most recent 30 from http://mathoverflow.net 2013-05-26T03:59:41Z http://mathoverflow.net/feeds/user/23241 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86102/quasi-nilpotent-trace-class-operators-as-limits-of-nilpotents/95313#95313 Paul Skoufranis 2012-04-26T23:57:43Z 2012-04-26T23:57:43Z

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