Cyclicity of the trace

Products are trace class. $ST$ and $TS$ are trace class by [A2] with $\mathrm{\parallel }ST{\mathrm{\parallel }}_1\le \mathrm{\parallel }S\mathrm{\parallel }\mathrm{\parallel }T{\mathrm{\parallel }}_1$ and $\mathrm{\parallel }TS{\mathrm{\parallel }}_1\le \mathrm{\parallel }S\mathrm{\parallel }\mathrm{\parallel }T{\mathrm{\parallel }}_1$.

Cyclicity for rank-one operators. Let $R:=⟨\cdot ,u⟩v$ for fixed $u,v\in H$, so that $R$ is trace class; then $SR=⟨\cdot ,u⟩Sv$ and $RS=⟨\cdot ,S^{\ast }u⟩v$ are nuclear representations with one term, so by [A1] $\mathrm{tr}(SR)=⟨Sv,u⟩$ and $\mathrm{tr}(RS)=⟨v,S^{\ast }u⟩$, and these are equal by the adjoint identity of [A4].

The Hilbert–Schmidt case. Write $e_k=k$ for $k\in E$, so that basis elements and indices are unambiguous. Let $A,B$ be Hilbert–Schmidt relative to the supplied basis $E$; by [A3] the products $AB$ and $BA$ are trace class. Since $A$ is continuous and $E$ is a Hilbert basis, $⟨ABe,e⟩={\sum }_{k\in E}⟨Be,e_k⟩⟨A{e}_k,e⟩$ for every $e$, and the double family ${\alpha }_{k,e}{\beta }_{k,e}$ with ${\alpha }_{k,e}:=⟨A{e}_k,e⟩$, ${\beta }_{k,e}:=⟨Be,e_k⟩$ has finite total, ${\sum }_{k,e}\mathrm{\mid }{\alpha }_{k,e}\mathrm{\mid }\mathrm{\mid }{\beta }_{k,e}\mathrm{\mid }\le \mathrm{\parallel }A{\mathrm{\parallel }}_{HS,E}\mathrm{\parallel }B{\mathrm{\parallel }}_{HS,E}$, by two applications of finite Cauchy–Schwarz, Bessel and Parseval [A5]; For detail, on each finite rectangle $K\times F\subseteq E\times E$, finite Cauchy–Schwarz bounds the absolute sum by $({\sum }_{k\in K,e\in F}\mathrm{\mid }{\alpha }_{k,e}{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}({\sum }_{k\in K,e\in F}\mathrm{\mid }{\beta }_{k,e}{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}\le \mathrm{\parallel }A{\mathrm{\parallel }}_{HS,E}\mathrm{\parallel }B{\mathrm{\parallel }}_{HS,E}$, using Bessel in the inner sums. Every finite set of pairs lies in a finite rectangle. The full absolute sum is therefore finite; outside a finite rectangle its tail is arbitrarily small by [A5], which proves that both iterated scalar sums have the same value as the double-family sum. Invoking the diagonal trace formula in [A1] for the trace-class products, $\mathrm{tr}(AB)={\sum }_{e\in E}{\sum }_{k\in E}{\alpha }_{k,e}{\beta }_{k,e}={\sum }_{k\in E}{\sum }_{e\in E}{\beta }_{k,e}{\alpha }_{k,e}=\mathrm{tr}(BA)$, the last equality by the symmetric computation for $BA$.

Cyclicity for general trace-class operators. By [A1] take a nuclear representation $Tx={\sum }_{j\ge 1}⟨x,u_j⟩v_j$ with $C={\sum }_{j\ge 1}\mathrm{\parallel }{u}_j\mathrm{\parallel }\mathrm{\parallel }{v}_j\mathrm{\parallel }<\mathrm{\infty }$ and partial sums $R_n={\sum }_{1\le j\le n}⟨\cdot ,u_j⟩v_j$ for $n\in \mathbb{N}$, so $R_0=0$ and $R_n\to T$ in operator norm. Then $S{R}_n\to ST$ and $R_{n}S\to TS$: both errors are at most $\mathrm{\parallel }S\mathrm{\parallel }\mathrm{\parallel }T-R_n\mathrm{\parallel }$ by the operator-norm bound. Their terms give nuclear representations $ST={\sum }_j⟨\cdot ,u_j⟩S{v}_j$ and $TS={\sum }_j⟨\cdot ,S^{\ast }{u}_j⟩v_j$. Their sums of norm products are at most $\mathrm{\parallel }S\mathrm{\parallel }C$ and $\mathrm{\parallel }{S}^{\ast }\mathrm{\parallel }C=\mathrm{\parallel }S\mathrm{\parallel }C$, respectively. Both products are already trace class by step 1.1, so [A1] gives $\mathrm{tr}(ST)={\sum }_j⟨S{v}_j,u_j⟩={\sum }_j⟨v_j,S^{\ast }{u}_j⟩=\mathrm{tr}(TS)$ by [A4], term by term in absolutely convergent series. No ambient Hilbert basis is used in this part.

Conclusion. The general cyclicity statement is [step 2.1] with [step 1.1], and the Hilbert–Schmidt statement is [step 1.3].