How to justify the conclusion $|E_{sq}(\rho)-E_{sq}(\sigma)|\le f(\epsilon)$, when proving the continuity of the squashed entanglement?

Question

I am following the paper by Christandl and Winter introducing squashed entanglement. My question is particularly on the continuity proof of squshed entanglement mentioned after conjecture 14 and before remark 15 (page 6 of the arXiv version).

To make the question more or less self-contained, squashed entanglement of a density matrix $\rho^{AB}$ is defined by $E_{sq}(\rho){=}\inf_E I(A:B|E)_{\rho}/2$, where the infimum of quantum conditional mutual information (QCMI) is obtained for all possible extensions of $\rho^{AB}$. More precisely, if a purification of $\rho^{AB}$ is given by $|\rho\rangle\rangle^{ABC_1}$, we have to take infimum over all possible CPTP maps $\Lambda^{C_1{\rightarrow}E_1}$ acting on the purified extension, $\rho^{ABE_1}{=}\Lambda^{C_1{\rightarrow}E_1}(|\rho\rangle\rangle\langle\langle\rho|^{ABC_1})$.

Now Christandl and Winter attempt to show the continuity of squashed entanglement, which asks if the trace distance between two density matrices is small, $\parallel\rho^{AB}-\sigma^{AB}\parallel_1{\leq}\epsilon$, does it imply $|E_{sq}(\rho)-E_{sq}(\sigma)|{\leq}f(\epsilon)$ with $\lim_{\epsilon {\rightarrow} 0}f(\epsilon){=}0$?

Brief proof-sketch: $\parallel\rho^{AB}-\sigma^{AB}\parallel_1{\leq}\epsilon$ implies the purifications are also close, $\parallel|\rho\rangle\rangle^{ABC_1}-|\sigma\rangle\rangle^{ABC_2}\parallel_1{\leq}2\sqrt{\epsilon}$. Now trace-distance is contractive under a CPTP map, i.e. for a CPTP map $\Phi(.)$, $\parallel \Phi(\rho)-\Phi(\sigma)\parallel_1{\leq}\parallel \rho-\sigma\parallel_1$. Hence if on the purified systems, same CPTP maps, $\Lambda^{C_1{\rightarrow}E_1}$ and $\Lambda^{C_2{\rightarrow}E_2}$ are applied, we have $\parallel \rho^{ABE_1}-\sigma^{ABE_2}\parallel_1{\leq}2\sqrt{\epsilon}$. From that continuity of QCMI follows (see corollary 1 of the Shirokov paper for a better bound).

However, I have trouble understanding the next part of the proof. The part where after getting $|I(A:B|E_1)_\rho-I(A:B|E_2)_\sigma|{\leq}f(\epsilon)$, they argued ``since this applies to any quantum operation $\Lambda$ and thus to every state extension of $\rho_{AB}$ and $\sigma_{AB}$ respectively, we obtain" $|E_{sq}(\rho)-E_{sq}(\sigma)|{\leq}f(\epsilon)$.

Question: Note the optimal extensions may have different dimensions, $\dim(E_1){\neq}\dim(E_2)$. Even if this is not the case, the extensions $\rho^{ABE_1}$ and $\sigma^{ABE_2}$ may be obtained by different CPTP maps (e.g., $\Lambda_1$ for $\rho$ and $\Lambda_2$ for $\sigma$), in which case we cannot apply the contractivity of trace distance. Given these issues, how can we justify the conclusion $|E_{sq}(\rho)-E_{sq}(\sigma)|{\leq}f(\epsilon)$?

Answer 1

Regarding the dimensions, this is not a problem because we can purify $\rho$ and $\sigma$ using the same system and then we anyway consider applying the same map to both systems and hence the dimensions also remain the same. (Actually it's a bit strange that you write different system labels for the same map). Regardless, if you encounter a situation where $\mathrm{dim}(E_2) > \mathrm{dim}(E_1)$ then you can just view $E_1$ as a subspace of $E_2$ and extend everything.

Now I'm assuming you're happy up to the statement that $\|\rho_{AB} - \sigma_{AB}\|_1 \leq \epsilon$ implies that $|I(A:B|E)_{\rho} - I(A:B|E)_{\sigma}|\leq \epsilon'$ when we apply the same extension map $\Lambda$ to the purifications of both $\rho$ and $\sigma$. Taken this as a given let's assume without loss of generality that $E_{\mathrm{sq}}(\rho) \leq E_{\mathrm{sq}}(\sigma)$. I'm also going to make the simplifying assumption that the infimum for $E_{\mathrm{sq}}(\sigma)$ is achieved and $\Lambda$ is the achieving map (a more rigorous analysis would need a limiting argument). More precisely, I'll assume there exists $\Lambda$ such that $$ E_{\mathrm{sq}}(\sigma) = I(A:B|E)_{\Lambda(|\sigma\rangle \langle \sigma|)}. $$

So then $$ \begin{aligned} |E_{\mathrm{sq}}(\rho) - E_{\mathrm{sq}}(\sigma)| &= E_{\mathrm{sq}}(\rho) - E_{\mathrm{sq}}(\sigma) \\ &= \inf I(A:B|E)_{\rho} - \inf I(A:B|E)_{\sigma} \\ &= \inf I(A:B|E)_{\rho} - I(A:B|E)_{\Lambda(|\sigma\rangle \langle \sigma|)} \\ &\leq I(A:B|E)_{\Lambda(|\rho\rangle \langle \rho|)} - I(A:B|E)_{\Lambda(|\sigma\rangle \langle \sigma|)} \\ &\leq \epsilon' \end{aligned} $$