Countable Kraus normalization and trace-norm convergence

[Norm identity, including the zero space] If $H=\{0\}$, then every $M_a$ and every trace-class input is zero, so the unique map on ${\mathcal{S}}_1(H)=\{0\}$ satisfies (a)-(c). Suppose henceforth that $H\mathrm{\ne }\{0\}$. For every $x$, ${\sum }_{a\le N}\mathrm{\parallel }{M}_{a}x{\mathrm{\parallel }}^2$ is the increasing sequence $⟨({\sum }_{a\le N}{M}_a^{\ast }{M}_a)x,x⟩$, so its supremum is $⟨x,x⟩=\mathrm{\parallel }x{\mathrm{\parallel }}^2$ by strong convergence and continuity of the inner product in $N$; by [F1] this supremum is ${\sum }_a\mathrm{\parallel }{M}_{a}x{\mathrm{\parallel }}^2$.

[Positive input] Let $\rho ={\sum }_n{w}_n{T}_{u_n,u_n}$ be the spectral expansion [F3]. The partial sums ${\mathrm{\Phi }}_N(\rho ):={\sum }_{a\le N}{M}_a\rho M_a^{\ast }$ are positive trace-class with $\mathrm{tr}{\mathrm{\Phi }}_N(\rho )={\sum }_{a\le N}{\sum }_n{w}_n\mathrm{\parallel }{M}_a{u}_n{\mathrm{\parallel }}^2$ by [F2]; the double sum increases with $N$, and its limit is ${\sum }_n{w}_n{\sum }_a\mathrm{\parallel }{M}_a{u}_n{\mathrm{\parallel }}^2={\sum }_n{w}_n\mathrm{\parallel }{u}_n{\mathrm{\parallel }}^2=\mathrm{tr}\rho$ by step 1.1 and Tonelli for nonnegative series. Hence the increasing positive sequence ${\mathrm{\Phi }}_N(\rho )$ has traces increasing to $\mathrm{tr}\rho$, so for $N\ge M$ we have $\mathrm{\parallel }{\mathrm{\Phi }}_N(\rho )-{\mathrm{\Phi }}_M(\rho ){\mathrm{\parallel }}_1=\mathrm{tr}{\mathrm{\Phi }}_N(\rho )-\mathrm{tr}{\mathrm{\Phi }}_M(\rho )\to 0$ ([F4] with positivity of the difference); by completeness of ${\mathcal{S}}_1(H)$ [F2] the series converges in trace norm to a trace-class $\mathrm{\Phi }(\rho )\ge 0$ (the trace-norm limit of positive operators is positive) with $\mathrm{tr}\mathrm{\Phi }(\rho )=\mathrm{tr}\rho$.

[Linear extension by nuclear convergence] For rank-one input, $M_a{T}_{u,v}{M}_a^{\ast }=T_{M_{a}u,M_{a}v}$ by [F2]. Step 1.1 and Cauchy–Schwarz give ${\sum }_a\mathrm{\parallel }{M}_{a}u\mathrm{\parallel }\mathrm{\parallel }{M}_{a}v\mathrm{\parallel }\le \mathrm{\parallel }u\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }$. Thus for a nuclear input $T={\sum }_j{T}_{u_j,v_j}$ the double series of Kraus rank-one images is absolutely summable in trace norm, with total norm at most ${\sum }_j\mathrm{\parallel }{u}_j\mathrm{\parallel }\mathrm{\parallel }{v}_j\mathrm{\parallel }$. Its sum is the trace-norm limit of ${\mathrm{\Phi }}_N(T)={\sum }_{a\le N}{M}_{a}T{M}_a^{\ast }$, by interchanging absolutely convergent sums and the bounded two-sided ideal action. This limit depends only on $T$, so defines a linear map $\mathrm{\Phi }$ extending the positive-input map of step 2.1. Taking the nuclear infimum (Nuclear series characterizes trace norm) gives the bound $\mathrm{\parallel }\mathrm{\Phi }(T){\mathrm{\parallel }}_1\le \mathrm{\parallel }T{\mathrm{\parallel }}_1$ for every input. For rank-one inputs, the trace sum is ${\sum }_a⟨M_{a}u,M_{a}v⟩=⟨u,v⟩$, by strong completeness and the adjoint identity; extending through the absolutely convergent nuclear sum proves $\mathrm{tr}\mathrm{\Phi }(T)=\mathrm{tr}T$. The positive trace-class operators linearly span trace class by the self-adjoint real/imaginary and positive/negative spectral decompositions in [F4], so no other linear extension agreeing with step 2.1 is possible.

[Norm bound and conclusion] For self-adjoint $\rho$ with spectral decomposition $\rho ={\rho }^{+}-{\rho }^{-}$ write $T=\rho$. Then $\mathrm{\Phi }(\rho )=\mathrm{\Phi }({\rho }^{+})-\mathrm{\Phi }({\rho }^{-})$ by construction, so $\mathrm{\parallel }\mathrm{\Phi }(\rho ){\mathrm{\parallel }}_1\le \mathrm{tr}\mathrm{\Phi }({\rho }^{+})+\mathrm{tr}\mathrm{\Phi }({\rho }^{-})=\mathrm{tr}{\rho }^{+}+\mathrm{tr}{\rho }^{-}=\mathrm{\parallel }\rho {\mathrm{\parallel }}_1$ using step 2.1 and [F4]. Positivity for arbitrary $\rho \ge 0$ is step 2.1, and trace preservation was shown in step 3.1; hence density operators map to density operators.