TITLE: Approximate eigenvalue and continuous spectrum
QUESTION [7 upvotes]: Let $\mathcal{H}$ be a Hilbert space and let $A: \mathcal{H} \rightarrow \mathcal{H}$ be a bounded operator. While studying different definitions of the continuous spectrum of $A$ (one using approximate eigenvalues) I wanted to prove the following equivalence:
Suppose that $\lambda \in \sigma(A)$. Then
$$
\mathrm{Im}(A-\lambda I) \text{ is dense in }\mathcal{H} \iff \lambda \text{ is an approximate eigenvalue of $A$.}
$$
However, I am having some difficulties with the "$\Rightarrow$" direction. Obviously I need to find a sequence $(v_n)$ of elements in $\mathcal{H}$ such that $\|v_n\| = 1$ and $\|Av_n - \lambda v_n\| \longrightarrow 0$, but I do not see how to start, so I would really appreciate any hint. Thanks in advance.

REPLY [5 votes]: Since $\lambda \in \sigma (A)$, the operator $A-\lambda I$ has no bounded inverse. Hence $$\inf_{|v|=1} |Av-\lambda v|=0.$$ Otherwise, you could boundedly define $(A-\lambda I)^{-1}$ on the dense subspace $\operatorname{Im}(A-\lambda I)$, and $\lambda \in \rho (A)$.