Constructing an approximation of an eigenvector 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.
 A: 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.
