Trace class operator

Assume the Axiom of Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)). Let $H$ and $K$ be Hilbert spaces over the same field $\mathbb{F}\in \{\mathbb{R},\mathbb{C}\}$ (Hilbert space) and let $T\in \mathcal{B}(H,K)$ be a compact operator (Compact linear operator) with zero-padded singular-value sequence $(s_n(T){)}_{n\ge 1}$ (Absolute value and singular values of a compact operator).

Trace class. The operator $T$ is trace class when ${\sum }_{n\ge 1}{s}_n(T):={\sum }_{m\in \mathbb{N}}{s}_{m+1}(T)<+\mathrm{\infty }\mathrm{.}$ Thus the series in the zero-based convention of Series and absolute convergence in a normed space is formed from the explicit sequence $(s_{m+1}(T){)}_{m\in \mathbb{N}}$. In that case its trace norm is $\mathrm{\parallel }T{\mathrm{\parallel }}_1:={\sum }_{m\in \mathbb{N}}{s}_{m+1}(T)\in [0,+\mathrm{\infty }),$ and $\mathrm{\parallel }T{\mathrm{\parallel }}_1$ is also written $\mathrm{\parallel }T{\mathrm{\parallel }}_{\mathrm{t}\mathrm{r}}$. The set of trace-class operators $H\to K$ is written ${\mathcal{S}}_1(H,K)$.

Immediate consequences. Since $0\le s_n(T)\le s_1(T)=\mathrm{\parallel }T\mathrm{\parallel }$ (Absolute value and singular values of a compact operator), the terms are nonnegative and $\mathrm{\parallel }T{\mathrm{\parallel }}_1\ge \mathrm{\parallel }T\mathrm{\parallel }\ge 0$ whenever $T$ is trace class; the zero operator is trace class with $\mathrm{\parallel }0{\mathrm{\parallel }}_1=0$. When $T$ has finite rank $r$ the sequence is zero-padded, the series is the finite sum ${\sum }_{n=1}^r{s}_n(T)={\sum }_{\lambda >0}\lambda {\dim }_{\mathbb{F}}{E}_{\lambda }(\mathrm{\mid }T\mathrm{\mid })$ over the finitely many positive eigenvalues of $\mathrm{\mid }T\mathrm{\mid }$ counted with multiplicity (Finite-dimensional vector space, and its dimension ${\dim }_{F}V$; infinite-dimensional means having no finite basis), and every finite-rank operator is compact (Bounded finite rank operators are compact) and therefore trace class; in particular every operator with finite-dimensional range and every rank-one operator is trace class. If $T$ has infinite rank then $s_n(T)>0$ for every $n$ and the series ${\sum }_{n\ge 1}{s}_n(T)$ converges in the summable case. By the necessary condition for convergence of a scalar series (If a series converges then its terms tend to $0$), a trace-class operator with infinite rank has $s_n(T)\to 0$, hence is a norm limit of finite-rank operators (Singular value decomposition for compact operators).

Choice accounting. Trace class is defined through the singular values of Absolute value and singular values of a compact operator, whose construction uses ${\mathrm{A}\mathrm{C}}_{\omega }$ through the countable selection of finite orthonormal bases of the eigenspaces of $\mathrm{\mid }T\mathrm{\mid }$ and the spectral theorem; no Hilbert basis of the ambient space, and no stronger choice, is used here.