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

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 $$|I_{\max}(\rho_{AB}) - I_{\max}(\sigma_{AB})|$$?

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.

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).
• @QuestionEverything How can it be? The above is an explicit counterexample. Maybe if you define it as $\min_{\sigma_B} D_{\max}(\rho_{AB} \|\rho_A \otimes \sigma_B)$ it will be continuous. Jul 25 at 21:28