Nuclear series characterizes trace norm

Suppose $T$ is trace class. For every $j\in J$ set $u_j=s_j{e}_j$ and $v_j=f_j$. Because $s_j$ is real, conjugate-linearity gives $⟨x,s_j{e}_j⟩=s_j⟨x,e_j⟩$. In finite rank put $u_j=v_j=0$ for $j>r$; in rank zero use zero families throughout. In infinite rank $J$ already consists of all positive integers, so no index shift and no vector $e_0$ is used. Orthonormality gives $\mathrm{\parallel }{u}_j\mathrm{\parallel }\mathrm{\parallel }{v}_j\mathrm{\parallel }=s_j$ for $j\in J$. The shifted zero-based sum is $\mathrm{\parallel }T{\mathrm{\parallel }}_1<\mathrm{\infty }$, and the zero-based partial-sum sequence with $R_0=0$ converges in operator norm by [A1].

Conversely assume given positive-integer-indexed families with $C={\sum }_{j\ge 1}\mathrm{\parallel }{u}_j\mathrm{\parallel }\mathrm{\parallel }{v}_j\mathrm{\parallel }<\mathrm{\infty }$ in the shifted sense just specified, and let the zero-based sequence $(R_m{)}_{m\in \mathbb{N}}$ converge to $T$ in operator norm. Each term is linear and bounded by $\mathrm{\parallel }{u}_j\mathrm{\parallel }\mathrm{\parallel }{v}_j\mathrm{\parallel }$ using [A5], so $R_m$ is bounded and its range lies in the finite span of $v_1,\dots ,v_m$ (the zero subspace when $m=0$). The given compactness of $T$ licenses [A1]; no new compactness theorem is needed.

Fix a finite $F\subseteq J$. Since $T{e}_k=s_k{f}_k$, put $S_F={\sum }_{k\in F}{s}_k={\sum }_{k\in F}⟨T{e}_k,f_k⟩$. For every $m\in \mathbb{N}$, finite rearrangement (with the sum empty at $m=0$) gives $a_m:={\sum }_{k\in F}⟨R_m{e}_k,f_k⟩={\sum }_{j=1}^m{\sum }_{k\in F}⟨e_k,u_j⟩⟨v_j,f_k⟩\mathrm{.}$ Finite Cauchy–Schwarz, applied to the vectors of absolute values in ${\mathbb{R}}^{\mathrm{\mid }F\mathrm{\mid }}$, and Bessel give $\mid {\sum }_{k\in F}⟨e_k,u_j⟩⟨v_j,f_k⟩\mid \le {({\sum }_{k\in F}\mathrm{\mid }⟨u_j,e_k⟩{\mathrm{\mid }}^2)}^{1\mathrm{/}2}{({\sum }_{k\in F}\mathrm{\mid }⟨v_j,f_k⟩{\mathrm{\mid }}^2)}^{1\mathrm{/}2}\le \mathrm{\parallel }{u}_j\mathrm{\parallel }\mathrm{\parallel }{v}_j\mathrm{\parallel }\mathrm{.}$ Thus $\mathrm{\mid }{a}_m\mathrm{\mid }\le C$. Also $\mathrm{\mid }{S}_F-a_m\mathrm{\mid }\le \mathrm{\mid }F\mathrm{\mid }\mathrm{\parallel }T-R_m\mathrm{\parallel }$ by [A3] and [A5], so the zero-based sequence $(a_m)$ converges to $S_F$ and $S_F\le C$. For empty $F$ this says $0\le C$; otherwise a hypothetical $S_F>C$ contradicts the displayed error bound for sufficiently large $m$.

Take $F=J\cap \{1,\dots ,n\}$ for each $n\in \mathbb{N}$. The zero-padded singular-value partial sums are nondecreasing, start at zero, and are bounded above by $C$ by step 2.1. Therefore [A7] makes their series converge to a value at most $C$. By [A2], $T$ is trace class and $\mathrm{\parallel }T{\mathrm{\parallel }}_1\le C$.

Steps 1.1 and 3.1 prove the equivalence. For trace-class $T$, the set of nuclear-representation sums is nonempty by step 1.1, every such sum is at least $\mathrm{\parallel }T{\mathrm{\parallel }}_1$ by step 3.1, and step 1.1 attains this bound. It is therefore the infimum by [A6]. All sequences used in the reverse implication were given; the forward implication spends only the Countable Choice already assumed by the SVD.