Continuity bounds on $D_{\max}(\rho_{AB}\|\rho_A\otimes\rho_B)$

Question

The max-relative entropy between two states is defined as

$$D_{\max }(\rho \| \sigma):=\log \min \{\lambda: \rho \leq \lambda \sigma\},$$

where $\rho\leq \sigma$ should be read as $\sigma - \rho$ is positive semidefinite. Consider the following quantity for a bipartite state $\rho_{AB}$ with reduced states $\rho_{A}$ and $\rho_{B}$.

$$I_{\max}(\rho_{AB}) = D_{\max}(\rho_{AB}||\rho_{A}\otimes\rho_{B})$$

I would like to know if this satisfies a continuity bound. That is, given $\rho_{AB}\approx_{\epsilon}\sigma_{AB}$ in some distance measure, can we bound $|I_{\max}(\rho_{AB}) - I_{\max}(\sigma_{AB})|$?

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Motivation for question: Recall the quantum relative entropy $D(\rho||\sigma) = \text{Tr}(\rho\log\rho - \rho\log\sigma)$ with the convention that $0\log 0 = 0$. Let us define the mutual information as follows

$$I(\rho_{AB}) = D(\rho_{AB}||\rho_{A}\otimes\rho_{B}) = -S(\rho_{AB}) + S(\rho_A) + S(\rho_B),$$

where $S(\rho) = -\text{Tr}(\rho\log\rho)$ is the von Neumann entropy. In this case, we may use Fannes inequality to find a bound on $|I(\rho_{AB}) - I(\sigma_{AB})|$ in terms of $\|\rho_{AB} - \sigma_{AB}\|_1$. I'm wondering if the move from $D(.||.)$ to $D_{\max}(.||.)$ can be made while still having some Fannes type bound.

Answer 1

Unfortunately $D_{\max}$ is not a continuous function and so functions built from it tend not to be continuous. For example consider consider the two states

$$ \rho_{AB} = |00 \rangle \langle 00|, $$ and $$ \tau_{AB}(\epsilon) = (1-\epsilon) |00 \rangle \langle 00 | + \epsilon | 11\rangle \langle 11 |. $$ A quick calculation gives $I_{\max}(\rho_{AB}) = 0$ and $I_{\max}(\tau_{AB}(\epsilon)) = \max\{- \log(\epsilon), -\log(1-\epsilon)\}$. So $I_{\max}(\rho_{AB}) \neq \lim_{\epsilon \rightarrow 0} I_{\max}(\tau_{AB}(\epsilon)) = \infty$.

However, some authors have studied a smoothed version of $I_{\max}$ (see https://arxiv.org/pdf/1308.5884.pdf).