# Spectrum equals eigenvalues for unbounded operator

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 $$n \to \infty$$.

• @Sascha: Why is this clear? What is a counter example? – Bas Winkelman Jan 11 at 12:45
• "Diagonalizable" means what? $D = U^* E U$ where $U$ is a (bounded) unitary operator and $E$ is a diagonal operator (with diagonal entries unbounded, but possibly not converging to $\infty$) ?? If that is the definition, then any limit point of the eigenvalues is in the spectrum, but possibly not an eigenvalue itself. – Gerald Edgar Jan 11 at 12:59
• I don't see that the counterexamples proposed in previous comments satisfy the requirements in the problem: "diagonalizable with all eigenvalues having finite multiplicity and growing toward infinity." – Andreas Blass Jan 11 at 14:30

## 2 Answers

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.

Let $$H$$ be an infinite dimensional Hilbert space. For each $$n\in\mathbb{Z}-\{0\}$$ let $$r_n = \begin{cases} -\frac{1}{n},\ &n<0\\ n,\ &n>0\end{cases}$$

Choose a countable orthonormal basis $$e_n$$ and define $$T$$ by $$Te_n = r_ne_n$$. Then $$T$$ is diagonalizable with eigenvalues $$r_n$$ of multiplicity one, which can be ordered so that they are a strictly increasing unbounded set (i.e. "growing to infinity").

Because the spectrum of $$T$$ is closed, it must contain zero, but zero is not an eigenvalue of $$T$$.

• I interpreted "growing toward infinity" in the question to mean that the eigenvalues form an ordinary sequence, of order-type $\omega$, tending to $\infty$. But I admit that your interpretation, allowing order-type $\omega+\omega$, is also consistent with the wording in the question. – Andreas Blass Jan 11 at 14:47