Trace of a trace class operator

Assume the Axiom of Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)). Let $H$ be a real or complex Hilbert space (Hilbert space), let $T\in \mathcal{B}(H)$ be a trace-class operator on $H$ (Trace class operator) and let $E$ be a Hilbert basis of $H$, supplied as data (Orthonormal families, complete orthonormal systems and Hilbert bases; existence of such a basis is not asserted here). The trace of $T$ relative to $E$ is ${\mathrm{tr}}_E(T):={\sum }_{e\in E}⟨Te,e⟩,$ the sum of the scalar family $(⟨Te,e⟩{)}_{e\in E}$ in the finite-subset-net sense of Square-summable families on an arbitrary index set and the space ${\mathrm{\ell }}^2(I)$.

The family is absolutely summable, uniformly in $E$. Let $T={\sum }_j{s}_j⟨\cdot ,e_j⟩f_j$ be the singular-value series (Singular value decomposition for compact operators, Absolute value and singular values of a compact operator) with $(e_j{)}_{j\in J}$, $(f_j{)}_{j\in J}$ orthonormal and ${\sum }_j{s}_j=\mathrm{\parallel }T{\mathrm{\parallel }}_1<+\mathrm{\infty }$ (Trace class operator). For every $e\in H$ the series $⟨Te,e⟩={\sum }_j{s}_j⟨e,e_j⟩⟨f_j,e⟩$ converges absolutely, because the moduli are bounded by $s_1{\sum }_j\mathrm{\mid }⟨e,e_j⟩⟨f_j,e⟩\mathrm{\mid }\le s_1\mathrm{\parallel }e{\mathrm{\parallel }}^2$ by Cauchy–Schwarz and Bessel (Cauchy–Schwarz: $\mathrm{\mid }⟨x,y⟩\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$, with equality exactly for dependent pairs, The finite Bessel inequality and best approximation by a finite orthonormal family). Consequently, for every finite $F\subseteq E$, using the interchange of finite sums with the nonnegative finite-subset supremum over $j$ and then finite Cauchy–Schwarz and Bessel twice, ${\sum }_{e\in F}\mathrm{\mid }⟨Te,e⟩\mathrm{\mid }\le {\sum }_{e\in F}{\sum }_j{s}_j\mathrm{\mid }⟨e,e_j⟩\mathrm{\mid }\mathrm{\mid }⟨f_j,e⟩\mathrm{\mid }={\sum }_j{s}_j{\sum }_{e\in F}\mathrm{\mid }⟨e,e_j⟩\mathrm{\mid }\mathrm{\mid }⟨f_j,e⟩\mathrm{\mid }\le {\sum }_j{s}_j({\sum }_{e\in F}\mathrm{\mid }⟨e,e_j⟩{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}({\sum }_{e\in F}\mathrm{\mid }⟨f_j,e⟩{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}\le {\sum }_j{s}_j=\mathrm{\parallel }T{\mathrm{\parallel }}_1,$ because for each $j$ the two finite coefficient sums are bounded by $\mathrm{\parallel }{e}_j\mathrm{\parallel }=1$ and $\mathrm{\parallel }{f}_j\mathrm{\parallel }=1$ (finite Bessel). Hence the finite subsums of $\mathrm{\mid }(⟨Te,e⟩){\mathrm{\mid }}_{e\in E}$ are bounded by $\mathrm{\parallel }T{\mathrm{\parallel }}_1$, the family is absolutely summable, and an absolutely summable family of scalars is summable in ZF with $\mathrm{\mid }{\sum }_{e\in E}⟨Te,e⟩\mathrm{\mid }\le \mathrm{\parallel }T{\mathrm{\parallel }}_1$ (Square-summable families on an arbitrary index set and the space ${\mathrm{\ell }}^2(I)$). This justifies the notation ${\mathrm{tr}}_E(T)$ before any basis-independence statement.

Choice accounting and status. Only a supplied basis $E$ and the singular values of $T$ are used; no Hilbert basis of $H$ is assumed to exist, and the next theorem proves that ${\mathrm{tr}}_E(T)$ does not depend on the supplied basis and equals the basis-free trace of $T$ (Trace is absolutely convergent and basis independent).