda)x\rVert \geqslant \delta \lVert x\rVert$$ for all $x \in \mathcal{H}$.

Next, this form of the condition immediately implies that $A - \lambda$ is injective, and it quickly implies that the range of $A - \lambda$ is closed. Conversely, the open mapping theorem implies that if $A - \lambda$ is injective with closed range, then such a $\delta$ exists.

It remains to be seen that for a self-adjoint $A$, if $A - \lambda$ is injective with closed range, then it is surjective. That follows since $A - \lambda$ is normal (and thus this criterion holds more generally for normal operators, since $A - \lambda$ is normal iff $A$ is normal). Thus suppose $x \in \bigl(\operatorname{im} (A - \lambda)\bigr)^{\perp}$. We then have $$\lVert (A - \lambda)x\rVert^2 = \langle (A - \lambda)x, (A - \lambda)x\rangle = \langle (A - \lambda)^{\ast}(A - \lambda) x, x\rangle = \langle (A - \lambda)(A - \lambda)^{\ast} x, x\rangle = 0$$ by definition of the adjoint and normality. Since by assumption $A - \lambda$ is injective it follows that $x = 0$, hence $\operatorname{im} (A - \lambda)$ is dense, and since it is by assumption closed, that $A - \lambda$ is surjective.

  • $\begingroup$ Amazing! Thank you so much! $\endgroup$
    – Idontgetit
    Aug 23, 2020 at 19:27

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