Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

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.

share|improve this answer
    
Thank you, that is exactly what I was looking for! –  kensentme Apr 24 '12 at 12:34
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.