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

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 $$|E_{sq}(\rho)-E_{sq}(\sigma)|{\leq}f(\epsilon)$$?

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}
• It simplifies the answer I wrote but it is not without loss of generality. In general there does not have to exist a point achieving the infimum. E.g., take $\inf x$ for $0 < x < 1$ the infimum is 0 but it is never achieved by any point in the considered domain. You'd thus need to adapt the agument to consider some limit of a sequence of extensions if you want a full proof. Jun 27 at 14:03