Trace class is a two sided Banach operator ideal

Operator norm dominated by the trace norm. For trace-class $T$, [A1] gives $\mathrm{\parallel }T\mathrm{\parallel }=s_1(T)\le {\sum }_n{s}_n(T)=\mathrm{\parallel }T{\mathrm{\parallel }}_1$.

Vector-space structure and triangle inequality. Let $S,T$ be trace class and $a,b$ scalars. Given $\epsilon >0$, [A1] and [A2] provide nuclear representations of $S$ and $T$ with sums $\le \mathrm{\parallel }S{\mathrm{\parallel }}_1+\epsilon$ and $\le \mathrm{\parallel }T{\mathrm{\parallel }}_1+\epsilon$; interleaving them, after multiplying the first by $a$ and the second by $b$, gives a nuclear series converging in operator norm to $aS+bT$ with sum $\le \mathrm{\mid }a\mathrm{\mid }(\mathrm{\parallel }S{\mathrm{\parallel }}_1+\epsilon )+\mathrm{\mid }b\mathrm{\mid }(\mathrm{\parallel }T{\mathrm{\parallel }}_1+\epsilon )$. Its partial sums have finite rank, so [A5] makes $aS+bT$ compact; [A1] now makes it trace class and bounds its trace norm by that nuclear sum. Letting $\epsilon \downarrow 0$ gives $\mathrm{\parallel }aS+bT{\mathrm{\parallel }}_1\le \mathrm{\mid }a\mathrm{\mid }\mathrm{\parallel }S{\mathrm{\parallel }}_1+\mathrm{\mid }b\mathrm{\mid }\mathrm{\parallel }T{\mathrm{\parallel }}_1$; homogeneity follows by also applying the bound to $a^{-1}(aT)$ when $a\mathrm{\ne }0$ (and is immediate for $a=0$), and the triangle inequality is the case $a=b=1$. The norm is definite: if $\mathrm{\parallel }T{\mathrm{\parallel }}_1={\sum }_n{s}_n(T)=0$ then $s_1(T)=\mathrm{\parallel }T\mathrm{\parallel }=0$ by [A1], so $T=0$; it is nonnegative by definition.

Two-sided ideal estimate. Let $T={\sum }_j⟨\cdot ,u_j⟩v_j$ be nuclear and let $A,B$ be bounded. Then for every $x\in H_0$, $ATBx=A{\sum }_j⟨Bx,u_j⟩v_j={\sum }_j⟨x,B^{\ast }{u}_j⟩A{v}_j$, and the finite-rank partial sums converge to $ATB$ in operator norm because composition is operator-norm continuous [A3]. Their nuclear sum satisfies ${\sum }_j\mathrm{\parallel }{B}^{\ast }{u}_j\mathrm{\parallel }\mathrm{\parallel }A{v}_j\mathrm{\parallel }\le \mathrm{\parallel }B\mathrm{\parallel }\mathrm{\parallel }A\mathrm{\parallel }{\sum }_j\mathrm{\parallel }{u}_j\mathrm{\parallel }\mathrm{\parallel }{v}_j\mathrm{\parallel }$ by [A3]. Thus $ATB$ is compact by [A5], and [A1] makes it trace class with trace norm bounded by this sum; taking the infimum over nuclear representations of $T$ gives $\mathrm{\parallel }ATB{\mathrm{\parallel }}_1\le \mathrm{\parallel }A\mathrm{\parallel }\mathrm{\parallel }T{\mathrm{\parallel }}_1\mathrm{\parallel }B\mathrm{\parallel }$.

Completeness. Let $(T_m)$ be $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_1$-Cauchy. Choose a subsequence $T_{m_1},T_{m_2},\dots$ with $\mathrm{\parallel }{T}_{m_{k+1}}-T_{m_k}{\mathrm{\parallel }}_1<2^{-k}$ [A4] and write $D_k:=T_{m_{k+1}}-T_{m_k}$. For $T_{m_1}$ and each $D_k$ choose by [A2] nuclear representations with sums at most $\mathrm{\parallel }{T}_{m_1}{\mathrm{\parallel }}_1+1$ and $\mathrm{\parallel }{D}_k{\mathrm{\parallel }}_1+2^{-k}$ respectively. Flattening these countably many positive-integer-indexed series by a fixed pairing of positive integers produces one nuclear series with total sum at most $\mathrm{\parallel }{T}_{m_1}{\mathrm{\parallel }}_1+1+{\sum }_{k\ge 1}(\mathrm{\parallel }{D}_k{\mathrm{\parallel }}_1+2^{-k})<+\mathrm{\infty }$. The nuclear-tail estimate makes its finite-rank partial sums converge in operator norm to a bounded operator $S$, and [A5] makes $S$ compact; [A1] therefore makes $S$ trace class. Absolute operator-norm convergence permits regrouping, and the grouped partial sums are $T_{m_1}+{\sum }_{k<K}{D}_k=T_{m_K}$, so $T_{m_K}\to S$ in operator norm. For every $K$ the tail representation made from $D_K,D_{K+1},\dots$ gives $\mathrm{\parallel }S-T_{m_K}{\mathrm{\parallel }}_1\le {\sum }_{k\ge K}(\mathrm{\parallel }{D}_k{\mathrm{\parallel }}_1+2^{-k})\to 0$ by [A1] and [step 1.2]; hence $T_{m_K}\to S$ in $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_1$, and by [A4] the original Cauchy sequence converges to $S$ in $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_1$.

Conclusion. Claim 1 is [step 1.1] and [step 1.2], claim 2 is [step 1.3], and claim 3 is [step 2.1]; together ${\mathcal{S}}_1(H,K)$ with $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_1$ is a normed space complete in its norm, that is a Banach space.