Let $A$ be a bounded self-adjoint (or normal) operator on Hilbert space $H$. Weyl's criterion states that $\lambda \in\sigma(A)$ iff there exists sequence $\{x_n\}\subset H$ such that $\|x_n\|=1$ and $\|Ax_n-\lambda x_n\|\rightarrow 0$.
What I'm looking for is "counterexamples" to this statement. By that I mean an example of not self-adjoin operator for which the statement does not hold. The structure of proof makes me think that this should be like that: we should observe that $\lambda\in\sigma(A)$ BUT the sequence above does not exist, but I struggle with constructing the example itself.
So, any hint appreciated!