About vectors that have bounded support representation on the spectrum of a self-adjoint operator Let $A$ be an unbounded operator densely defined on an Hilbert space $X$.  Let $A^*$ be its adjoint. Furthermore, suppose that the commutator satisfies $[A,A^*]=1$.
If $\psi$ is an $\lambda$-eigenvector of $N:= A^* A$, then  $A\psi$, is an $(\lambda-1)$-eigenvector, indeed 
$$N A\psi = A N\psi - A\psi = \lambda A \psi- A\psi = (\lambda -1) A\psi.$$
Therefore since $N\geq0$ its spectrum $\sigma(N) \subset [0, +\infty)$ and  there exists some $n\geq 0$ such that $A^n \psi = 0$.
This is all clear.
Now, the author of the book that I am reading says that 

"The same result is still valid if instead of an eigenvector we consider a vector $\psi$ such  that it has a bounded support representation on the spectrum of $N$ ". What does this means  and how we can show that $A^n\psi = 0$ for some $n\in \mathbb{N}$?

I assumed that he means that $\chi_{[0,L]}(N) \psi = \psi$  where $\chi_{[0,L]}$ is the characteristic function of $[0,L]$ and $\chi_{[0,L]}(N)$ is the operator obtained through the Borelian Calculus ($N$ is symmetric, but I suppose maybe it can be shown that it is also self-adjoint).
From this I don't know how to conclude.  I tried to prove -without success-   that $\chi_{[0,L-1]}(N) A\psi = A\psi$.
 Maybe my interpretation is wrong.
 A: "Vector $\psi$ has a bounded support representation on the spectrum of $N$" means that $\psi$ belongs to the range of the spectral projection $E_N(\Omega)$ where $\Omega\subseteq\mathbb R$ is a bounded Borel set.
More generally, the equation $NA=A(N-I)$ implies $$f(N)A=Af(N-I).$$ 
This equation can be easily proven for polynomials, then for continuous functions and even for Borel functions $f$. Since the spectral projection $E_N(\Omega)$ of $N$ is the Borel function $\mathbf 1_\Omega(N)$ we have 
$$E_N(\Omega)A=AE_{N-I}(\Omega)=AE_N(\Omega+1),$$ or equivalently
$$AE_N(\Omega)=E_N(\Omega-1)A,$$ and $$\ A^nE_N(\Omega)=E_N(\Omega-n)A^n,\ n\in\mathbb N$$
Positivity of $N$ implies that the support of the spectral measure $E_N$ is contained in $\mathbb R_+$. Hence $E_N(\Omega-n)=0$ for a bounded set $\Omega$ and $n$ large enough. It implies $A^n\psi=0$ for $\psi$ in range of $E_N(\Omega)$ and $n$ large enough.
Edit. Here is the (sketch of the) proof of the equality $E_N(\Omega)A=AE_{N-I}(\Omega)$. Write the polar decomposition $A=UC$. Then $AA^*=I+A^*A$ implies 
$$UC^2U^*=I+C^2,\ C^2=N$$ or equivalently 
$$U(N-I)U^*=N.$$ 
Consider $U^*$ as a unitary operator from $(\ker A^*)^\perp$ to $\ker A^\perp$. Then the last equality can be viewed as a unitary equivalence of (restrictions of) operators $N$ and $N-I$. The same unitary equivalence holds also for spectral projections, i.e.
$$E_N(\Omega)=UE_{N-I}(\Omega)U^*\iff E_N(\Omega)U=UE_{N-I}(\Omega).$$
Now multiply both sides by $C$ and you get using $E_{N-I}C=CE_{N-I}$
$$E_N(\Omega)A=AE_{N-I}(\Omega).$$
