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There's a theorem I should prepare for the oral exam, however I cannot find it anywhere in the recommended literature.

I should prove the following:

Let X be a complex Hilbert space and $A \in B(X), \lambda \in \sigma_c(A)$. Then, there exists a sequence $(x_n)_{n \in \mathbb{N}}$ in X such that $\lim_{n \to \infty}(\lambda I - A) x_n = 0$ and $|| x_n|| =1$ for every $n \in \mathbb{N}$.

I'd approach this by constructing such a sequence, but I'm not sure how to.

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That's the definition of $\sigma_c(A)$. $\lambda$ is in the continuous spectrum if $\mathcal{N}(\lambda I-A)=\{0\}$, but the inverse operator is not bounded. Because the inverse is not bounded, there is a sequence $\{ x_n \}$ of unit vectors in the domain of $\lambda I-A$ such that $(\lambda I-A)x_n\rightarrow 0$.

Example: Let $X=L^2[0,1]$ and let $A : X\rightarrow X$ be the multiplication operator $A : X \rightarrow X$ defined by $(Af)(x) = xf(x)$. Then the spectrum of $A$ is $[0,1]$, and it is entirely continuous spectrum. If you choose $\lambda \in (0,1)$, then you can see that you nearly have an eigenvector $x_{\lambda,\epsilon}$ with eigenvalue $\lambda$ defined by $$ x_{\lambda,\epsilon}=\chi_{[\lambda-\epsilon,\lambda+\epsilon]}. $$ This is because $$ \|(A-\lambda I)x_{\lambda,\epsilon}\|^2 = \int_{0}^{1}(x-\lambda)^2\chi_{[\lambda-\epsilon,\lambda+\epsilon]}(x)^2dx \\ \le \epsilon^2\int_0^1\chi_{[\lambda-\epsilon,\lambda+\epsilon]}(x)^2 dx = \epsilon^2\|\chi_{\lambda,\epsilon}\|^2. $$ That is, $$ \|(A-\lambda I)x_{\lambda,\epsilon}\| \le \epsilon\|x_{\lambda,\epsilon}\|. $$ So every $\lambda\in[0,1]$ is an approximate eigenvalue, but not an actual one. I'll let you consider the case of other Lebesgue measures, not just absolutely continuous ones.

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  • $\begingroup$ We've defined the continuous spectrum a little bit different (using closure). Also, had no idea about the sequence of unit vectors. I'll try to prove these two you've stated, and hopefully that would solve my issue. Thanks $\endgroup$ Aug 5 at 21:26
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    $\begingroup$ @NikolaBurbakić : I made a correction to my post. I wrote that the unit vectors were in the range. I meant that they were in the domain, so that $(\lambda I-A)x_n$ makes sense. This is sometimes referred to as having an approximate eigenvalue $\lambda$. $\endgroup$ Aug 5 at 22:03
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    $\begingroup$ Yes, indeed, for self-adjoint (bounded) operators, or at least "normal" operators, so there's no (annoying) spectrum beyond eigenvalues and "continuous" spectrum, every point in the continuous spectrum can be approximated by "approximate eigenvalues", with "approximate eigenvectors". I think this was discovered by H. Weyl. $\endgroup$ Aug 5 at 22:06

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