Let $D$ be an unbounded densely defined operator on a separable Hilbert space $H$. If $D$ is diagonalisable with all eigenvalues having finite multiplicity and growing towards infinity, does it follow that the spectrum of $D$ contains only eigenvalues?


EDIT: Diagonalisable means the existence of a countable ONB $e\_n$ (orthogonal set with dense span) such that $D(e\_n) = \alpha\_n e\_n$, for some complex number $\alpha\_n$. 


Growing towards infinity means that $\alpha\_n \to \infty$ as $n \to \infty$.



Spectrum equals eigenvalues for unbounded operator
I agree with Andreas that the obvious straightforward interpretation of "the eigenvalues grow to infinity" is that the sequence of eigenvalues $(\lambda\_n)$ increases to infinity. (And, counter to Sascha's interpretation, that "diagonalisable" means that the eigenvectors form a basis for $H$.) Under this interpretation the answer is yes. It is standard that the spectrum of the normal operator $M\_f$ (multiplication by $f$, on any $L^2$ space) is the essential range of $f$. On a discrete space, this would simply be the closure of the range of $f$, and the hypothesis given here assures that this range has no cluster points.


