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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 $n \to \infty$.

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  • $\begingroup$ @Sascha: Why is this clear? What is a counter example? $\endgroup$ – Bas Winkelman Jan 11 at 12:45
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    $\begingroup$ "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. $\endgroup$ – Gerald Edgar Jan 11 at 12:59
  • $\begingroup$ 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." $\endgroup$ – Andreas Blass Jan 11 at 14:30
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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.

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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 $T$.

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    $\begingroup$ 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. $\endgroup$ – Andreas Blass Jan 11 at 14:47

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