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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.

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up vote 2 down vote accepted

So we assume that the range of $A-\lambda I$ is dense, and we can assume that $\lambda$ is not an eigenvalue (since otherwise the result is trivial). But then $A-\lambda I$ has trivial kernel, and it is not invertible -- because $\lambda\in\sigma(A)$ --, so we conclude that $A-\lambda I$ is not bounded below; this is exactly what you are looking for.

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Thank you, that is exactly what I was looking for! – kensentme Apr 24 '12 at 12:34

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 $\lambda \in \rho (A)$.

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