# Proof of Weyl's criterion not using essential spectrum

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?

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.