Does absolutely convergent trace series imply that the operator is of trace class? [duplicate]

Let $\mathscr{H}$ be a separable Hilbert space and let $T\in B(\mathscr{H})$ be such that $\displaystyle \sum_{j=1}^\infty\langle T\xi_j,\xi_j\rangle$ is absolutely convergent for any choice of orthonormal basis $\{\xi_j\}$. Does this imply that $T$ is trace-class?

The converse is true and easy to prove. It is a question from Conway's book on Operator theory.

• A real series for which all rearrangements converge is absolutely convergent, and you can look separately at real and imaginary parts of $\langle T\xi_j,\xi_j\rangle$. Nov 30 '16 at 0:13
• The assertion is false for real Hilbert spaces.
– Zero
Dec 27 '16 at 10:21
• @Anton What is a counter-example? Dec 27 '16 at 17:35
• On $\ell^2({\mathbb N})$ its the operator that maps $e_j$ to $T(e_j)=\frac{(-1)^{j+1}}{j} e_{j+(-1)^{j+1}}$.
– Zero
Dec 27 '16 at 19:39

For convenience of language, let us say that an operator that satisfies your hypothesis is traceable, that is, a bounded linear map $T\colon V\rightarrow V$ on a separable infinite-dimensional Hilbert space $V$ is traceable iff $\sum _{m\in \mathbb{N}}\left| \langle e_m|T(e_m)\rangle \right| <\infty$ for all orthonormal bases $\{ e_m:m\in \mathbb{N}\}$ of $V$.
Assume we've proven the result for self-adjoint operators. If $T$ is traceable, then so are $\Re (T)$ and $\Im (T)$, and hence if we've proven the result for self-adjoint operators, we will have that $\Re (T)$ and $\Im (T)$ are trace-class, and hence $T=\Re (T)+\mathrm{i}\Im (T)$ is trace-class. So, without loss of generality assume that $T$ is self-adjoint.
We next reduce to the case of invertible self-adjoint operators. Assume we've proven the result for invertible self-adjoint operators. As $V=\operatorname{Ker}(T)\oplus \operatorname{Ker}(T)^{\perp}=\operatorname{Ker}(T)\oplus \operatorname{Cls}(\operatorname{Im}(T^*))=\operatorname{Ker}(T)\oplus \operatorname{Cls}(\operatorname{Im}(T))$, $T$ restricts to an invertible self-adjoint operator on $\operatorname{Ker}(T)^{\perp}=\operatorname{Cls}(\operatorname{Im}(T))$. This restriction is traceable, and hence trace-class. If $\{ e_m:m\in \mathbb{N}\}$ is an orthonormal basis for $\operatorname{Ker}(T)^{\perp}$ and $\{ f_m:m\in \mathbb{N}\}$ is an orthonormal basis for $\operatorname{Ker}(T)$, $\{ e_m,f_m:m\in \mathbb{N}\}$ is an orthonormal basis for $V$, and we have $$\sum _{m\in \mathbb{N}}\langle e_m||T|(e_m)\rangle +\sum _{m\in \mathbb{N}}\langle f_m||T|(f_m)\rangle =\sum _{m\in \mathbb{N}}\langle e_m||T|(e_m)\rangle <\infty .$$ So, without loss of generality assume that $T$ is invertible self-adjoint.
Let $P_T$ be the spectral measure of $T$ and define $P_{\pm}:=P_T(\operatorname{Spec}(T)_{\pm})=P_T(\operatorname{Spec}(T_{\pm}))$, where $\mathbb{R}\ni x\mapsto x_+\in \mathbb{R}$ is the continuous function defined by $x_+:=\begin{cases}x & \text{if }x\geq 0 \\ 0 & \text{otherwise}\end{cases}$, and similarly for $x\mapsto x_-$. Note that, as $T$ is invertible, $\operatorname{Spec}(T)_+\cap \operatorname{Spec}(T)_-=\emptyset$, and so $P_+P_-=0=P_-P_+$. Thus, $V=V_+\oplus V_-$ is an orthogonal direct sum, where $V_{\pm}:=P_{\pm}(V)$. Let $\{ e_{\pm m}:m\in \mathbb{N}\}$ be an orthonormal basis of $V_{\pm}$, so that $\{ e_{+m},e_{-m}:m\in \mathbb{N}\}$ is an orthonormal basis of $V$ such that $T_{\pm}(e_{\mp m})=0$ for all $m\in \mathbb{N}$. Thus, $$\sum _{m\in \mathbb{N}}\langle e_{+m}||T|(e_{+m})\rangle +\sum _{m\in \mathbb{N}}\langle e_{-m}||T|(e_{-m})\rangle =\sum _{m\in \mathbb{N}}\langle e_{+m}|T_+(e_{+m})\rangle +\sum _{m\in \mathbb{N}}\langle e_{-m}|T_-(e_{-m})\rangle =\sum _{m\in \mathbb{N}}\left| \langle e_{+m}|T(e_{+m})\rangle \right| +\sum _{m\in \mathbb{N}}\left| \langle e_{-m}|T(e_{-m})\rangle \right| <\infty ,$$ and so $T$ is trace-class, as desired.
• This argument makes no sense. Over an infinite-dimensional Hilbert space, there are no invertible trace-class operators, as they are all compact. Also, $T|_{(\ker T)^\perp}$ is not onto $\overline{\operatorname{Im}(T)}$. For an easy example of an injective trace-class operator, take $T$ acting on some orthonormal basis $\{e_n\}$ by $Te_n=\tfrac1{n^2}\,e_n$. The closure of its range is all of $H$. Jan 5 '19 at 3:08