In as yet unwritten work with T. Figiel and A. Szankowski we make an observation in a Banach space context that for Hilbert spaces reduces to:

If $T$ is a quasi-nilpotent (i.e., has only $0$ in its spectrum) trace class operator on a Hilbert space, then $T$ is the limit, in the trace class norm, of finite rank nilpotent operators.

Is this a known result? The proof is simple but employs tools that I think are not generally used when studying nilpotents, so the observation might have been overlooked. A Google search produced nothing promising. My colleagues R. Douglas, C. Foias, and C. Pearcy, as well as G. Weiss, did not know the result, but that is not conclusive as the first three are even older than I and Gary never worked on quasi-nilpotent operators.

The general topic of quasi-nilpotents as limits of nilpotents has been studied a lot. For a start, click



1 Answer 1


This question appears as the third exercise in the third chapter of Kenneth Davidson's textbook Nest Algebras. Based on the material presented in the third chapter, here is the intended proof (with the numbers referencing the text).

Let $Q$ be a quasinilpotent trace class operator. Since $Q$ is compact, there exists a maximal nest $\mathcal{N}$ of invariant subspaces of $Q$ (Corollary 3.2). Let $\mathbb{A}$ be the collection of (one-dimensional) atoms of $\mathcal{N}$. If $A \in \mathbb{A}$, the map $T \mapsto P(A)T|_A$ is an algebra homomorphism of the nest algebra $\mathcal{T}(\mathcal{N})$ into $\mathbb{C}$ and thus $P(A) Q|_A$ is the zero scalar for all $A \in \mathbb{A}$ as $Q$ is quasinilpotent (see 3.3).

By the Erdos Density Theorem (Theorem 3.11) there exists a net $(R_\lambda)$ of finite-rank contractions in $\mathcal{T}(\mathcal{N})$ that converge to the identity in the strong-$\ast$ topology. Since $Q^\ast$ is also a trace class operator, it is easy to verify that $R^\ast_\lambda Q^\ast$ converges to $Q^\ast$ in the trace class norm (Proposition 1.18). Thus $QR_\lambda$ converges to $Q$ in the trace class norm.

Since each $QR_\lambda$ is a finite-rank operator, it suffices to show that each $QR_\lambda$ is nilpotent. Since each $R_\lambda$ is a finite-rank element of $\mathcal{T}(\mathcal{N})$, each $R_\lambda$ is a finite sum of rank one operators of the form $xy^*$ where $y \in N^\bot$ and $x \in N_+$ for some element $N \in \mathcal{N}$ ($N_+$ being the successor of $N$) (see 3.7 and 3.8). If we multiply $R_\lambda$ by $Q$, we obtain a similar decomposition of $QR_\lambda$ as a sum of such operators (where, if $x \in N_+$, $Qx \in N_+$ as $Q \in \mathcal{T}(\mathcal{N})$). However, if $N_+ \neq N$, $N_+ \ominus N$ is an atom so if $x \in N_+$, $Qx \in N_+$ and thus $Qx \in N$ as $Q$ is the zero scalar on all atoms. Hence each $QR_\lambda$ is a finite sum of rank one operators of the form $xy^*$ where $y \in N^\bot$ and $x \in N$ for some element $N \in \mathcal{N}$. Since $\mathcal{N}$ is a nest, it is then easy to see that each $QR_\lambda$ is a nilpotent operator.

  • 2
    $\begingroup$ Thanks, Paul. That is exactly what I was looking for. This is similar to the proof we came up with, except we avoid using any special properties of Hilbert space, so that the proof works in Banach spaces that satisfy the conclusion of the Erdos Density Theorem (there are some, including all weak Hilbert spaces). $\endgroup$ Apr 27, 2012 at 15:51

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