Let us focuse on the set of positive operators of trace 1 acting on some infinite dimensional Hilbert space, call it $S(H_{1})$ where $H$ is the mentioned Hilbert space. Let $S(H_{2})$ be another such space and now let $\rho \in S(H_{1})$ and $\sigma \in S(H_{2})$. The tensor product operator $\rho\otimes\sigma \in S(H_{1})\otimes S(H_{2})$. Now, let $U$ be some unitary map that mapping, i.e. an automorphism on $S(H_{1})\otimes S(H_{2})$, and $\Lambda$ a completely positive map acting on $S(H_{1})\otimes S(H_{2})$ acting non trivially only on the subspace $S(H_{2})$ of $S(H_{1})\otimes S(H_{2})$. i.e. for $\rho\otimes\sigma \in S(H_{1})\otimes S(H_{2})$
$$\Lambda(\rho\otimes\sigma) = \rho\otimes\Lambda_{2}\big(\sigma\big)$$
where $\Lambda_{2}$ is acompletly positive map.
I would like to know if there are any helpful results that help in the treatment of trace distances of the following sort. I know that the trace has a nice seperability property i.e. $Tr(A\otimes B) = Tr(A)Tr(B)$ and the trace distance also shares this $\|A \otimes B \|_{1} = \|A\|_{1}\|B\|_{1}$. I am wondering if there is anything, some separability or linearity property, that might render the following into a product of traces which are not traces of tensor product states.
$$\| U\Big(\rho\otimes \sigma\Big) - \Lambda\Big( U\Big(\rho\otimes \sigma\Big)\Big)\|_{1}$$
where the trace distance is defined as follows
$$ \|A-B\|_{1} = \frac{1}{2}Tr\{\sqrt{(A-B)^{\dagger}(A-B)}\}$$
Thanks in advance.